How to prove continuity of $e^x$. I simply want a proof that $e^x$ is continuous.

I have never really been able to find something satisfying these points:

$e$ is defined to be the limit $\lim_{n\to\infty}\left(1+{1\over n}\right)^n$. This is not essential, and another reasonable definition may be used if that significantly simplifies things. EDIT: In fact, it does not really matter what the base is taken to be. Any number, or a general $a^x$, would be fine.
For positive integers, we define $a^n:=a\times a\times\cdots\times a$ where there are $n$ multiplications. We then let $a^0=1$ and $a^{-n}=1/a^n$. For roots, $a^{1/n}$ is the (principal) number $x$ such that $x^n=a$. For general rationals, $a^{b/c}=(a^{1/c})^b$ is the $c$'th root brought to the $b$'th power. And for irrational numbers $x$, $a^x=\lim{a^n}$ where $n$ is a sequence of rational numbers which converge to $x$.

In other words, I want a proof of the exponential's continuity derived from the "arithmetic" definition itself. The reason I ask is because as a student I never had an intuition for why a number brought to two close exponents would have to yield similar answers. For example, why would $$2^{1/2}\approx 2^{47/99}$$ I thought? After all, one has a $99$'th root to a $47$'th power, and the other is just a square root.

I now know that I was looking for a proof of continuity... and I'm still looking. Please do not use anything advanced - the kid in me still groans when I see people 'define' $e^x$ as a power series, because I know students everywhere (like I was) are getting confused.

Also, I want the proof to be convincing, as in I can sleep well at night, but it need not be headache-inducingly thorough. For example, I didn't derive some of the laws of exponents. I didn't explain what I mean by an irrational number, or prove that the limit must exist and be unique, etc. These you should take as granted - what I want is a straightforward proof of continuity based on the "actual" (!) definition that could be explained to an advanced calculus student frustrated about this question and ready to push themselves to learn - in other words, to my past self.

Fix $a>1$ (the case $a=1$ is trivial, and $0<a<1$ can be handled by switching to $a^{-1}>1$). I'll assume standard facts about $n\mapsto a^n$ for $n\in\Bbb Z$ are known (it is increasing and satisfies the law of exponents). Now to argue the continuity of $\frac bc\mapsto a^{b/c}=(\sqrt[c]a)^b$ on the set of rational numbers $\frac bc$, which is dense in$~\Bbb R$, one can proceed by the following steps.

$a^{b/c}$ is well defined, namely independent of the representation of the rational number $\frac bc$. This means that whenever $\frac bc=\frac pq$, which by definition means $bq=cp\neq0$, one should have $(\sqrt[c]a)^b=(\sqrt[q]a)^p$.
Taking positive integer powers is an injective operation ($x^n=y^n$ implies $x=y$), so we raise both sides to the power $cq$ (assumed positive), so that we must prove $(\sqrt[c]a)^{bcq}=(\sqrt[q]a)^{pcq}$; now we can cancel each root against part of the exponent, and there remains to prove $a^{bq}=a^{cp}$, which is clear.
The function $\frac bc\mapsto a^{b/c}$ is strictly increasing: if $\frac bc<\frac pq$ then $a^{b/c}<a^{p/q}$. Given the previous point, one may assume the fractions have a common denominator $c=q$. Raising both sides to this power (which is also a strictly increasing operation) means it suffices to show $a^b<a^p$; again this is clear.
This increasing function has no jump discontinuities*: for any nonempty open interval$~I$ of positive real numbers, there is some $\frac bc$ with $a^{b/c}\in I$ (a jump discontinuity would give such interval without images). If $I=(x,y)$ put $\gamma=y/x\in\Bbb R_{>1}$, and let $n\in\Bbb N$ be sufficiently large that $\gamma^n>a$. Then $y^n>ax^n$ so there is some integer power $a^m$ of $a$ with $x^n<a^m<y^n$. Taking $n$-th roots (again a strictly increasing operation) it follows that one has $x<a^{m/n}<y$, in other words $a^{m/n}\in I$.
Any increasing function$~f$ defined on a dense set and without jump discontinuities is continuous. Given $x$ in the domain of $f$, and $\epsilon>0$, find values $x_<,x_>$ with $f(x_<)\in(f(x)-\epsilon,f(x))$ and $f(x_>)\in(f(x),f(x)+\epsilon)$ by point$~$3, and take $\delta=\min(x-x_<,x_>-x)$. Then for all $x'$ with $|x-x'|<\delta$ one has $x'\in(x_<,x_>)$ so $f(x')\in(f(x_<),f(x_>))$ by the increasing property of$~f$, and $(f(x_<),f(x_>))\subset(f(x)-\epsilon,f(x)+\epsilon)$, so that $|f(x)-f(x')|<\epsilon$.

This does not yet entirely answer your question, for which it is necessary to show that any function like in 4. can be uniquely interpolated to a continuous increasing function defined on $\Bbb R$. For $x_0$ not in the domain $D$ of $f$ (here an irrational number) one wants to have $f(x_0)=\lim_{x\to x_0, x\in D}f(x)$. To show that the limit exists, one uses that $\sup\{\,f(x)\mid x\in D_{<x_0}\,\}=\inf\{\,f(x)\mid x\in D_{>x_0}\,\}$ (by the absence of jump discontinuities) which is going to be the value of the limit; now the proof that the limit converges can be done by finding values in $D$ sufficiently close to the limit value and some easy but boring estimations, which I will not do here (as per the headache-avoiding request). Also the proof that the extended function is still is continuous and increasing is straightforward.

*This terminology is is not quite proper, as the function on $\Bbb Q
\to\Bbb R$ sending $x\mapsto x+[x^3>2]$ (the brackets meaning $1$ if the condition holds, $0$ otherwise) is continuous on its domain, so it is somewhat unfair to accuse it of having a jump discontinuity, as the above implies. The proper term for what I call "function without jump discontinuities" would be something like "function the closure of whose image is connected" (assuming an increasing function defined on a dense set in both cases), not very attractive.

$f$ is said to be continuous at $a$ if $\lim_{h\to 0}f(a+h)=f(a)$.

In our case:

$\lim_{h\to 0}f(a+h) =\lim_{h\to 0}e^{a+h} = \lim_{h\to 0}e^ae^h = e^a$

This follows because $e^0=1$.

so $e^x$ is continuous at $a$ for all $a \in \mathbb R$

Doesn't get any simpler than that.


Finding k in a polynomial The equation in x

$3x^4 - 16x^3+18x^2+k=0$

has real solutions when.....

I need to find when this equation has four real solutions. I've tried

$x^2(3x^2-16x+18) +k=0$

and using $b^2-4ac$

$256-216 = 40$

I don't know where to go from here....

Given $$3x^4-16x^3+18x^2+k=0\Rightarrow 3x^4-16x^3+18x^2 = -k$$

Now Let $$f(x) = 3x^4-16x^3+18x^2 = -k$$

So we will draw the graph of $f(x) = 3x^4-16x^3+18x^2$ and $f(x)=-k$

Now for Skeching graph of $$\displaystyle f(x) = 3x^4-16x^3+18x^2 $$

Now Using Second Derivative test.

$$f'(x) = 12x^3-48x^2+36x=12x(x^2-4x+3) = 12x(x-1)(x-3)$$

Now for Max. and Min., Put $f'(x) =0$

So we get $x=0,1,3$

Now $$f''(x) = 36x^2-96x+36 = 12(3x^2-8x+3)$$

Now at $x=0;,$ We get $f''(0) = 36>0$

So $x=0$ is a point of local Minimum

Now at $x=1;,$ We get $f''(1) = -24<0$

So $x=1$ is a point of local Maximum

Now at $x=3;,$ We get $f''(3)>0$

So $x=3$ is a point of local Minimum

So The graph of the function is like this.

So we get $f(0) = 0$ and $f(1) = 5$

So for $4$ real solution  line $f(x)=-k$ intersect the graph of $f(x)=0$ at exactly $4$ points

So from graph $0<-k<5\Rightarrow -5 <k<0\Rightarrow k\in (-5,0)$

The answer is $k\in (-5,0)$. Critical points are comming from the square root $\sqrt{-k^3 + 22 k^2 + 135 k}=0$ which gives $-5,0,27$ as suspicious points. Then you check in between them and find the unique interval. You can also find the suspicious points from the inverse function theorem - find the values of $k(x)$ where $k'(x)=0$.


A confusing question on probability In a race, the probabilities of A and B winning the race are
  $\frac{1}{3}$  and $\frac{1}{6}$  respectively. Find the probability
  of neither of them winning the race.

I solved the question in the following manner-

Since A and B are running in a race, probability of neither of them winning is $$1-\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}$$

However, all the solution books that I refer to are solving it in the following manner

$$\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)=\frac{5}{9}$$

Now this does not make any sense to me since the events of A winning and B winning are not independent. I was pretty confident of my answer but even the official answer key of the test has given $\frac{5}{9}$ as the answer.

Where exactly am I wrong? How can the winning of A and B be independent of each other since given that one does not win, the winning chances of the other increases?

As has been discussed in the comments, your interpretation of the question makes much more sense than one that might lead to these events being viewed as independent. Even if this were a race that can be won by more than one participant, it would be a strange race indeed if the chance of winning it were independent of someone else’s chance of winning it.

I did find the wrong answer that you found online, but your correct answer is also online here.

Important
Since $A$ and $B$ are not opponents (this race is not zero-sum),then the probabilities of winnings do not sum up to $1$ and it is possible for both $A$ and $B$ to win the race simultaneously.

Your mistake in summing up the probabilities is that you have considered $$1-[P(A)+P(B)]$$instead of $$P(A'\cap B')=1-[P(A\cup B)]$$i.e. you have assumed that it is impossible for both $A$ and $B$ to win the race.
Remark
The case of not winning one of them, does not increase the chance for the other one's winning. You can equivalently assume that neither of them is aware of the state of the other one and the race is simply a math competition during limited time.


Is $x^n$ Cauchy in $(C[0, 1], \|\cdot\|_{\infty})$? Consider the sequence of functions

\begin{equation}
  f_n(x) = x^n, \quad x \in [0, 1].
\end{equation}

Is this sequence Cauchy in $(C[0, 1], \|\cdot\|_{\infty})$?

The pointwise limit is not continuous because we have

\begin{equation}
  f(x) := \lim_{n \to \infty} f_n(x) =
  \begin{cases}
    0 & \mbox{if $x \in [0, 1)$} \\
    1 & \mbox{if $x = 1$}
  \end{cases}
\end{equation}

Moreover we have

\begin{align}
  \|x^n - x^m\|_{\infty} &= \sup_{x \in [0, 1]} |x^n - x^m| \\
  &\leq \sup_{x \in (0, 1)} |x^n| + |x^m| \\
  &\leq 2 \varepsilon.
\end{align}

The first inequality uses the triangle inequality and the fact that $x^n - x^m = 0$ for $x \in \{0, 1\}$. The second inequality uses the fact that for $x^n \to 0$ for $x \in (0, 1)$ so we can find an $N$ such that $x^n < \varepsilon$ for any $n > N$.

Thus $f_n$ is Cauchy in $(C[0, 1], \|\cdot\|_{\infty})$, which is complete. Hence $f_n \to f \in (C[0, 1], \|\cdot\|_{\infty})$. But we already showed that $f$ is not continuous. Contradiction.

Can anyone tell me where things are going wrong?

We have $\sup_{x\in (0,1)}|x|^n=1$, so the argument in the OP doesn't work. An other reason it that the sequence $\{f_n\}$ is not Cauchy for the supremum norm. Indeed, if it was the case, then $\sup_{x\in (0,1)}|x^{2n}-x^n|$ would be $0$. Let $g_n(x):=x^n-x^{2n}=x^n(1-x^n)$. If $x^n=1/2$, $x=2^{-1/n}$  so $\sup_{x\in [0,1]}|x^{2n}-x^n|\geq \frac 14$ (it's actually equal but we don't need that).

The Cauchy sequence in C[0,1] is not point wise convergent to a continuous function in C[0,1]. Thus, C[0,1] is NOT complete under point-wise convergence.


Prove that the Hamming distances between three n-tuples cannot be 6,2,7 Let $x,y,z \in \{0,1\}^n$, and let $d_H(x,y)$ be the Hamming distance between codes x and y.

Prove
$d_H(x,y) = 6$,
$d_H(y,z) = 2$,
$d_H(x,z) = 7$
cannot happen.

Notice that $d(x,y)=w(x+y)$   where the sum is modulo $2$, and $w$ is the weight function (it counts the number of ones).

Then $d(x,z) = w(x+z) =  w(x+ y + y+ z) = w(a +b)$  where $a=x+y$ and $b=y+z$

Now $w(a+b) = w(a)+ w(b) - 2 w(c)$ where $c=a.b$ is the pointwise product (or bitwise AND function - or the number of ones in common). Then, in general

$$d(x,z)=d(x,y)+d(y,z)-2 k$$

for some integer $k$.

Or, equivalently, $d(x,z)+d(x,y)+d(y,z)$ is even.

Hence, it can never happen that, of the three mutual distance, two distances are even and one is odd. Either we have three even distances, or two odd and one even.

BTW 1: This problem has nothing to do with Hamming (or linear) codes, the statement refers to arbitrary binary tuples.

BTW 2: Next time you ask a question, please try to add some of your thoughts or what have you tried, to show us that you are not simply looking for someone else to do your homework. That's not what this website is for. And if the people here suspect you are doing that, they might downvote, close or refuse to answer your questions.

BTW 3: Welcome to MSE

Welcome to MSE! I've looked into the triangular inequality: $d(x,z)\leq d(x,y)+d(y,z)$. But it doesn't help here: $2+6\geq 7$, $2+7\geq 6$, $6+7\geq 2$.


Locus of centre of variable circle I am not able to figure out this question

What is the locus of the centre of a circle which touches a given line and passes through a given point, not lying on the given line?

I think it's a parabola but I am not able to prove it mathematically

Without loss of generality, we may assume that the line is the $x$-axis, and that the point is the point $(0,a)$, where $a$ is positive.

A circle with centre $(p,q)$ that touches the $x$-axis must have radius $|q|$. So it has equation
$$(x-p)^2+(y-q)^2=q^2.$$
The circle goes through $(0,a)$. It follows that
$$(0-p)^2+(a-q)^2=q^2.$$
Simplify. We get
$$p^2+a^2-2aq=0.$$
We can rewrite this as
$$q=\frac{1}{2a}p^2 +\frac{a}{2},$$
indeed the equation of a parabola.

Assume the line is y-axis and the point is $(a ,0)$.the centre of circle is $(h,k)$ then the equation of circle:

$$(x-h)^2 +(y-k)^2 =r^2 \qquad (1)$$

where $r=$radius of  circle
but $r=k$
putting this value in equation $(1)$
we get

$$(x-h)^2 +(y-k)^2 =k^2 \qquad (2)$$

but the circle is passing through the point $(a,0)$ therefore putting
$x=a, y=0$ in equation $(2)$ we get

$$(a-h)^2 +(0-k)^2 =k^2 $$

simplify:

$$a^2 +h^2 =2ah \qquad (3)$$

equation $(3)$ is the locus of centre of circle touching one axis and passing through a point not lying on the given line.


Group of order 15 is abelian How do I prove group of order 15 is abelian?

Is there any general strategy to prove that a group of particular order(composite order) is abelian?

Here is a 2000 paper of Pakianathan and Shankar which gives characterizations of the set of positive integers $n$ such that every group of order $n$ is (i) cyclic, (ii) abelian, or (iii) nilpotent.

Say that a positive integer $n > 1$ is a nilpotent number if $n = p_1^{a_1} \cdots p_r^{a_r}$ (here the $p_i$'s are distinct prime numbers) and for all $1 \leq i,j \leq r$ and $1 \leq k \leq a_i$, $p_i^k \not \equiv 1 \pmod{p_j}$.  Also, let us say that $1$ is a nilpotent number.

(So, for instance, any prime power is a nilpotent number.  A product of two distinct primes $pq$ is a nilpotent number unless $p \equiv 1 \pmod q$ or $q \equiv 1 \pmod p$.)

Then, for $n \in \mathbb{Z}^+$:

(i) (Pazderski, 1959) Every group of order $n$ is nilpotent iff $n$ is a nilpotent number.
(ii) (Dickson, 1905) Every group of order $n$ is abelian iff $n$ is a cubefree nilpotent number.
(iii) (Szele, 1947) Every group of order $n$ is cyclic iff $n$ is a squarefree nilpotent number.

For example, if $n = pq$ is a product of distinct primes, then $n$ is squarefree, so every group of order $n$ is nilpotent iff every group of order $n$ is abelian iff every group of order $n$ is cyclic iff $p \not \equiv 1 \pmod q$ and $q \not \equiv 1 \pmod p$.  In particular, every group of order $15$ is cyclic.

Addendum: This 2006 paper of T. Müller is a natural followup.  Rather than describing it myself, let me quote the MathSciNet review.

  It is a popular problem to find for which positive integers n do all groups of order n have a given property (e.g., cyclicity, are abelian, etc.). The article under review contains a contribution to this problem which seems to have escaped notice.
     Define a multiplicative function $\psi$ on the positive integers by letting $\psi(1)=1$, and $\psi(p^ν)=(p^{ν}−1)(p^{ν−1}−1)\cdots(p−1)$ if $p$ is a prime and $ν\geq 1$. The author proves that every group of order $n$ is nilpotent of class at most $c$ if and only if $\operatorname{gcd}(n,\psi(n))=1$ and $n$ is $(c+2)$-power free. Setting $c=\infty$ yields a result of G. Pazderski [Arch. Math. 10 (1959), 331--343; MR0114863 (22 #5681)] describing the case of nilpotency; and setting $c=1$ yields the classic result of L. E. Dickson [Trans. Amer. Math. Soc. 6 (1905), no. 2, 198--204; MR1500706] describing the case of abelianness. (Reviewed by Arturo Magidin)

Since $G$ contains an element $x$ of order $3$ and an element $y$
of order $5$.  The  map $f: \mathbb{Z_3} \times  \mathbb{Z_5} \rightarrow  G $defined by $f((i, j)) = x^i y^j$ is an isomorphism, provided $xy = yx$,
and so $\mathbb{Z_3} \times \mathbb{Z_5} \cong \mathbb{Z_{15}}$


Cardinality of Cartesian Product where (a,b) are elements of A x B I'm having issues with a question regarding the cardinality of a cartesian product.

Question:
Let $A=\{0, 1, 2, 3, 4, 5, 7\}$ and $B=\{0, 2, 4, −1, 12\}.$

How many elements are in $\{(a, b) ∈ A × B \; |  \; a < 7 \text{ and } b < 4\}$?

I want this to be $6 \cdot 3 = 18$. but that seems to be wrong.

You may begin by omitting all $a \in A$ such that $a\ge 7$ and all $b \in B$ such that $b \ge 4$ because we know ordered pairs with said elements will not be included in the Cartesian product we're looking for. So, after omission, we have $A = \{0,1,2,3,4,5\}$ and $B=\{0,2,-1\}$. Now, the cardinality of the Cartesian product between sets $A$ and $B$, denoted as $|A$ x $B|$, is given by $|A$ x $B|$ = $|A| \cdot |B| = 6 \cdot 3 = 18$. So you are correct! If your book says differently, then I highly suspect your book contains a mistake.

If I've understood your question correctly, you'll begin by taking the appropriate subsets of A and B.

Since A = {0,1,2,3,4,5,7} and B = {0,2,4,−1,12}, and you're restricting your subsets to a < 7, and b < 4, this leaves the sets C = {0,1,2,3,4,5} and D = {0,2,-1}.

If you allow for repetition, which your question seems to assume, then there are 6 ways to select the first element, and 3 ways to select the second, implying a total of 18 ways to construct ordered pairs on the subsets.

Note that your textbook might be looking for un-ordered subsets of size 2, which is a different question.


Is $0^i$ indeterminate? According to Euler's identity

  $$e^{ix}=\cos{x}+i\sin{x}$$

Using this identity

  $$0^i=e^{i\ln{0}}=\cos\ln0+i\sin\ln0=\cos{-\infty}+i\sin{-\infty},\because \ln0=-\infty$$

But $\cos{-\infty}$ and $\sin{-\infty}$ are indeterminate Does it mean that $0^i$ is indeterminate.
Please Explain.

The critical point comes from the trigonometric functions. Instead, start from the equality
$$
e^{\ln x} = x
$$
which holds for every positive real $x$. Then $\lim_{x \to 0^+} x = 0$, $\lim_{x \to 0^+} = -\infty$ and $\lim_{x \to 0^+} e^{\ln x} = 0$, since $x \to e^x$ is continuous on $\mathbb{R}$ and tends to $0$ when $x$ tends to $-\infty$. Thus in this case, $e^{-\infty} = 0$ is an acceptable convention, which is actually frequently used in mathematics.

However, in your case, neither $\cos x$ nor $\sin x$ have a limit when $x$ tends to $-\infty$. I hope this explains the difference between the two cases.

As you point out, one can not define $0^i$ via the identity of Euler. The question is whether we can find any other good properties of $0^i$. For me good properties would be that the function $f:x\mapsto x^i$ is continuous. This function is basically defined for $\mathbb{C}\backslash\mathbb{R-}$, where the $log function is not defined.

Can we extend the definition of the function to $0$?
For this we see
$$\lim_{x\in\mathbb{R},x\rightarrow 0} x^i =$$
$$=\lim_{x\in\mathbb{R},x\rightarrow 0} e^{i\cdot log(x)}$$

This limit does not converge, so we do not have any continuous extension to $0$


The set of all things. A thing itself? If the universe is the set of all things. Does it contain itself? In other words is it a thing itself?

I know its a stupid question, but it really grinds my gears.

Thanks!

Edit 8.12

Okey, someone here said that it cant exist. So what if it would be a proper class does, that change anything? And when I say "thing" I mean to be distinguishable.

Hope I didnt confuse someone even more!

Typically the way you define this is you let $\Omega$ be the set of all elementary things, and then you pick from the set of all subsets of $\Omega$ (a.k.a. the power set of $\Omega$, denoted either as $2^\Omega$ or $\mathcal{P}(\Omega)$.

Then, $\Omega \in 2^\Omega$ and $2^\Omega \not \in 2^\Omega$, so this setup causes no casuistic problems.

However in your case, defining $\Omega$ as the set of all things you must first define is a set of things is a thing itself or not.

The set of all things (physical or abstract) cannot exist. Put another way, every set must exclude something.

Suppose the set $U$ of all things did exist.

$\forall a: a\in U$

Then, there would have to exist a subset $r$ of $U$ that is the set of all things that are not elements of themselves (the so-called Russell set).

$\forall a:[a\in r \iff a\in U \land a\notin a]$

In the manner of Russell's Paradox, since $r\in U$, we could then obtain the the contradiction

$r\in r \iff r\notin r$

Therefore, $U$ as defined here cannot exist.


Differentiability and continuity of a trig function Here's a problem I'm having a lot of trouble with:

  We have the following function:

  $f(t) = t^2\cos(\dfrac{1}{t})$ for $t \neq 0, f(t) = 0$ for $t = 0$.

  Show $f$ is continuous everywhere, use the squeeze theorem for $t=0.$
  Show $f$ is differentiable in $ t\neq 0$, and find the derivative $f'(t)$.
  Show (using the definition of differentiability: $f'(t) = \lim_{h \to 0}\dfrac{f(t+h) - f(t)}{h}$) that $f$ is differentiable in $t=0$ and find the derivative $f'(0)$. Show that the derivative $f'$ is not continuous in $0$.
  Look at $f$ on the domain $[0, \infty]$. Show that $f$ has no local maximum or minimum in the boundary point $0$. You can do this by showing that for every number $a>0$ there are values with $0<t<a$ for which $f(t) >0$ and also values $t$ with $0<t<a$ for which $f(t) < 0$.

Here's my attempt (and my difficulties) at these:

I don't exactly know how to show that a function is continuous everywhere, but from what I looked up online this is what I tried:

$t^2\cos(\dfrac{1}{t})$ has a domain $\mathbb{R}$, except at $t=0$, where it is undefined. However, we have $f(0) = 0$, so to show $f(t)$ is continuous at $t=0$ we have to show that $\lim_{t \to 0} f(t) = f(0) = 0$. We can do this by the squeeze theorem:

On the interval $[-1,1]$ we have $$-1 \leq \cos(\dfrac{1}{t}) \leq 1$$   $$-t^2 \leq t^2\cos(\dfrac{1}{t}) \leq t^2$$

We know that $\lim_{t \to 0} -t^2 = 0$, and  $\lim_{t \to 0} t^2 = 0$, so  $\lim_{t \to 0} t^2\cos(\dfrac{1}{t}) = 0$, by the squeeze theorem.

Here my biggest confusion lies in the fact that you have to "show" that $f$ is differentiable for $t \neq 0$, before actually differentiating it. This implies that the two are not equal to each other, which I didn't expect. My logic tells me that if you're able to find a function's derivative, you've shown that it exists. But I think what they want here is for you to find the derivative using the following method:

At $ t \neq 0$, we have $f(t) = t^2\cos(\dfrac{1}{t})$. This function is differentiable if the following limit exists: $$\lim_{h \to 0}\dfrac{(t+h)^2\cos(\dfrac{1}{t+h}) - t^2\cos(\dfrac{1}{t})}{h}$$

Sadly, I don't know how to evaluate this limit, and you're not allowed to use L'Hospital because we haven't had that in class yet. Would anyone know how to do this using more "elementary" methods (meaning mostly algebraic stuff)?

Of course, I can find the derivative using traditional means, it is $2t\cos(\dfrac{1}{t}) + \sin(\dfrac{1}{t})$, but I don't know if that's sufficient because of the way the question is phrased.

The first part of this question is a mystery to me, here's what I tried:

Use the definition of differentiability, and fill in $t=0$ in the equation. We get $$ f'(0) = \lim_{h \to 0}\dfrac{h^2\cos(\dfrac{1}{h}) }{h}$$

Of course this makes no sense, because we're dividing by zero and stuff, and even if we remove that term as I did above, the limit still doesn't exist. So I have no clue how to tackle this problem otherwise. As for the continuity of $f'$ in 0:

$ \lim_{t \to 0} 2t \cos(\dfrac{1}{t}) + \sin(\dfrac{1}{t})$ doesn't exist, hence it is not continuous at $0$.

I understand this conceptually, but I have no idea how to tackle it. I understand that as $t$ goes to $0$, that the amplitude of $t^2\cos(\dfrac{1}{t})$ goes to $0$ because $t^2$ goes to $0$ and I understand that the period will also go to $0$ because $\dfrac{1}{t}$ goes to infinity. But I don't know how to prove that there will always exist a number arbitrarily close to $t=0$ for which $f(t)>0$ or $f(t)<0$.

A partial answer. First, regarding your confusion in step (2):

The product rule and the chain rule are not just formulas, but also logical statements:

If $f$ and $g$ are differentiable on an interval $I$, then their product $fg$ is differentiable on $I$. (And its derivative is given by the following formula: ...)
If $f$ is differentiable on an interval $J$, and $g$ is differentiable on an interval $I$, and $g$ maps $I$ into $J$, then the composition $f \circ g$ is differentiable on $I$. (And its derivative is given by the following formula: ...)

Now you can argue like this: the function $x(t)=1/t$ is differentiable on interval $I=\{t>0\}$ and maps $I$ into the interval $J=\{x>0\}$, and $\cos x$ is differentiable on $J$; hence the composition $\cos(1/t)$ is differentiable on $I$. And $t^2$ is differentiable on $I$, hence the product $t^2 \cos(1/t)$ is differentiable on $I$. Similarly for the interval $\{ t<0 \}$. Q.E.D.

In practice, of course, this kind of argument works for any function where you are actually able to compute its derivative using the product and chain rules, so usually one doesn't bother to write it out. And it's perfectly fine for you to compute the derivative using the rules in this kind of exercise; there is no need to fall back to the definition in step (2) (but in step (3) you will need to do that, of course).

Regarding (3):

The limit of
$\frac{h^2\cos(1/h) }{h}$ (as $h \to 0$) is zero since you can write the expression as
$$
\underbrace{h}_{\to 0} \cdot \underbrace{\cos(1/h)}_{\text{bounded}}
$$
(the cosine of anything is always between $-1$ and $+1$),
and there's a theorem about limits that says that if $f\to 0$ and $g$ is bounded, then $fg\to 0$.

For (2) use the Chain Rule:

$$\left(t^2\cos\frac1t\right)'=2t\cos\frac1t+t^2\left(-\frac1{t^2}\right)(-\sin\frac1t)=2t\cos\frac1t+\sin\frac1t$$

For (3) use the definition given there:

$$\frac{f(t)-f(0)}t=t\cos\frac1t\xrightarrow[t\to 0]{}0$$

the derivative function isn't continuous at $\;t=0\;$ since $\;\lim_{t\to 0}\sin\frac1t\;$ doesn't exist.

For (4) just show that there arbitrarily many points in any neghborhood of zero where $\;\cos\frac1t>0\;$ and many with $\;\cos\frac1t<0\;$

By the way, note that (1) now follows from (2)+(3) since a differentiable function is continuous.


Commutator of generalized Gell-Mann matrices Is there an explicit formula for commutator of generalized Gell-Mann matrices? For example, pauli matrices (generalized Gell-Mann in dimensions $ d = 2 $):
$$ [\sigma_a, \sigma_b] = 2 i \varepsilon_{abc} \sigma_c $$

Yes, there is such a thing.  For the Gell-Mann matrices of $\mathrm{SU}(N)$
$$\begin{align}
    \left[ \lambda_i, \lambda_j \right] &= 2 i \sum_k f_{ijk} \lambda_k, \\
    \{ \lambda_i, \lambda_j \} &= \frac{4}{N} \delta_{ij} I + 2 \sum_k d_{ijk} \lambda_k,
\end{align}$$
with the structure constants
$$\begin{align}
    f_{ijk} = -\frac{1}{4} i \operatorname{tr}(\lambda_i [ \lambda_j, \lambda_k ]), \\
    d_{ijk} = \frac{1}{4} \operatorname{tr}(\lambda_i \{ \lambda_j, \lambda_k \}),
\end{align}$$
where $\operatorname{tr}(\cdot)$ is the trace and $I$ is the identity matrix.


Finding Tangent Line to Graph Find a vector equation for the tangent line to the curve of intersection
of the cylinders

$\ x^2 + y^2 = 25$
and
$\ y^2 + z^2 = 25$

at the point (3,4,2).

I don't understand the answer key. I've explained my interpretations below. Could someone clarify:

Why is the projection of C onto the xy plane:$\ x^2 + y^2 = 25$.  I understand that a projection in 3D space is the "shadow" of the graph onto one plane, but I don't quite see that C's projection can be modeled as given.

Why is z >= 0?

The intersection is a subset (or, if you prefer, contained within) the cylinder given by $x^2+y^2=25$, right? That equation is not sufficient to determine the points on the curve, but they all do fulfill it, because one thing the intersection is is exactly being part of that cylinder.

Now, if you take any point on the cylinder and project it onto the $x/y$ plane, what do you get? You get a point with $z=0$ and $x^2+y^2=25$. Try with the given point, $3^2+4^2=25$.

As for $z>0$, you are in the vicinity of $(3,4,2)$, and for that point, $z=2>0$.

The projection of the cylinder $x^2 + y^2 = 25$ onto the $xy$ plane is the circle $x^2 + y^2 = 25,\,z = 0$. Any intersection with the cylinder will lie on the cylinder. Thus the projection of the intersection will be some part of the projection of the cylinder.

$z$ is close to 2 near the point $(3,\,4,\,2)$ and is thus more than 0.


Get an alternate form of an equation I am trying to prove the following limit value:

$$\lim_{n\to \infty} \frac{n^3 + 1}{n^2 +1} = \infty$$

But how does

$$\frac{n^3 + 1}{n^2 +1}$$

become this

$n + \frac{1 - n}{n^2 +1}$?

by using the long division ..................


Can you help me to solve the recurrence relation $T(n) = T(\sqrt n) + 1 $? I have this recurrence relation to solve :

$T(n) = T(\sqrt n) + 1 $

I have tried to expand the recursion but I stopped here:

\begin{align}
T(n) &= T(n^{\frac12})+1\\
     &= T(n^{\frac14})+1+1\\
     &\text{after $i$ replacements I have}\\
     &= T(n^{\frac1{2^i}}) + i\\
\end{align}
I know that $T(1) = 1$

And now? How can I get to the solution?

If it's related to algorithm complexity, then in this context $\sqrt{n}$ is likely to mean $\lfloor\sqrt{n}\rfloor$ (or $\lceil\sqrt{n}\rceil$). So assume that $T(1) = 1$ and $T(n) = T(\lfloor\sqrt{n}\rfloor)+1$ for $n>1$. Consider $n=2^{2^k}$:
$$T(2^{2^k})=T(\lfloor\sqrt{2^{2^k}}\rfloor)+1 = T(2^{2^{k-1}}) +1 = T(2^{2^{k-2}}) + 2 = \dots=T(2^{2^0}) + k = T(2) + k$$
Now $T(2) = T(\lfloor\sqrt{2}\rfloor) + 1=T(1) + 1 = 2$. So we ended up with
$$T(2^{2^k})=k+2$$
Notice that $T(2^{2^{k+1}}-1)=T(2^{2^k})$ and for $n\leq m$ we have $T(n)\leq T(m)$ (easy to prove by induction). This gives us $T(n)=\lfloor\log_2{(\log_2{n})}\rfloor+2$ for $n>1$.

Let n= 2^m
sqrt(n) = 2^(m/2)
m = logn

replacing T(2^m)= T(2^(m/2)) + 1

now take s(m)= T(2^m)
therefore, s(m)= s(m/2) + 1
now
applying Master rule : n^log(base2)1 = n^0 = 1 equals to f(n)=1
Therefore, t(n)=s(m)= 1.logm = log m =log(log n). [Ans.]


Triangular vs Normal distribution I'm trying to approximate a standard normal distribution with a triangular distribution. What parameters of the triangular distribution (min, max and mode) are more suitable? Thank you

The best approximation in the $L^2$ sense is given by the value of $\alpha\in\mathbb{R}^+$ for which:

$$ \frac{d}{d\alpha}\left(\frac{1}{2\pi}\int_{|x|\geq \alpha}e^{-x^2}\,dx + \int_{-\alpha}^{\alpha}\left(\frac{1-|x/\alpha|}{\alpha}-\frac{e^{-x^2/2}}{\sqrt{2\pi}}\right)^2\,dx\right)=0,$$
i.e. by minimizing the $L^2$ norm of the difference between the pdf of a standard normal distribution, $\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$, and a distribution supported on $[-\alpha,\alpha]$ having pdf $\frac{1}{\alpha}\left(1-\left|\frac{x}{\alpha}\right|\right).$

Numerically, it is $\color{red}{\alpha\approx 2.297}$. Here there are the two distributions:

$\hspace2in$

from your Triangle Distribution you have a (min), (max), and (mode).
in a Normal Distribution you have a (mean) and (standard deviation) abbreviated as (sd) .
in the Normal Distribution the (mean) is equal to the (mode) due to the symmetry of the model.
to move from a Triangle model to a Normal model we need to estimate the (mean) and the (standard deviation) from the Triangle models parameters
for the (mean):
it is reasonable to use your (mode) of the Triangle Distribution as your (mean) for your Normal Distribution because they are equal under the Normal model.  NOTE at this point one should be aware they maybe breaking a symmetry assumption in their normal model.
a second reasonable estimate for the (mean) can come from the (midpoint) of the (min) and the (max) where (midpoint)=((max)-(min))/2.
the (mean) of the Normal distribution represents the central tendency. a (midpoint) and a (mode) are also measures of central tendency thus they are candidates that we have readily available from the Triangle Distribution.
a third method will to be use the (mean) of the Triangle Distribution and use it to estimate the mean a Normal Distribution.  the (mean) of a triangle distribution is (mean_tri)=((min)+(max)+(mode))/3
method 1 (central tendency):
(mean)=(mode)
or
method 2 (central tendency):
(midpoint)=((max)-(min))/2
(mean)=(midpoint)
or
method 3 (match moments):
(mean_tri)=((min)+(max)+(mode))/3
(mean)=(mean_tri)
for the (standard deviation):
a Triangle distribution gives us a (min) and a (max).  We can take the difference to find a range value (range)=(max)-(min).  a function of the range can provide an estimate of the standard deviation of a Normal distribution f(range)=(sd). It is up to the user to determine a reasonable function at this point.  example: four standard deviation covers 95% of a Normal Distribution's data and would result in an estimated standard devastation of (sd)=(range)/4, while a six standard deviation range covers 99% and would result in an estimate of (sd)=(range)/6.
Another method will be to take the (standard deviation) of the Triangle Distribution and use that to estimate the (standard distribution) of the Normal Distribution. the (sd) of a triangle distribution is (sd_tri)=sqrt(((min)^2+(max)^2+(mode)^2-(min)(max)-(min)(mode)-(max)(mode))/18)
method 1 (range coverage):
try setting
(range)=(max)-(min)
and using:
(sd)=(range)/2  (65% normal coverage)  (fattest of the three models)
or
(sd)=(range)/4  (95% normal coverage) (middle width of the three models)
or
(sd)=(range)/6  (99% normal coverage) (skinniest of the three models)
method 2 (match central moments):
(sd_tri)=sqrt(((min)^2+(max)^2+(mode)^2-(min)(max)-(min)(mode)-(max)(mode))/18)
(sd)=(sd_tri)
when estimating your (mean) and (standard deviation) you may want to consider how much confidence you had in your Triangle's coverage of the population and if the original distribution was skewed.  Did the (min) and (max) result from a very small sample or a very large sample?  If i am skewed, do i want to focus in on the mode or do i want to go for more coverage?


No. of four-digit even numbers formed by digits 0 1 2 3 4 5 without repetition I started from the units.
There can only be $3$ possibilities $0,2,4$.
Then there are $5$ possibilities for the tenth place.
THen there are $4$ possibilities for the hundredth place.
Then only $3$ possibilities remain.
Thus the permutation is $3 \times 4 \times 5\times 3 = 180$.
Is it correct?
Need guidance.

4 digit even numbers(including $0$ at the beginning) = $3* 5* 4* 3 = 180$

for choosing the last digit we have 3 ways {0,2,4}. Now we are left with 5 numbers, for rest 3 places  $5*4*3$ ways

Now we will subtract those 4 digit even numbers with $0$ as the first digit = $2 * 4 * 3= 24$.
First digit is fixed = 0, 1 way. Now we are left with 2 choices for last digit (2,4). For choosing last digit 2 ways.
Now we have 4 digits and 2 places, $4*3$ ways.

Req.number of numbers = $180 - 24 = 156$

case(1): If repetition is not allowed,
1’s place can be filled by the nos (0,2 or 4) in 3P1 ways = 3 ways
10’s, 100’s and 1000’s places can be filled by 5 x 4 x 3 = 60 ways
Therefore, total number of arrangements = 60 x 3 = 180.
If ‘0’ comes in the 1000’s place, the number becomes 3 digits. Such 3 digits numbers =1 x 3 x 4 x 2 = 24.
Total number of 4 digit even numbers = 180  24 =156.


Routine Probability Question, Computational Issue, Alternate Way to Solve? I feel like my approach and reasoning for the question provided below are good. Considering it is from a multiple choice exam with a strict time limit, I am not sure what other approach one might take. However, once I derive my final expression (which I believe to be accurate), I am unable to compute its decimal value for computation reasons. Please help me to see if either:

1) I am wrong

2) There is an alternative approach that will avoid the computational complexity

3) There is a trick to computing the expression I am stuck on.

The question is:

  An investor buys \$100 of worth of a stock. Each month, the investment
  has probability .5 of increasing \$1.10 and probability .5 of
  decreasing \$0.90. Changes in price each month are mutually
  independent. What is that probability that after 100 months the value
  of their stock is worth more than \$91?

How I reasoned through this was to picture the different possibilities each month as a tree - each node has the two options: increase 1.10 or decrease .90. For month 100 there are $2^{100}$ nodes. Since mutually independent, the probabilities along these  nodes are all $(.5)^{100}$. Now we just need to count all the branches that give a final value greater than \$91. If there are $k$ of them, the final probability we seek is $k \cdot (.5)^{100}$.

From here I proceed by considering $$100 +m \cdot (1.10) - (100 - m) \cdot (.9) > 91$$ where $m$ represents the number of months the stock increased, and $(100-m)$ is the number of months it decreased. Solving yields $m = 40.5$ but we want integers so $41 \leq m \leq 100$. This gives 60 different combinations $(\text{increase, decrease}) = (41,59), (42, 58), \dots , (100,0)$ that result in a success branch.

However, for each combination we need to account for all the permutations. I.e, for say $(41,59)$ there would be ${100\choose 41}$ arrangements in which this success branch occurs. This gave me my final expression which should evaluate to the desired probability, $$P[X \geq 91] = (.5)^{100} \sum_{j=41}^{100} {100 \choose k}.$$

This is not easily computable, even with a calculator. So now I'm not sure what to do. I also don't see a very clean way to apply the Central Limit Theorem, which would be my usual go to in situations like this.

The options were:

a) .63

b) .75

c) .82

d) .94

e) .97

I dare you to ask me ‘what have you tried’ :)

As mentioned, this question comes from a multiple choice exam where time is crucial. You are expected to spend 4 - 8 minutes per problem, so there must be some reasonably efficient approach to this.

After leaving this question unanswered for quite some time, I stumbled upon it today and decided to see what I could do. I have found a way to compute the expression I derived above, but I am still not satisfied with this method because I know they wouldn't expect this solution on a timed exam, so I will also present an approximation.

Exact Computation:

As I mentioned we have 61 ordered pairs (decrease, increase) that result in a success branch. However, we can exploit the symmetry of the binomial coefficient. For the first 41 ordered pairs (0,100), (1, 99),$ \dots $, (40,60) that result in a success branch, there are corresponding ordered pairs that give a fail branch (100,0), (99,1),$ \dots $, (60,40). By the symmetry of the binomial coefficient, the corresponding ordered pairs give the same number of arrangements, so the probabilities that corresponds to each of these two sets of outcomes is equal, call it $a$. By definition $a$ is the probability of the compliment event P[X $\leq$ 91], but we have found a portion of the original event that shares this probability.

For the event we are interested in, the total probability is then $$P[X > 91] = a + (1/2)^{100} \cdot \sum_{k = 41}^{59} {100 \choose k}.$$ Call the term on the right $c$ so that P[X > 91] = $a + c$. Then from above, we also know that $2a + c = 1$. Since we have complimentary events. Thus if we can get a value for $c$ we can determine $a$ which gives the desired probability.

For computing $c$, we exploit the symmetry further since 100 - 41 = 59, 100 - 42 = 58, etc. With this, we can compute $c$ as $$c = (1/2)^{99} \sum_{k = 41}^{49} {100 \choose k} + (1/2)^{100} {100 \choose 50}.$$

Although this is still not exactly "nice" or "easy" to compute, it is computable. I was able to plug this into my simple TI-30 calculator in just a minute or two and get a value of $c \approx .943111$. With this once can determine $a$, then we have $$c + a = P[X > 91] \approx .9715$$ which is indeed one of the options for answers.

But, all of this was a lot of work for a question that should be done in 6 - 8 minutes. So, there is also this;

Approximation via C.L.T

Let $X_i$ represent the change in value at the end of any particular month, so $$X_i:= \begin{cases} 1.1, \hspace{2mm} P = .5\\ -.9, \hspace{2mm} P = .5 \end{cases} $$
Then the total change at the end of the hundred month is $$X = \sum_{i = 1}^{100} X_i$$ and we have $Z = 100 + X$ represents the value of the stock after the 100th month. Then $$P[Z > 91] = P[100 + X > 91] = P[X > -9].$$

The expected value and variance of $X_i$ can easily be computed so that by the Central Limit Theorem $X \sim N(10,100)$. Employing the standard normalization we have $$P[X > -9] = P[Z > -1.9] = P[Z < 1.9] \approx .9713 \hspace{2mm} (\text{via a Z-table}).$$

This answer is consistent with what we found previously.

I suppose this method is perhaps what joriki was suggesting in his answer, though a lot more specific, and I wasn't sure what values he was suggesting I take logarithms.

Take logarithms and approximate the resulting sum by a normal distribution with mean and variance given by the sums of the corresponding quantities of the individual changes.


Prove that $\sqrt 2 + \sqrt 3$ is irrational I have proved in earlier exercises of this book that $\sqrt 2$ and $\sqrt 3$ are irrational. Then, the sum of two irrational numbers is an irrational number. Thus, $\sqrt 2 + \sqrt 3$ is irrational. My first question is, is this reasoning correct?

Secondly, the book wants me to use the fact that if $n$ is an integer that is not a perfect square, then $\sqrt n$ is irrational. This means that $\sqrt 6$ is irrational. How are we to use this fact? Can we reason as follows:

$\sqrt 6$ is irrational

$\Rightarrow \sqrt{2 \cdot 3}$ is irrational.

$\Rightarrow \sqrt 2 \cdot \sqrt 3$ is irrational

$\Rightarrow \sqrt 2$ or $\sqrt 3$ or both are irrational.

$\Rightarrow \sqrt 2 + \sqrt 3$ is irrational.

Is this way of reasoning correct?

If $\sqrt{2} + \sqrt{3}$ is rational, then so is $(\sqrt{2} + \sqrt{3})^2 = 5 + 2 \sqrt{6}$. But this is absurd since $\sqrt{6}$ is irrational.

As already pointed out, the sum of
two irrational numbers can be rational,
so your proof is invalid.

This is even true if both numbers
are positive, as the following shows:

  Let $a = 0.12112111211112...$
  and form $b$ by changing every $1$ in $a$
  to a $2$ and every $2$ to $1$.

  So $b = 0.21221222122221...$

Clearly $a$ and $b$ are irrational,
but $a+b = 0.33333... = \frac 13$,
which is a rational number.


Finding probabilities given a small amount of data in the situation A survey found that 79% of children like chocolate, 63% like cheese and 59% like chicken.

a) What is the highest possible value for the percentage of children that don't like any of the three?

b) What is the lowest possible value for the percentage of children that like all three of the foods?

c) What is the highest possible value for the percentage of children that like exactly one of the foods?

I have absolutely no idea how we would work these out with the information given. I don't feel like I've been given enough information.

I've taken a look at this question and believe I know the answers to part a and part b. However I'm stuck on how to calculate part c.

Part a) Consider this as three Venn diagram bubbles, the totals of which are constant. Then the figures are saying that 79% of children must like chocolate. As the other Venn bubbles are smaller percentages then it is possible that these bubbles could fit entirely within the category of children liking chocolate. This would give an answer of 21%.

Part b) This can be solved using the Inclusion Exclusion principle provided in the following wikipedia article:

http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

We know the individual percentages for each category. For each paired category, as we are looking for the minimum percentage of people who like all three then we want minimum overlapping for the three pairings. This will leave us an equation to solve for children who like all three. I believe this equals 1%.

I hope this information is helpful

There must be an overlap of at least 42% between those who like chocolate and those who like cheese, otherwise the total percentage would be greater than 100%.  The maximum overlap between these to is 63%, obtained if everyone that likes cheese also likes chocolate. Reason similarly with the other cases.


Find out the primitive polynomial GF(3) 1.) $x^2 + 2x$
2.) $x^2 + 1$
3.) $x^2 + 2$
4.) $x^2 + 2x$
5.) $x^2 + 2x + 1$
6.) $x^2 + 2x + 2$
7.) $x^2 $
8.) $x^2 + x + 2$
9.) $x^2 + x + 1$

Can any one help me in listing out primitive polynomials and tell me why is it a primitive polynomial please.

I don't really know much about about primitive polynomials
But You can read about it  here

Following are primitive polynomials
$$x^2+2x+2$$
$$x^2+x+2$$

and This Page Generates Primitive polynomials

$$x^2 + 1,
x^2 + 2x + 2,
x^2 + x + 2,$$
are primitive polynomials over $gf(3)$.
Because they are irreducible and monic polynomials
Monic means coefficient of highest exponent of $x$ is $1$.
And irreducible means can not be further factorise into small polynomials.

Let's take first case $x^ 2 + 2x$ is further factorise as $x(x + 2)$.
Third case $x^2 + 2$ is monic but it is factorise as $(x-1)(x+1)$.
Because in $gf(3), 2 = -1$ and hence $x^2+2 = x^2 -1 = (x-1)(x+1)$.
And you can also check it by taking $x = 1$.

$1^2 + 2 = 3$ which equal to $0$ ( because $3 \pmod3 = 0$) so $x-1$ is factor of $x^2 + 2$  in $gf(3)$.
Now check other polynomials by yourself and get answer.


Are the definite and indefinite integrals actually two different things? Where is the flaw in my understanding? Some context: I'm an engineer, and I tend to have a rather unusual way of understanding and thinking about things, most likely related to my being autistic. I found this question on the HNQ and upon reading it I felt rather confused.

I have always understood that the indefinite and definite integrals are two sides of the same underlying concept, that there is no meaningful difference between them other than that one evaluates to a function and the other to a quantity. Essentially, the definite integral is what you get from running the numbers on the result of an indefinite integral, and saying it's unrelated to the definite integral is like saying that $\sin(x)$ is fundamentally different from $\sin(a)$, assuming $x$ is a variable and $a$ a constant, for no reason other than that $x$ is a variable and $a$ a constant.

To me, whether the definite and indefinite integrals are two sides of the same thing, or two closely related but different things, seems more like a question of philosophy than mathematics.

Where is the flaw in my understanding? Is there one?

The point is, there are three slightly different concepts here:

1) If $f$ and $F$ are functions with the same domain, and $F'(x)=f(x)$ on that domain, then we say that $F$ is an antiderivative of $f$.

2) If $f$ is a function on the interval $a \le x \le b$, then the definite integral of $f$, $\int_a^b f(x) \, dx$, is (loosely speaking) the area under the graph of $y=f(x)$ for $a \le x \le b$, or (more precisely speaking) a limit of Riemann sums. This definition does not involve any derivatives.

3) If $f$ is a function on the interval $a < x < b$ and $c$ is a point in that interval, then the function $G(x)=\int_c^x f(t) \, dt$ could be called an indefinite integral of $f$. This definition of the function $G$ comes directly from the definition in 2) of the definite integral, so it also does not involve any derivatives.

You are saying that you don't see any difference between 2) and 3). You are correct! If your mental definition of "indefinite integral" is 3), then an indefinite integral is just a definite integral with some unknown limits.

The question you're linking is saying that many calculus books use 1) as the definition of an "indefinite integral": that is, they say that if $F$ is an antiderivative of $f$, then $\int f(x) \, dx = F(x)+C$ is the indefinite integral of $f$. They can get away with this because of the fundamental theorem of calculus, which says (in part) that

$$
\frac{d}{dx}\int_c^x f(t) \, dt=f(x) \, .
$$

That is, the function $G(x)$ defined in 3) is an antiderivative of the function $f$, so these two notions of "indefinite integral" are closely related.

But closely related does not mean identical! K B Dave's answer is giving you some examples of situations where the two definitions 1) and 3) diverge slightly. Specifically:

We gave the definition 3) on an interval $a < x < b$, but it doesn't work very well on a more complicated domain. This means that definition 1) is more robust when $f$ has singularities.
The fundamental theorem of calculus says that every "indefinite integral" in the sense of 3) is an antiderivative. It does not say that every antiderivative is an indefinite integral, which in fact is false; you can have functions $f$ and $F$ where $F'(x)=f(x)$, but there is no way to write $F(x)=\int_c^x f(t) \, dt$.

(there are several ways to define integrals, and they are not 100% equivalent, so I'll go the elementary way to try to be clear; for the same reason, I will only discuss continuous functions)

We define the integral of the continuous function $f$ on the interval $[a,b]$ as
$$\tag1
\int_a^b f(t)\,dt=\lim_{n\to\infty}\sum_{j=1}^n f(x_j)\,\Delta_j,
$$
where $x_j=\frac{j(b-a)}n$ and $\Delta_j=1/n$ (more generally, one uses arbitrary partitions, but it is the same for continuous funcions).

If you think about it, you will note that there is no trace of a derivative nor  anti-derivative in $(1)$. An integral is an integral, it has a million applications, and if $f$ is nice enough it can be approximated very well by taking $n$ big enough on the right-hand-side of $(1)$ (and maybe tweaking the $x_j$, like using the mid-point rule).

The relation with derivatives comes from Newton's wonderful Fundamental Theorem of Calculus: if $f$ is a continuous function, then the function
$$\tag2
F(x)=\int_0^x f(t)\,dt
$$
satisfies $F'(x)=f(x)$. Because of $(2)$, the notation $\int f(t)\,dt$ for an antiderivative of $f$ is so widespread. The fact that any two antiderivatives of $f$ differ by a constant leads to Barrow's Rule:
$$\tag3
\int_a^b f(t)\,dt=F(b)-F(a).
$$
So $(3)$ is the reason why we use antiderivatives to calculate integrals (when possible).

To emphasize the fact that integral means $(1)$, and not $(2)$ nor $(3)$, consider the Error Function
$$
\operatorname{erf}(x)=\frac1{\sqrt\pi}\int_{-x}^x e^{-t^2}\,dt,
$$
which is one of many important functions defined by an integral. The function $\operatorname{erf}(x)$ is a multiple of an  anti-derivative of $f(t)=e^{-t^2}$, and no explicit expression for it  exists.


Field $\mathbb F_2$, Reducible and irreducible polynomials In a lecture on ring theory, we stated that if $K$ is a field, then $K[X]$ is factorial and then looked at examples of polynomials in $K[X]$ (for instance for $K=\mathbb C$). Then we chose $K=\mathbb F_2$. We wrote down: $$(X-a)(X-b) = \begin{cases}X^2, \text{if} \ a=0=b, \\ X^2+X, \text{if} \ a=1, b = 0 \ \text{and vice versa}, \\ X^2+1, \text{if}\ a = 1 = b.\end{cases} $$

We continued by stating that $X^2+X+1$ would be irreducible over $F_2$, since this polynomial doesn't appear in the above list. But if $a=1$ and $b=0$, then the factor $(X-0)$ is a unit, or I am mistaken? Therefore - at least for me - the above list does not necessarily state all the reducible polynomials with degree $2$ ...

Clarification would really be appreciated!

Kind regards,
MathIsFun

$X = X-0$ is not a unit.  if it were, there would be an $f(X) \in \Bbb{F}_2[X]$ such that $X f(X) = 1$, but this product always produces a polynomial one degree higher than that of $f$.

For a general field of coefficients, one should take care of units, which in $F[x]$ are the nonzero elements of $F$, again by arguing through degrees.  However when the field is $\Bbb{F}_2$, there is only one unit and it is adequately represented in the list of possible factorizations you have written.


Prove that $ \sum \frac{2^k}{k}$ is divisible by $2^M$ For each integer $M > 0$, there ${\bf exists}$ an $n$ such that
$$ \sum_{k=1}^n \dfrac{ 2^k}{k} $$
is divisible by $2^M$
${\bf try}$ Im struggling a bit to visualize this exercise. So, I tried to see for concrete number, for instance take $M=1$, then $n=2$ works: as
$$ 2 + \dfrac{2^1}{2} = 2^1 (1 + 2 )$$
Now, take $M=2$ and factor
$$ 2^{2} \underbrace{ \left( \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{2}{3} + \dfrac{2^4}{4} ... + \dfrac{2^{n-2} }{n} \right) }_{(*)}$$
now, we need to choose $n$ so that $(*)$ is an integer. Im unable to do so. Any help?

First, an important definition

Definition the $2$-adic valuation $v_2$ is given by
$$v_2\left( 2^n\frac{p}{q} \right) = n$$
where $p, q$ are odd (not divisible by $2$).

For instance $v_2(12/5) = 2$ since we can write $12/5$ as $2^2 \frac{3}{5}$.
In this language, what we want to show is that
$$\lim_{n \to \infty} v_2\left( \sum_{k = 1}^n \frac{2^k}{k} \right) = \infty. \tag{$*$}$$

In my opinion the best/most conceptual/requires the least tricks way to prove this is to identify the infinite series
$$ \sum_{k = 1}^\infty \frac{2^k}{k} $$
as the $2$-adic logarithm of $-1$. That is, it is $-\log(1 - 2) = -\log(-1)$. Recall that the logarithm is given by the series
$$-\log(1 - x) = \sum_{k = 1}^\infty \frac{x^k}{k}. $$
Given this, and assuming that basic properties of logarithms still hold in the $2$-adics, we have
$$\log(-1) + \log(-1) = 2 \log(-1) = \log((-1)^2) = \log(1) = 0. $$
Therefore $\log(-1) = 0$ as a $2$-adic number.
Well the series converges to $0$ in the $2$-adics if and only if the partial sums grow to $\infty$ in $2$-adic valuation. I.e. The series converges to $0$ if and only if $(*)$ holds.
This requires some work to understand:

what does it mean to converge in the $2$-adic valuation
do these series converge?
is $\log(xy) = \log x + \log y$ still true and when?

For all these questions, I must refer to Keith Conrad's notes. The identity $(*)$ is discussed at the top of page 29 (Example 8.10) and I'm not going to fit 29 pages of exposition in here.
However, I will say that this approach is much nicer than the non-logarithm approach. Because that approach requires several very hard-to-find tricks. If I were locked in a room with no internet or any other references and I had to prove this fact, I would pick the p-adic logarithm approach because at least then I have some idea what needs to be done.

I'll try to find a "more elementary" approach in a separate answer (this one's getting sort of long). But I'll warn you now: don't expect it to be easy to understand or enlightening.

This certainly doesn't seem true
If any $n > 2$ there is a prime $p$ so that $2<p \le n < 2p$ (Bertands postulate) and so
And so if $2^m| \sum_{k=1}^n \frac {2^k}k$ then $\sum_{k=1}^n \frac {2^k}k=W$ is as an integer.  And so $W-\frac {2^p}p$ is not an integer.
But $W - \frac {2^p}p= \frac {pW - 2^p}p$ is a fraction in lowest terms.
Which means $\sum_{k=1;k\ne p}^n \frac {2^k}k = \frac {pW -2^p}p$.  But be figuring out the least common denominator of $\frac {2^k}{k}; k\ne p$ we get the lets common denominater is $\operatorname{lcm}(k; k\ne p)$.  But that will not reduce so $\frac{pW-2^p}p$ unless $p$ divides into $\operatorname{lcm}(k; k\ne p)$.
Which it clearly can't as $\operatorname{lcm}(k; k\ne p)$ does not have $p$ as a prime divisor of any of the $k$ terms.
SO.... unless I'm making a bonehead error, this is not possible.
Not only is this not true  $\sum_{k=1}^n \frac {2^k}k$ is never an integer if $n > 2$.
..... unless I'm making a boneheaded error.


Intuitive/direct proof that a rectangle partitioned into rectangles each with at least one integer side must itself have an integer side A challenge problem asked to show that if rectangle $R$ with axis-parallel sides is partitioned into finitely many subrectangles $R_1,R_2,\ldots,R_n$ (also with axis-parallel sides), such that each $R_i$ has at least one integer side length, then $R$ must also have at least one integer side length. The only proof I've seen, although simple, is quite unintuitive. Namely you look at the integral
$$ \int_R e^{2 \pi i (x+y)} dx dy $$
and note that the integral is $0$ if and only if $R$ has an integer side length, and then note that by assumption the integral is $0$ over all the $R_i$, hence summing the integral over all $R_i$ gives that the integral is also $0$ over $R$. Can someone give a more direct/intuitive argument that doesn't use voodoo like a complex-valued integral?

Make the plane into a giant checkerboard grid with squares of size $\tfrac{1}{2}\times\tfrac{1}{2}$. In this problem we consider only rectangles with sides parallel to the axes. It is (intuitively) clear that such a rectangle with a side of integer length covers equal amounts of black and white area.

Let $R$ be a rectangle of size $a\times b$ partitioned into finitely many subrectangles, each having a side of integer length. Note that $R$ then covers equal parts black and white, because each of the smaller rectangles does. Without loss of generality we may assume that the lower left corner of $R$ is at the origin.

Divide $R$ into two rectangles of sizes $\lfloor a\rfloor\times b$ and $(a-\lfloor a\rfloor)\times b$, with the first rectangle having its lower left corner at the origin. The first covers equal amounts of black and white. The second can again be subdivided into two rectangles of sizes
$$(a-\lfloor a\rfloor)\times\lfloor b\rfloor\qquad\text{ and }\qquad(a-\lfloor a\rfloor)\times(b-\lfloor b\rfloor),$$
with the former having one side adjacent to an axis. This rectangle again cover equal parts black and white. The latter rectangle must then also cover equal parts black and white.

Note that this smaller rectangle is positioned in the lower left corner of a unit square, covered by precisely two black and two white squares. For it to cover equal parts black and white, one of its sides must have integer length. But for $a-\lfloor a\rfloor$ or $b-\lfloor b\rfloor$ to be an integer, it must be zero. We conclude that either $a$ or $b$ is an integer, and hence that $R$ has a side of integer length.

I can prove it assuming that all side lengths of subrectangles are rationals of the form $q/p$ where $p$ is a fixed prime. In that case, if we scale the rectangle and subrectangles by a factor of $p$ then we get that all side lengths are integers, and all subrectangles have a side length divisible by $p$, and we want to show that the overall rectangle also has a side length divisible by $p$. But this is trivial: every subrectangle has area divisible by $p$, so the total area of the overall rectangle is also divisible by $p$, and hence at least one side of the total rectangle must have length divisible by $p$.


Is there $z \in \mathbb{C}$ with $\mathrm{Re}(z) > 0$ such that $e^{2z}+\frac{z+1}{z-1} = 0$? Let $z$ be a complex number with positive real part. Is it possible that $e^{2z}+\frac{z+1}{z-1} = 0$?

Define:
$$f:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto(z-1)e^{2z}+(z+1),$$
$$g_1:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto|f(x+iy)|^2,$$
$$g_2:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto g(x,y)e^{-2x},$$
$$h:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto(x-1)e^{2x}+1.$$
Now prove this:
$$(\forall x>0)\frac{\partial g_1}{\partial x}>0\Rightarrow(\forall x>0)g_1(x,y)>0,$$
$$(\forall x>0)\frac{\partial g_2}{\partial x}>0\Rightarrow(\forall x>0)\frac{\partial g_1}{\partial x}>0,$$
$$(\forall x>0)\frac{\partial g_2}{\partial x}\geqslant 2x(h(x)+h(-x)),$$
$$(\forall x\neq 0)h(x)+h(-x)>0.$$

Near $z=1$, the second term dominates the first; far from $z=1$, the first dominates the second.  By numerical methods, the curve in the right half plane where $\left|e^{2z}\right| = \left|(z+1)/(z-1)\right|$ looks like this:

and nowhere on that curve (except the origin) is there a zero of your function.


Is $(\frac 1{n^2 \sin n })$ convergent ? If so , what is the limit? Is the sequence  $\left(\dfrac 1{n^2 \sin n }\right)$ convergent ? If it does, with what limit?

If it converges to a not 0 then from 1/n^2->0 and by division we get 1/sinn-->0 or sinn-->infinity which is absurd.
If it converges to 0 then 1/(nsinn) -->0 and so n.sinn-->infinity. Thus there is N such that n>N implies that n.sinn is positive. This is absurd too,
The sequence does not converge.


Proof $(P \land S) \rightarrow \lnot q$ in hilbert system How can I proof $(P \land S) \rightarrow \lnot q$ using this principles :

$p \rightarrow (q \lor r) $
$q \rightarrow \lnot r $
$r \leftrightarrow s  $

in Hilbert system and modus ponens?

You don't need principle 1, just start with the following propositional tautology:

$\fbox{4}$ $(q \rightarrow \lnot~r)$ $\rightarrow$ $((r \leftrightarrow s)$ $\rightarrow$ $((p \land s) \rightarrow \lnot~q))$ $\color{gray}{(propositional~tautology)}$

$\fbox{5}$ $(r \leftrightarrow s)$ $\rightarrow$ $((p \land s) \rightarrow \lnot~q)$ $\color{gray}{(from~2,4~by~modus~ponens)}$

$\fbox{6}$ $(p \land s) \rightarrow \lnot~q$ $\color{gray}{(from~3,5~by~modus~ponens)}$

Axiom 2 is (q→¬r).  Substituting q with (P $\land$ S) yields 4

4 [(P $\land$ S)→¬r].

Substituting  r with q in 4 yields 5

5 [(P $\land$ S)→¬$\lnot$q].


Geometrical proof required regarding midpoints $ABCD$ is a convex quadrilateral with $W, X, Y, Z, M$ and $N$ as the midpoints of $AB, BC, CD, DA,$ the diagonals $AC$ and $BD$ respectively. [Then, $WXYZ$ is a parallelogram with $K$ as the intersection point of its diagonals by Varignon's Theorem]

The 1st question is: “Will $K$ be the center of the circle using $MN$ as diameter?”

Line $BP$ is now drawn parallel to $AC$ cutting $DC$ produced at $P$. Let $Q$ be the midpoint of $DP$. $MN$ is produced to cut $BC$ at $R$.

The 2nd question is: “Is $RQ \parallel  AD$?”

PS: The red and green lines are not perpendicular to each other.

Picture added to reflect @Aretino 's answer to the first question.

To the first question: yes, because:
$$
K={1\over2}(W+Y)=
{1\over2}\left({A+B\over2}+{C+D\over2}\right)=
{1\over2}\left({A+C\over2}+{B+D\over2}\right)=
{1\over2}(M+N).
$$
More geometrically.

In triangle $CAD$, $M$ and $Y$ are midpoints of two sides, so we have $MY=AD/2$ and $MY\parallel AD$. In the same way, in triangle $ABD$ we have $NW=AD/2$ and $NW\parallel AD$. It follows that $MY=NW$ and $MY\parallel NW$, but then $MYNW$ is a parallelogram and its diagonals $MN$ and $WY$ have the same midpoint $K$.

Regarding the second question, note that the answer is no, in general, due to the following remark:

Two vectors $u, v$ are such that $u, v$ have the same length if and only if the vectors $u+v, u-v$ are orthogonal.

The result follows from calculating $(u+v)\bullet (u-v)$ right away.

Taking $u=\overrightarrow{KW}, v=\overrightarrow{KZ}$ one has: $u+v=\overrightarrow{XW}$, and $u-v=\overrightarrow{YX}$, so the only possibility that your parallelogram is inscribed in a circle centred at $K$ is for it to be a rectangle.

I'm sure that you can draw a rhombus or some other parallelogram, and then find a convex quadrilateral so that it all fits your picture, then you'll have the desired counterexample.  In fact, it suffices to take the above example in the figure and to apply an affine transformation which does not preserve angles!

The proof that your figure is a parallelogram uses simply baricentric coordinates, or vectors, so it is an affine concept which is therefore preserved by affine transformations, which in coordinates have the form

$$x'=ax+by+e, y'=cx+dy+f,$$

but the condition that the figure be a rectangle is easily broken by a convenient affine transformation, if we place the axes so that none of them is parallel to the sides of your rectangle and then take, say $x'=3x, y'=y.$

Addition: It is known that for every two bases of $\mathbb{R}$, $u, v$ and $u', v'$, there is a $2\times 2$-matrix $A$ such that $Au=u', Av=v'.$

This clearly means that we can send two perpendicular vectors to a non-perpendicular pair holding an arbitrary angle between them.  Hence, the affine-geometric construction shown will certainly have an arbitrary angle between vectors $u, v$ defined as above.

Hope this is clear enough.


Uniform convergence of $\frac{n^2\sin(x)}{1+n^2x}$ My goal is to show that

$$\frac{n^2\sin(x)}{1+n^2x}$$

does not converge uniformly on $S=(0,\infty)$ but does so on any compact subset of $S$. First, we find the limit function

$$\frac{n^2\sin(x)}{1+n^2x} \to \frac{\sin(x)}{x} \qquad \text{ as } n \to \infty$$

Now, to show that this doesn't uniformly converge on the whole interval I am really only interested in the boundary points (since any compact subset supposedly makes it uniformly convergent). As $x \to 0$ our function goes to $1$and as $x \to \infty$ our function goes to $0$. I am unsure of how to calculate

$$\left|\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right|\right|_S$$

and, moreover, show that it is $0$. From there, on any compact subset of $S$ I am guessing that

$$\left|\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right|\right|=0$$

Will be zero since if we consider $[a,b] \subset S$ we have

$$\left|\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right|\right| \leq \left|\frac{n^2\sin(a)}{1+n^2a}\right|+\left|\frac{\sin(a)}{a}\right|=0$$

Thanks for your help!

For $0 < a \leqslant x$ we have

$$\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right| = \left|\frac{\sin (x)}{n^2x^2 + x}\right| \leqslant \frac{1}{n^2a^2 + a} \to 0
$$

and convergence is uniform on $[a, \infty)$.

For $x > 0$

we have

$$\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right| = \left|\frac{\sin (x) /x}{n^2x + 1}\right|
$$

Choose a sequence $x_n = 1/n^2$ to show that the convergence is not uniform on $(0,\infty)$.

Here we have $\sin (x_n) /x_n \geqslant \sin 1$ and

$$\left|\frac{n^2\sin(x_n)}{1+n^2x_n}-\frac{\sin(x_n)}{x_n} \right| \geqslant \frac{\sin 1}{2} $$

Be consider to following sequences x_n=1,x_m=mπ in this case this sequential function does not satisfy in Cauchy condition.lim n^2/1+n^2=1and lim n^2sinnπ/1+n^2=0 and we can choose  epsilon=1/2  |x_n-x_m|>1/2 so that it is not uniform convergence.


If $AB=I$ show that $A$ has full rank. Let $A,B\in \mathbb R^{n\times n}$. Show that if $AB=I$ then $B$ has full rank.

In fact, I show that $A$ has full rank, which is quite obvious but I really have difficulties to show that $B$ has full rank. I tried by contradiction : there is $u\in \mathbb R^n$ s.t. $Bv\neq u$ for all $v\in \mathbb R^n$. So $v=ABv\neq Au$ which contradict $A$ has full rank.

It doesn't look correct. What do you think ?

There is a missing thing. How do you know that $Bv\neq u\implies ABv\neq Au$? At this point, you need to use the fact that $A\text{ surjective}\implies A\text{ injective}$. Then, since $AB=\operatorname{Id}$, your argument is correct.

$AB = I$ means $B = A^-1$, so $A$ is invertible. A matrix is only invertible if it is full rank, otherwise when you are in the process of inverting it, you will eventually find a row or colum of zeros.


Is the inner angle of a circle is 180? Is this true that polygon with infinitely many sides (which is a circle)
all inner angles have 180 degrees (each) ?
 by this argument:

$$\frac{(m-2)\cdot180}{m}\space\space\space\Rightarrow\space\space\space180\space-\space\frac{360}{m}\space\space\space\space{\overrightarrow{_{m\rightarrow\infty}}}\space\space\space\space180$$

Is this true?

Technically not exactly 180, as a circle is a curve which doesn't technically have an angle, the best we can do is measure tangent lines. However the limit you have derived I agree with, and it makes sense. $180^0$ is a straight line, and for a circle these lines are hardly noticed because individually they all seem straight, just ever so slightly off the $180^0$ - this causes it to curve around in the fashion we see it do.

There's no polygon with infinitely many sides. A circle is not a polygon. It has no rectilinear parts, and hence no angles.


$N$ perfect logicians wearing hats I once came across the following riddle: (assume $N$ to be extremely large)

  There are $N$ perfect logicians arranged in a vertical row. They are
  allowed to strategize before the game, during the game they are barred
  from communicating. In the game, the lights are switched off, and a
  single hat, either red or blue, is placed on the heads of each of the
  logicians. The lights are turned on; each logician can see the hat
  colours of all the people in front of him, he can't see the colour of
  his own hat, or the people behind him. Start with the person at the
  back, each logician has to call out a colour (heard by all). If his
  own hat is of that colour, he is spared, otherwise, silently killed.
  The same is done by the next person, and so on, right till the first
  person. How many logicians can you guarantee to save, irrespective of
  the initial colours or any probability?

The answer was simple. The first logician would sacrifice his life. He would call out red if there are an even number of red hats in front of him, and black if there an odd number of red hats. After him, each logician can correctly tell the colour of his own hat.

I decided to give the question a twist. Suppose there is a blind man among the logicians. His life is worth as much as anyone else, and he is placed randomly in the queue. How many logicians can you guarantee to save?

The same approach wouldn't work because, if the blind man gave the wrong colour, not only him, but all the people in front of him will give the wrong colour as they depend on his information also.

After a lot of thinking, I finally came up with this solution.

Find the smallest $n$ such that $3(n+1)+2^n \geq N$. The first $3(n+1)$ people (from the back) will be the ones sacrificing their lives (Group A), and the rest ($2^n$) will try to figure out their own hat colours (Group B). Starting with the back of Group B, assign each logician an ($n+1$)-digit binary number, starting with a string of zeroes, and adding $1$ to get the next logician's ID. Note that the first digit will always be a $0$.The first Group is divided into $n+1$ subgroups of $3$ logicians each. Each subgroup is assigned a normal decimal number $1,2,3,\cdots$ and so on (again starting from the back). Let the subgroup number of a logician be $k$

When the game starts, each Group A logician will look at all the people in Group B, whose $k^{th}$ binary digit is $1$. Of these people, if there are an even number of red hats, he will say red, otherwise he will say blue. The other $2$ logicians in his subgroup will give exactly the same information as the first one. This is necessary to confirm that he is saying the truth, if there is a discrepancy among the $3$ answers, the information given by $2$ is correct, and the third one is the blind one. The same is done by everyone in Group A.

The people of Group B only try to figure out their own hat colours. Each logician has $n+1$ ways of figuring out his own hat colour. If all of the given data doesn't lead to the same answer, he knows that the blind man is behind him and has already called out a lie. He can then use the first logician's info (first in Group A and the queue itself) to figure out his own colour, since the blind man has for sure been counted by that person.

Is my approach correct and is there a faster one? I hope my question is clear enough.

Note:
The blind man knows where is placed, but nobody else knows his position.

I think the Hamming code with extra parity provides a solution.  If there are $2^n$ logcians, this requires sacrificing $n$ of them (each with chance $1/2$).  Let the rear $n$ be the parity bits of the code, called out by those logicians.  All the rest can run the error detection scheme.  If no bits are wrong yet, either because the blind logician has not spoken or because he was right, each person can find the hat color that has no errors.  If one bit is wrong already, one of the hat choices will make two errors, which are detectable by the code.

If the ones behind the blind man can see where he is, you can save all but maybe 2.  The rear logician announces the color of the blind man's hat.  The next one announces the parity of all the hats he can see.  Everybody else is saved by the original solution.  If the blind man is at the back, it is not a problem.  If the blind man is second, the first announces the color he should say for the parity.

Your approach is correct but not optimal. For large $n$, it requires the sacrifice of $3 \log_2 n$ logicians, while my first paragraph requires the sacrifice of $\log_2 n$.  As you say, you are repeating every bit three times, which guarantees the correct answer will be a majority if only one bit is flipped.  This exploits the fact that the Hamming distance between $000$ and $111$ is three.  Flipping one bit will leave you distance $1$ from the correct string and distance $2$ from the incorrect string.  The code I cited at the start does the same thing with $n$ bit code words.  All the acceptable ones are at least a distance $4$ from any other, so you can correct single bit errors and detect two bit errors.

Added:  We can do it with at most three being sacrificed.  You do the classic solution that does not involve the blind logician. The rear one guesses the color that would make the number of red hats even, and is sacrificed with probablity $1/2$. Each one in turn adds the number of red hats he sees to the number of red guesses he has heard. If even, he guesses blue. If odd, he guesses red. All (except maybe the first) will be right. Now we add in the blind logician, who guesses randomly.  If the blind man is wrong, the next after him will be wrong, but that restores the parity and the rest are saved. We now lose at most three (the first, the blind man, and the next).

as logicians are allowed to discuss before the game begins, they decide that every alternate man will shout the color of the logician just before him, now if the color of the logician's hat who is being questioned is the same as the one of the hat of the logician in his front, he will be saved too otherwise the logician just before him is saved. but if they can varry their shouts that is if anybody shouts for long it means that the color of the hat of the logician before the logician before him is same and one shouts for a shorter time the colors are different in this case at least n-1 will be saved


Intuition behind the $cl(cl(A))=cl(A)$ Kuratowski closure axiom in general topology? If we use the real numbers as a model of topology (with the standard topology), then it is obvious why $cl(cl(A))=cl(A)$ should hold, given the definition of "closure of a set" in the context of the real numbers.

But what a good intuition for why this axiom should hold in the context of general topology?

I have an intuition for all 3 other axioms:

$cl(\emptyset)=\emptyset$ (intuition: no objects touch the empty set, because you can't touch a set consisting of no objects)
$A\subseteq cl(A)$ (intuition: every element of $A$ touches $A$, because elements "touch themselves").
$cl(A\cup B) = cl(A)\cup cl(B)$ (intuition: if an object touches a set consisting of two parts, then it must at least touch one of those parts, because if you don't touch either of two things, you can't suddenly touch the combination of the two).
$cl(cl(A))=cl(A)$ (intuition: if a set of objects $B$ all touch $A$, then any object that touches $B$ must also touch $A$, but this doesn't feel right to me. it feels too restrictive).

What is a good intuition for why nr 4 should hold for topologies, given that topologies formalize the notion of "touch-relations"?

$\DeclareMathOperator\cl{cl}$In terms of metric spaces, the statement "$x\in\cl(A)$" that you want to think of as "$x$ touches $A$" becomes "the distance of $x$ to $A$ is $0$". Here the distance $x$ has to $A$ is defined as
$$
\operatorname{dist}(x,A) = \inf_{a\in A} d(x,a).
$$
Now take $x\in\cl(\cl(A))$ and let's figure out what this means in terms of distance. The point $x$ has a distance of $0$ to the set of all points having distance $0$ to $A$. Now in general for three points $x,y,z$ we have
$$
d(x,z) \le d(x,y) + d(x,z).
$$
So if $x$ has distance $0$ to all points $y$ having distance $0$ to $A$, these distances will at most sum up and $x$ has distance at most $0+0=0$ to $A$. We conclude that $x$ is indeed touching $A$ and hence
$$
\cl(\cl(A)) \subseteq \cl(A).
$$
The other inclusion is given by property 2.

Property 4 is called idempotence. It means after one application, it doesn't matter how many more is applied. You are providing "closure" axioms. Therefore once a subset is closed, it should not change after more "closure" applications


Results on intersections of compact sets in non-metrizable spaces We know that if $X$ is a Hausdorff topological space and $C_0 \supseteq C_1 \supseteq ...$ are nested, non-empty compact sets, then $\bigcap_{i \in \mathbb{N}} C_i \neq \emptyset$, this is Cantor's Intersection Theorem. However, this gives us no information about the "size" of $\bigcap_{i \in \mathbb{N}} C_i$. For instance, we could take $X= \mathbb{R}$ and take $C_i= [1- \frac{1}{i+1}, 1+ \frac{1}{1+i}]$ and get $\bigcap_{i \in \mathbb{N}} C_i= \left\{1 \right\}$, but we could also give an expression for $C_i$ such that $\bigcap_{i \in \mathbb{N}} C_i$ yields the Cantor Set and hence is uncountable. I am interested in statements of the following form:

"For a non-metrizable topological space $X$ and compact, non-empty $C_i \subseteq X$, if $C_i$ each have Property (1), then $\bigcap_{i \in \mathbb{N}} C_i$ contains exactly one element."

I emphasize "non-metrizable" here because there are analogous statements for metrizable spaces and the specific spaces I'm looking at are never metrizable. Does anyone know of any theorems of the above form?

Also, obviously if $|C_i|=1$ for each $i$ then the statement holds, so anything other than that would be better.

We have the follow related theorems:

Compact Subspace of a Hausdorff space is Closed: https://proofwiki.org/wiki/Compact_Subspace_of_Hausdorff_Space_is_Closed

Closed Subset of Compact Space is Compact https://proofwiki.org/wiki/Closed_Subspace_of_Compact_Space_is_Compact

Not a complete answer, but you can probably used these to help you with this question.


How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am not sure how this is useful.

You know that $2^{36}\equiv 1$, since $2$ is a primitive root.  Hence you may conclude that $(2^{12})^3\equiv 1$, and also $(2^{24})^3\equiv (2^{36})^2\equiv 1^2$.  These correspond to $26$ and $10$ respectively.  Of course $2^{36}=1$.

Now, suppose that $(2^a)^3\equiv 1$.  Then $2^{3a}\equiv 1$, so $36|3a$, or $12|a$.  That doesn't leave many choices for $a$, since $1\le a\le 36$...

Just for fun another way.

Because of $\mathbb F_{37}$ is a field the equation in $\mathbb F_{37}[x]$ $$x^3=1\iff (x-1)(x^2+x+1)=0$$

must have three roots so, since $1$ is clearly root, one can find out the other two ones solving $$x^2+x+1=0$$ We have
$$x+y=-1=36\\xy=1$$ Exploring $$35+1\\34+2\\33+3\\32+4\\31+5\\30+6\\29+7\\28+8\\27+9\\26+10$$ We stop because $26\cdot10=260\equiv 1\pmod{37}$, thus the roots are $1,10$ and $26$.


about the smallest normal subgroup containing some generators Assume $f:G\to K$ is a group homomorphism, and $S\subset G$ is a set of elements such that $f|_S$ is trivial. Denote by $N\leq G$ the smallest normal subgroup such that $S\subset N$. Is it obvious that $f$ is trivial on all of $N$?

Silly to give yet another answer for such a triviality, but none of the existing ones seem quite right.

By definition, $N$ is the intersection of the normal subgroups containing $S$. Since $\ker(f)$ is a normal subgroup and $S\subset\ker(f)$, this says that $N\subset\ker(f)$.

Tip: show that $S$ is a normal subgroup of $G$, so that $S=N$.

(In fact this is equivalent to the exercise, right?)


Convolution of measures We say that a family of measures $\mu_{t}\to \mu$ weakly if for any $g\in C_{0}$, $\int g d\mu_{t} \to \int g d\mu$. Show that if $\mu_{t}\to \mu$ weakly, then $\nu*\mu_{t}\to \nu*\mu$ weakly, where $\nu*\mu$ denotes the convolution of the measures $\mu$ and $\nu$.

If the measures are finite, then for $g\in C_0$
$$\int g(x)\,(\nu*\mu_t)(dx)=\int\int g(x+y)\,\mu_t(dy)\,\nu(dx).$$
For each $x$, the weak convergence $\mu_t\to \mu$ gives
$$\int g(x+y)\,\mu_t(dy)\to \int g(x+y)\,\mu(dy).$$
Then by the dominated convergence theorem,
$$\int\int g(x+y)\,\mu_t(dy)\,\nu(dx)\to\int\int g(x+y)\,\mu(dy)\,\nu(dx).$$
By definition, this is the weak convergence $\nu*\mu_t\to \nu*\mu$.

For a counterexample, consider the shifted $\delta$-masses on $\mathbb{R}$:
$$ \int f(x) \ \delta_t(x) = f(t).$$
It's clear that $\delta_{t} \to 0$ vaguely as $t \to \infty$, while
$\delta_t * dx = dx$ as measures by the translation invariance of Lebesgue measure.


Multiplication, What is It? What is multiplication? Upon review logarithms, and square roots, I realized that I have no intuitive grasp of multiplication-well no more so than I have for addition. Is it simply another thing we need to memorize? I understand that things like $\sqrt2$ could just be memorize as the thing, that when applied to itself, gives two... But this makes a lot less sense to me then thinking about things like $2+2=4$. Logarithms are a function that express a number as a power of some base, but when you get fractions for the logarithms, I don't really know what that means-What is $2^{1.2}$, or any decimal power for that matter.

Upon rereading the link posted below, I think I can repose my question a bit. If these operations are axiomatic, as I felt they were, the idea of a square root is the inverse of exponentiation. How do I grasp exponentiation intuitively? Multiplication and addition have natural roots in our minds, if exponentiation is also fundamental, what is it? Also when is exponentiation not just repeated multiplication?

First, you shouldn't find expressions like $\sqrt{2}$ weird.  Saying that $\sqrt[a]{b}$, where $a$ and $b$ are positive integers, is the number whose $a^\text{th}$ power is $b$ is perfectly fine.  Just think about the graph of $x^a$.  This is fudging a little bit, but we know $x^a$ is continuous, meaning it's a solid line, and we know it goes up to $\infty$ in the positive direction.  So we know that the lines $y = x^a$ and $y = b$ intersect at some point.  When you write down $\sqrt[a]{b}$ that's just notation for the $x$-coordinate of that point of intersection.

Now if you're fine with $\sqrt[a]{b}$ then you should be fine with raising a number to a fraction.  When we write $x^\frac{a}{b}$ we just mean $\sqrt[b]{(x^a)}$ or $(\sqrt[b]{x})^a$ (they're the same).

After you're fine with raising numbers to fractions you might ask what does $x^a$ mean if $a$ is $\pi$ or some other number that's not a fraction.  This is maybe farther than you want to go at the moment, so I won't give you a detailed answer (I'm sure this questions exists elsewhere on the site).  I'll just say that $x^a$ is the number that $x^b$ gets closer to as you choose fractions $b$ that get closer and closer to $a$.

Exponents are a way of showing multiplication when what is being multiplied is being multiplied by itself.  Saying $4^{1.2}$ is just saying that you are multiplying $4$ by itself $1.2$ times.


Lines parallel to a given line I have to show that any line parallel to $Ax+ By + c =0$ is of the form $Ax + By + k =0$

How do I show this?

Thank you!

Examples don't show it in any case. You cannot prove that something holds for infinitely many things by doing nothing except checking finitely many examples.

What you need to do is remember that two lines are parallel if and only if they have the same slope or they are both vertical (of the form $x=k$ for some constant $k$). Then figure out what the slope of an arbitrary line $\alpha x+\beta y+\gamma=0$ would be in terms of $\alpha$, $\beta$, and $\gamma$ (careful when dividing by constants to consider what happens if the constant you are dividing by is equal to zero). Then see what this means if you want to find all lines $\alpha x+\beta y+\gamma = 0$ that are parallel to a given line $Ax+By+c=0$.

You can solve the following system of equations:

$$
\begin{cases}
Ax+By+c=0,\\
Ax+By+k=0
\end{cases}
$$

If you subtract the second line from the firs, you obtain $c-k=0$. Then, if $c\ne k$, you have no solution: the lines are parallel.


Does notation ever become "easier"? I'm in my first semester of college going for a math major and it's pretty great. I'm doing well, however, there seems to be huge gap between how difficult /complex an idea is and how convoluted it is presented.

Let me make an example: In Analysis we discussed the Bolzano Weierstrass theorem and one of the lemmas showed that every sequence in $\mathbb{R}$ has a monotone subsequence. The idea behind the proof with the maximum spots ( speaking colloquially here ) is super simple and pretty elegant if you asked me, but I spent a significant amount of time trying to understand the notation of the professor until I went to this site to read a "proper explanation" of the proof, which had much simpler notation in it.

Extracting the idea of the proof took me lots of time because of the strange notation, but once you understand what is going on, it is really easy. Most of the time spent studying lectures is about digging through the formalities.

Do I just have to spend more time really going through all the formal details of a proof to become accustomed to that formality? Or do more advanced mathematicians also struggle to extract the ideas from the notation?

I'd assume there will come a point, where the idea itself is the most complex part, so I do not want to get stuck at the notation, when that happens.

( Proof Verification - Every sequence in $\Bbb R$ contains a monotone sub-sequence if you are interested )

As others have pointed out, it gets much better if that's your first semester. But in my experience, there is not much relief between, say, years 2 and 4 of your studies. Sure, you get more mature, but the material gets more difficult too.

To address your question whether "more advanced mathematicians also struggle to extract the ideas from the notation", I'd like to quote V.I. Arnold, since I think it's exactly in the spirit of your frustration.

  It is almost impossible for me to read contemporary mathematicians who, instead of saying "Petya washed his hands," write simply: "There is a $t_1<0$ such that the image of $t_1$ under the natural mapping $t_1 \mapsto {\rm Petya}(t_1)$ belongs to the set of dirty hands, and a $t_2$, $t_1<t_2 \leq 0$, such that the image of $t_2$ under the above-mentioned mapping belongs to the complement of the set defined in the preceding sentence.''

The trade-off is clear: without rigor math would've been quite a mess. But if rigor is the only way math gets communicated to someone, this person simply won't have time to get far in math.

The answer is that these things build one upon the next. It is pretty unusual encountering math books where the majority is presented in notation but you learn basics and work your way up, as with anything else. It becomes easier over time in other words. Trying to read a topology text without a foundations course and some analysis would be damn near impossible. So take heart in that it will get easier.


prove combinatorics theorem I am self-studying Combinatorics and come across this theorem. Can anyone show me how to prove the theorem. I like to understand why it is that way.

  n identical balls are put into $r$ labeled boxes and the number of balls in each box is not limited. The number of ways of doing this is $$\binom{n+r-1}{n}$$ or$$\binom{n+r-1}{r-1}$$

Regard the choice of the balls for each of the $r$ bins as determined by lining the balls up, and putting a hashmark in between the last ball to go in the $k$th bin and the first ball to go in the $k+1$th bin. That is, for example, let

$$ \cdot \cdot \cdot / \cdot \cdot / \cdot \cdot \cdot \cdot  $$

correspond to putting $3$ balls in bin $1$, $2$ balls in bin $2$, and $4$ balls in bin $3$, where $n = 9$ and $r = 3$.

Clearly, the number of ways to divide up the balls is the same as the number of ways you can choose where to put the hashmarks. But there are $n-1$ spaces in between two balls where you can put the $r-1$ hashmarks, so this gives ${n-1}\choose{r-1}$. However, this assumes that no bin gets $0$ balls. But each way you can put $n$ balls into $r$ bins, with bins possibly empty, corresponds one-to-one to the number of ways you can put $n+r$ balls into $r$ bins, giving each bin at least one ball - which we know by the above is ${n + r -1}\choose{r-1}$.

$\binom nk= \frac{n!}{k!(n-k)!}$

$\binom n{n-k} =\frac{n!}{k!(n-k)!}$

So, $\binom n{n-k} = \binom nk$


if $f(\frac{x+y}{2}) =\frac{f(x)+f(y)}{2}$ then find $|f(2)|$ if the following functional equation  $$f\bigg(\frac{x+y}{2}\bigg) =\frac{f(x)+f(y)}{2} \quad \text{ holds for all real }~ x  ~\text{ and }~ y$$  If$f'(0)$ exists and equals to $-1$ then find $|f(2)|$.
My work
I tried to find $f'(x)$ so that i can intregrate it and then find $f(x)$
$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
which gave me this equation:
$f'(x)=\lim_{h \to 0} \frac{f(2x)-2f(x)+1+f(2h)-1}{2h}$
I could not proceed further
and the answer is $2$

If the brackets are not the greatest integer function, then, It's Jensen's Functional Equation and the solution is $f(x) = ax + b~$ for real $a~$ and $~b.$
Therefore from the given, we have
$$ f(0) = b = 0 ~\text{ and }~ f'(x) = a = - 1$$
Therefore the function is
$$ f(x) = -x \implies \lvert f(2)\rvert = 2 $$


Free variables, parameters, dummy names -- what policy to choose? A long lead-up (sorry!) to a short question. Start with a semi-formal argument in Loglish:

(1) $\quad$ for every $x$, $x$ is $F$
(2) $\quad$ for every $x$, ($x$ is $F$ $\to$ $x$ is $G$)
Arbitrarily choose some object in the domain, a: then we have
(3) $\quad a$ is $F$
(4) $\quad a$ is $F$ $\to$ $a$ is $G$
(5) $\quad a$ is $G$
But a was arbitrarily chosen; so
(6) $\quad$ for every $x$, $x$ is $G$

Obviously, something we have to explain to beginners is: What is '$a$' doing here? This plainly isn’t an ordinary proper name, attached once-and-for all to a determinately given individual. On the other hand, it isn’t an anaphoric pronoun either. It is a sort of ‘dummy name’, or ‘temporary name’, or ‘ambiguous name’, or ‘arbitrary name’ – perhaps none of those labels is entirely happy, though all are in use. Let's say it is a ‘parameter’.
When we go fully formal wanting to regiment such arguments into an official first-order language, we need some symbols to play the role of parameters, as informally played by '$a$'. There seem to be three policies on the market, used by different textbooks:

(A) We can use the same symbols both as variables-tied-to-quantifiers and as stand-alone parameters (in conventional jargon, the same symbols can appear as both ‘bound variables’ and ‘free variables’).
(B) We can use the same symbols both as proper names and as parameters/dummy names (with just some of these symbols getting a fixed denotation in a given language).
(C) We can use distinctive symbols for proper names, for bound variables, and for  parameters/dummy names.

Each policy can be made to work. Policy (A) has historically been the most common one among logicians. And it does conform best to the practice of mathematicians who casually move between using letters as dummy names (‘Suppose $m$ is the number of positive roots . . . ’) and using letters of the same kind as quantified variables (‘For any integer $n, (n + 1)(n − 1) = n^2 − 1$’), letting context make it clear what the symbols are doing. However, when we go formal, overloading symbols like this does cause complications: we have to spell out very careful rules for their double use if we are to avoid getting into tangles.
Policy (B) is easier to work with, but hasn’t anything special to recommend it. Once we have distinguished, at the semantic level, the fixed-denotation names built into a language from its further supply of dummy names, why not highlight the distinction by using syntactically different symbols for the two styles of name?
Policy (C), then, may be less economical but it does make it easier to keep track of what is going on. Variables only appear bound, parameters only appear free. It conforms to the good Fregean principle of marking important differences of semantic role by using different styles of symbols. It has distinguished antecedents, e.g. in  Gerhard Gentzen. It's the policy in e.g. Richmond Thomason's impressive textbook.
Which is the easiest policy for the beginning student to understand? I'd say (C). But -- and here at last is the question --

Is there any strong reason -- issues of ease of understanding apart -- for preferring the more common policy (A) to the less common policy (C)?

The only thing I can think of is that by adopting policy (A), which makes it ok to use free variables in the statements, we can use it more naturally as a system that can prove properties and relations of formulas in general. For example, you can prove that $P(x)$ and $\neg \neg P(x)$ are equivalent wff's.

But you obviously you can only do those kinds of things if you have to be really careful about how the variables are used, i.e. whether it's one of those 'temporary' variables or not. And certainly for beginning students we just work with sentences, rather than formulas in general. So, policy (C) has a lot be said for it, I agree.

Not sure what you mean by a "parameter" here. Assuming a "dummy" variable is the same as a bound variable.

Start with a semi-formal argument in Loglish:
(1) $\quad$ for every $x$, $x$ is $F$

This is an assumption. $x$ is a bound (dummy?) variable here.

(2) $\quad$ for every $x$, ($x$ is $F$ $\to$ $x$ is $G$)

This is an assumption. $x$ is a bound (dummy?) variable.

Arbitrarily choose some object in the domain, a: then we have
(3) $\quad a$ is $F$

This is an assumption. $a$ is a free variable.

(4) $\quad a$ is $F$ $\to$ $a$ is $G$

Applied Universal Specification here.

(5) $\quad a$ is $G$

Applied Detachment (MP) here

But a was arbitrarily chosen; so
(6) $\quad$ for every $x$, $x$ is $G$

Applied Universal Generalization here. $x$ is a bound (dummy?) variable.


Taking log we have: $\dfrac{\log(3x+7)}{x^2-4}$, and you can use L'hospitale rule for this.


Applications of Countable Infinite Sets and Power Sets What are the possible applications of Countable Infinite Sets and Power Sets in areas that are not strictly mathematical?

Also I want to know the significance they carry. What was not possible before the concepts of countability and power sets were introduced and what became possible afterwards? How did the introduction of these concepts change our thinking and outlook and the areas they affect?

Please explain in plain language, possibly with examples, as I am from a non-mathematical background. Thank you in advance.

PS: The applications need not be in strictly practical fields either (as Asaf Karagila keeps pointing there aren't any). Please help!

The notion of cardinality as formalized and explored by Cantor introduced a whole new set of questions about the reals that wouldn't have occurred to mathematicians before Cantor.  Before Cantor, it was well-known that the reals and the rationals were very different infinite sets, but the difference was generally described by the fact that the reals have a completeness property (loosely, convergent sequences of reals converge to another real -- not so in the rationals, it isn't too difficult to construct a sequence of rationals that converges to $\sqrt{2}$).  Once you can distinguish countable and uncountable sets (so after Cantor), the whole picture changes.

For example, in terms of the first-order theory of linear order types (in particular, ignoring algebraic properties), the reals can be completely characterized by the following properties:

1) uncountable

2) no endpoints

3) dense in itself (between any two distinct reals there is another real)

4) no uncountable strictly monotonic sequences

5) contains a countable dense subset (the rationals -- i.e. between any two distinct reals there is a rational)

(If you're wondering why completeness isn't on this list, completeness is a second-order property, quantifying over sets of reals -- "all sets which have an upper bound have a least upper bound.")

Note, though, how many of these properties rely on cardinality.  The next interesting question is whether all of these properties are necessary.  In particular, consider dropping the last property.  Is there a linear order type which satisfies the first 4 properties but contains no subset order-isomorphic to the reals?  The answer turns out to be yes -- first construct an uncountable tree with no uncountable branches or levels (actually a counterexample to what would be a natural generalization of Konig's lemma), then look at the lexicographic order on nodes of that tree.  The result is a linear order as described above, known as an Aronszajn line (or Specker type).  This whole construction can be done in ZFC.

There are other interesting questions here.  Consider whether there are alternatives to the last property above.  Perhaps it is enough to consider that the linear order has a countable chain condition -- i.e. there are no uncountable collections of pairwise disjoint open intervals.  Does this give order types which contain a suborder isomorphic to the reals?  This question turns out to be independent of ZFC -- the counterexample is called a Souslin line, and whether such a thing exists depends on the model of ZFC you are using (some have them, some don't).

Again, the point is that these questions about the reals, and possible alternatives to them, would not have occurred to anyone before the development of a formal definition of cardinality for infinite sets, in terms of bijections.  Cantor's ideas brought on something of a crisis in our understanding of the reals, and the results (mostly independence results lately) continue to flood in.