Are there unsolved indeterminate limits? I find the question itself is hard to put precisely. I apologize in advance. A simple version could be:

Let $\mathcal{F}$ be the set of functions obtained via elementary binary operations (sum, product, power, end their inverses) and composition, including polynomials and sinusoidals ($cos(x)$ and $sin(x)$). For example $f(x)=e^{e^x}+cos(x)$. Basically the functions one encounters in basic calculus. The question is the following:

Are there $f,g\in \mathcal{F}$ such that:
$$\lim_{x\to \infty} f(x)=\infty,$$
$$\lim_{x\to \infty} g(x)=\infty,$$
but such that the limit $$\lim_{x\to \infty}\dfrac{f(x)}{g(x)}$$ is unsolved? What I mean by this is that no one in the mathematical community knows if the limit exists, or even if existence is guarranteed, no one knows if the limit is finite on infinit.

Given access to these elementary functions, it's fairly easy to convert statements about, say, rational approximations to $ \pi $ into the existence of a particular limit.

For example, it's unknown if the irrationality measure of $ \pi $ is equal to $ 2 $ or not (in the sense of measure theory, almost all irrational numbers have irrationality measure $ 2 $). It's easy to see that this statement is equivalent to saying that for any $ \epsilon > 0 $,  $ \sin(n) $ is never within $ O(1/n^{2 + \epsilon}) $ of $ 0 $ for all but finitely many values of $ n $, which in turn is equivalent to saying that

$$ \lim_{n \to \infty} \exp(-n^{2 + \epsilon} |\sin(n)|) $$

is equal to $ 0 $ for all $ \epsilon > 0 $, so for instance if you pick $ \epsilon = 1 $ it's unknown if this limit exists or not. If it exists, it's easy to see it must be equal to $ 0 $. If you don't like absolute values, you can instead formulate an equivalent statement as

$$ \lim_{n \to \infty} \exp(-n^{4 + \epsilon} \sin^2(n)) = 0 $$

This is not exactly in the form you want, but you should be able to put it into that form if you multiply and divide by some appropriate function diverging sufficiently quickly to infinity, say $ \exp(\exp(n)) $. This just obscures the basic idea of the construction, however.

There are a finite number of mathematicians.  Each mathematician has thought of only a finite number of functions in $\mathcal F$.
 But there are infinitely many many members of $\mathcal F$.
So there are infinitely many $f/g$ that no mathematician has ever
even thought of, therefore doesn't know the limiting behaviour of.


Solving Probability Density Function for continuous random variable The probability density of a random variable $x$ is $$f(x)=a\ \cdotp x^2\ \cdotp \mathrm{e}^{−kx}\ (k>0,\ 0\leq x\leq \infty)$$ Then, the coefficient $a$ equals
$$(i)\frac{k^3}{2}\ \ \ \ (ii)\ k^3 \ \ \ \ (iii)\ k^2 \ \ \ \ (iv)\ k \ \ \ \ (v) \frac{2\ \cdotp k}{\pi}$$

  Here is what I've tried:

  $$\int_0^\infty f(x)\mathrm{d}x = 1$$ $$\int_0^\infty\left(a\ \cdotp x^2\ \cdotp \mathrm{e}^{−kx}\right)\mathrm{d}x = 1$$

  solving by parts gives result as

  $$\left[-\frac{a\cdotp\mathrm{e}^{-kx}}{k}\cdotp\left(x^2+\frac{2x}{k}+\frac{2}{k^2}\right)\right]^\infty_0 = 1$$

  Now, here I'm stuck. Can't find a way to solve $a$.

What you need instead is
$$
a=\frac{1}{\int_{0}^{\infty}x^2 e^{-kx}dx}=\frac{k^3}{2}
$$


Find number of flips of transposition $(i \ j)$. Suppose that $(i \ j) \in S_n$ is a transposition and that $i < j$.

How would it be possible to find an expression for the number of flips of $(i \ j)$?

If I understand correctly your definition in the comments, the number of flips of an element $\sigma \in S_n$ is $nf := |\{i, j < n| i < j, x(j) < x(i)\}|$. In that case, the question seems to have an easy solution:

Suppose $(ij)$ is your transposition ($i < j$). It swaps $j$ and $i$ and leaves all other elements on their place. Now, suppose $p$ and $q$ are such numbers, that $p < q$ and $(ij)(p) > (ij)(q)$. This definitely can not happen in case, when neither $p$ nor $q$ belong to $\{i, j\}$. Neither can they in case, when $p, q \leq i$ or $p, q \geq j$. Now, suppose $p = i$ and $q \neq j$. Then $i < q < j$. Suppose $q = j$, $p \neq i$. That also results in $i < p < j$. Those cases and $p = i$, $q = j$ are the only possible ones. So $nf((ij)) = 2|\{p|i < p < j\}| + 1 = 2(|j - i| - 1) + 1 = 2|i - j| - 1$

The inversion number (what you call the number of flips) is the number of crossings in the arrow diagram of the permutation. This is $1$.

See here for more.


Why is $F[x,y,z]/(x^2-yz)$ a domain? How can I show that the ring $R=F[x,y,z]/(x^2-yz)$ is an integral domain? (Here $F$ is a field.)

I tried to prove this by contradiction. Suppose there are $\bar f, \bar g \in R-\{0\}$ such that $\bar f\bar g =0 $ in $R$, i.e. $fg$ is divisible by $x^2-yz$ in $F[x,y,z]$. But I don't know how to proceed.

Consider $p(x,y,z):=x^2-yz$ as a polynomial in $R[x]$, where $R:=F[y,z]$ is a unique factorization domain.  If we can show that $p(x,y,z)$ is irreducible in $R[x]=F[x,y,z]$, then we are done.  First, note that $y$ is a prime element of $R$.  The constant term of $p(x,y,z)$ as a polynomial in $x$ is $-yz$ which is not in the ideal generated by $y^2$ of $R$.  The coefficient of $x$ in $p(x,y,z)$ is $0$, which belongs in the ideal generated by $y$ of $R$.  Using Eisenstein's Criterion for integral domains,  we conclude that $p(x,y,z)$ is irreducible in $R[x]=F[x,y,z]$.  Since $F[x,y,z]$ is a unique factorization domain, $p(x,y,z)=x^2-yz$ is a prime element.  That is,
$$F[x,y,z]/\langle x^2-yz\rangle$$
is an integral domain.

$x^2-yz$ is irreducible, so $x^2-yz$ is prime, so $F[x,y,z]/(x^2-yz)$ is integral domain.


Calculating rectangle corners location in rotating around a pivot line I'm trying to transform shapes in an Android app by rotating them around a pivot line (using View.setRotationY). As you see in the below images, I rotate the shape around the green line which is parallel to the vertical axis and crossed the center point of the rect. Looking at its 2D drawing makes sense that the left edge comes forward and the right edge goes backward.
Now I want to formulate it. I mean the length of the edges based on rotation angle as well as their movement.

Any suggestions are really appreciated.

Rotation Angle: 0°

Rotation Angle: 15°

Rotation Angle: 30°

Rotation Angle: 45°

Rotation Angle: 60°

Rotation Angle: 75°

Rotation Angle: 90°

Assuming a square of sides 2 units, the spatial coordinates of the four vertices are

$$(x,y,z)=(r\cos\theta,s,r\sin\theta)$$ where the $r$ and $s$ are $\pm1$ independently, and $\theta$ is the rotation angle around the vertical axis ($y$).

Now their perspective projection, assuming the viewpoint at $f$ units in the direction of the $z$ axis is

$$(X,Y)=\left(\frac{fx}{z+f},\frac{fy}{z+f}\right)=\left(\frac{fr\cos\theta}{r\sin\theta+f},\frac{fs}{r\sin\theta+f}\right).$$

Assume the edge length at zero angle is $a$ and the edge length at 90-degree angle is $b$, you may formulate the edge length as function of the rotation angle $\theta$ as,

$$L(\theta) = b\sin\theta + a(1-\sin\theta)$$

Such a relationship is based on the assumption that the edge length varies linearly with respect to the front-back distance.


Closure of $(a,b)$ is [$a,b$] I'm trying to write more formal proofs for rather obvious/intuitive things. I wanted to know whether my proof for this question is good enough.

  Prove that the closure of the set $(a,b)$ is the set [$a,b$].

From a well known property we know that $Cl(A)=A \cup \operatorname{acc}(A)$

Then it is enough to prove that $a$ and $b$ are accumulation points of $(a,b)$.

I will be using the following definition for accumulation point :

  $x$ is an accumulation point of the set A if $B(x,r)\cap(A \setminus \{x\}) \neq \varnothing \ \forall r>0$.

Now to check whether $a$ and $b$ are accumulation points we need to check if there exists $r>0$ such that $B(a,r)\cap ((a,b) \setminus \{a\}) = \varnothing$.

$B(a,r)=(a-r,a+r)$

It is obvious that we can always find a  point such that belongs in the second half of the interval that is $x \in (a,a+r)$ and also belongs in $(a,b) \setminus \{a\}$.

So indeed $a \in \operatorname{acc}((a,b))$.

Similarly $B(b,r)=(b-r,b+r)$ and there exists point $x$ that belongs in the first half of the interval and also belongs in $(a,b) \setminus \{b\}$

So these two points are indeed accumulation points. It follows that $Cl((a,b))=
[a,b]$

Is this solution correct? Also is there perhaps a simpler approach ?

EDIT: After reading through the answers I realized that we also need to prove that there are no other accumulation points outside the interval $(a,b)$. This can be done by picking suitable radius. There were couple of them suggested in comments/answers.

The solution is correct and very simple as far as it goes.

However, it is not complete. You also have to prove that no point in $\mathbb R - [a,b] = (-\infty,a) \cup (b,\infty)$ is an accumulation point.

Put $$a_n=a+\frac{1}{n}$$
and
$$b_n=b-\frac{1}{n}$$

for large enough $n$,

$$(a_n,b_n)\in (a,b)^2,$$

and
$$\lim_{n\to+\infty}a_n=a$$
$$\lim_{n\to+\infty}b_n=b.$$

on the other hand, $(a_n)$ and $(b_n)$ are not stationnary sequences, thus $a$ and $b$ are accumulation points.


How do you describe your mathematical research in layman's terms? "You do research in mathematics!  Can you explain your research to me?"

If you're a research mathematician, and you have any contact with people outside of the mathematics community, I'm sure you've been asked this question many times.  For years now, I've struggled to find a satisfying answer.  I think an ideal answer to this question should:

be accessible to someone who hasn't studied math since high school
build intrigue and wonder
honestly, albeit vaguely reflect your research
only require a few sentences

(Of course, these guidelines will change depending on the audience and venue.  For example, speaking with an engineer over a meal allows more time and technical language than would speaking with a stranger on a bus.)

I study the representation theory of algebraic groups and Lie algebras over fields of positive characteristic, so I usually say something along the following lines:

  I work with two algebraic objects that are closely related called algebraic groups and Lie algebras.  These objects can act on spaces (like three-dimensional space) by transforming them in a nice way, and I study these actions.  One aspect of my work that is especially challenging is that I use number systems in which a chosen prime number is equal to zero.

Honestly, based on my guidelines above, I think this response is poor, but with so much to communicate in such limited terms with such limited time, the task seems nearly impossible.

Using my guidelines, how would you describe your own field of research? Or, if my guidelines are too strict, how would you deal with this question?

I'm a little disappointed by the comments. Granted, it's hard to explain mathematics, but having the attitude that you're not even going to try is not doing mathematics PR any favors. We can't in good faith expect the public to fund our research if we're not even going to try to tell them what we're doing with their money.

First, I hope you won't take this the wrong way, but I'd like to spend a bit talking about why I agree that your proposed answer is not good:

  I work with two algebraic objects that are closely related called algebraic groups and Lie algebras. These objects can act on spaces (like three-dimensional space) by transforming them in a nice way, and I study these actions. One aspect of my work that is especially challenging is that I use number systems in which a chosen prime number is equal to zero.

The general problem is that you are trying much too hard to be accurate. (In other words, I think specification #3 is the least important of your specifications and should mostly be discarded.) The technical terms you're using mean nothing to a layperson, and depending on the layperson, "algebraic object," "transforming," and "prime number" could all be technical terms.

I would aim much lower than where you're trying to aim, and settle for communicating some intuitive aspect of the already exciting general idea of group theory, symmetry, and representation theory. I would not even try to mention positive characteristic without more time. I would try to relate the ideas to concrete experiences the layperson has had or at least familiar ideas from other areas. For example:

  I study special kinds of symmetries. For example, think of the symmetries of a sphere [here I would pretend to hold a sphere in my hands while rotating it around]; I study symmetries that are like these but more complicated. The idea of symmetry has applications all across mathematics and physics; in the case of symmetries of a sphere, you can use these symmetries to predict certain properties of the periodic table.

Despite not being a mathematician myself, I would like to contribute a bit as in my field I share the same problem often (depending on the project I am working on). What I have learned is that what works best is to bypass the entire description of what you're actually doing and just jump straight ahead to applications even if you will never get anywhere close to them.

So, if for example you're doing anything with number theory you can just say 'I am contributing to theories which one day might be used by others to improve the security of facebook or your bank for example' or with lot's of mathematics one might be contributing to the systems/engines on which you will play their games, or maybe for astronomers to understanding the universe and how it functions, or for electrical engineers to make your next mobile phones, or it can be something that will help economist to understand (model) the economy. Or any crazy combination of those :D . Like, I dunno, from as far as I know mostly all math has some application nowadays, especially with computers being as complex and diverse as they are, so just find something cool :P .

Now, I am perfectly aware that you might not even be remotely aware of what your research will be used for, so in those cases I would just describe your sub field in general or make an educated guess, but in the end I do think this qualifies with all your guidelines. Sure, from the point of an insider spec 3 won't be met, but you have given an outsider far more of an understanding than with your description.


Why can't the Polynomial Ring be a Field? I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.

The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?

Hint $\rm\quad\rm  x \, f(x) = 1 \,$ in $\,\rm R[x]\ \Rightarrow \ 0 = 1 \, $ in $\,\rm R, \,  $  by evaluating at $\rm\ x = 0 $

Remark $\ $ This has a very instructive universal interpretation: if $\rm\, x\,$ is a unit in $\rm\, R[x]\,$ then so too is every $\rm\, R$-algebra element $\rm\, r,\,$ as follows by evaluating $\ \rm x \ f(x) = 1 \ $ at $\rm\ x = r\,.\,$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $\rm R$-algebra. $ $ A natural choice is the nonunit $\,\rm 0\in R,\,$ which yields the above proof.

Let $R$ be a field. If $f$ is a polynomial whose degree is at least 1 in $R[x]$, then $f$ cannot have an inverse.

For if $f(x)g(x) = 1$, where $f$ and $g$ are polynomials whose degree is at least 1 in $R[x]$, then the leading coefficient of $g$ would have to be $0$, which is
impossible.


Solve the equation without using square both sides Solve the equation without using square both sides:

$$(x+1)(2x^{3}-5)^{1/2}= 2x^{3}- 3x^{2}+x-2$$

I try to use Wolfram Alpha and here is the the answer:

https://www.wolframalpha.com/input/?i=(x%2B1)+(2x%5E3-+5)%5E(1%2F2)%3D+2x%5E3-3x%5E2%2Bx-2

I can' t continue. Help me!

The solution is $x=3$ and can be found as follows:

$2 x^3-5$ must be perfect square among cubic of few numbers we only have:

$2\times 3^3-5=54-5=49=7^2$

We check this:

$(3+1)\times 7= 28$

$2\times3^3-3\times 3^2+3-2=28$

Now suppose $(2x^3-5)^{1/2}=a$ then we have:

$a(x+1)=a^2-3x^2+x+3$.. or $a^2-(x+1)a-3x^2+x+3=0$

The condition for having solution in R is:

$\Delta=13x^2-2x-11=m^2; m ∈ R$

$2x^3-5=k^2; k∈ R$

The solution of question leads to the solution of this Diophantine system of equations.One solution is $x=3,..k=7,..m=10$


Find the initial direction and time of flight of a basketball, given initial speed and distance A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?

I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.

If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $x\in \left[0,\frac\pi2\right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u \cos(x) t= d$ and $u\sin(x) t-\frac12gt^2=0$.  Solve for $x$ to get $\sin(2x)=\frac{gd}{u^2}$.  There are two possible values for $x$, whence also two values for $t=\frac{d}{u\cos(x)}$.

Is the ball is thrown along a parabolic path at an angle $ \alpha $ with speed $v$, the time is

$$ \frac{distance }{v \cos \alpha} $$

So $\alpha$ should also be known.


Evaluating $\lim_{\alpha\to 1}\frac{x_0^{1-\alpha}x^{\alpha}-x}{1-\alpha}$ I have an engineering equation:

$$y=\frac{x_0^{1-\alpha}x^{\alpha}-x}{1-\alpha}$$

where $x_0$ is a constant and $\alpha$ is a parameter.

A potential singularity arises for $\alpha =1$ because then:

$$y=\frac{x-x}{1-1}=\frac{0}{0}$$

Can anyone show me how to determine:

$$\lim_{\alpha \to 1}y$$

Thank you.

You have already shown that you have an indeterminate form

Applying L'Hopital's rule:

The derivative of the numerator $\frac {d}{da} e^{(1-a)\ln x_0+a\ln x} - x = x_0^{1-a}x^a(\ln x-\ln x_0)$

as $a$ approaches $1,$ we get: $x(\ln x - \ln x_0)$

and the derivative of the denominator is $-1$

$x(\ln x_0 - \ln x)$

The notation is quite confusing and I think it is simpler if you write the problem as

$f(x) = \frac{ac^x-b}{1-x}$, where $a=x_0, b=x, c=\frac{x}{x_0}.$

$$\lim_{x\to 1} f(x)$$ now looks elementary(doesn't it?).


Find $\lim _{x\to \infty }\left(x\left(\ln\left(x+1\right)-\ln x\right)\right) $ Find

$$\lim _{x\to \infty }x\Big(\ln\left(x+1\right)-\ln x\Big)$$

Here's how I do it:

$$x\Big(\ln(x+1)-\ln x\Big) = x\Bigg(\ln(x(1+\frac{1}{x}) - \ln x\Bigg)$$
$$x\Big(\ln x + \ln(1+\frac{1}{x}\Big)-\ln x =x\ln\Big(1+\frac{1}{x}\Big)$$
$$\ln\Bigg( \Big(1+\frac{1}{x}\Big)^x \Bigg) \rightarrow \ln 1^\infty = 0$$

What am I doing wrong? The answer is supposed to be $1$, but I get $0$.

$1^\infty\neq 1$. It is a case you have to investigate. Here,
$$
\lim_{x\to +\infty}\bigl(1+1/x\bigr)^x = e.
$$
This should be derived in your book.

look at $$x\left(\ln(x+1) - \ln(x)  \right) = x \ln\left(\dfrac{x+1}{x}\right) =
x\left( \ln(1 + 1/x) \right) = x\left(1/x -1/(2x^2) + \cdots \right) = 1$$ for $x \to \infty$


Find the no of digits in the sum of all integral values of a in $[1,100]$ for which following condition satisfies. Find the no of digits in the sum of all integral values of a in $[1,100]$ for which the equation $x^2-\left(a-5\right)x+\left(a-\dfrac{15}{4}\right)=0$ has atleast one root greater than zero.

My attempt is as follows:-

$$D\ge0$$
$$(a-5)^2-4\left(a-\dfrac{15}{4}\right)\ge0$$
$$a^2+25-10a-(4a-15)\ge0$$
$$a^2-14a+40\ge0$$
$$(a-4)(a-10)\ge0$$
$$a\in(-\infty,4]\cup[10,\infty)$$

Case $1$: Both roots are greater than $0$

$f(0)>0, \dfrac{\alpha+\beta}{2}>0$

$$a>\dfrac{15}{4}$$
$$a-5>0\implies a>5$$

Hence intersection of all conditions would give $a\in[10,\infty)$

Case $2$: One root is greater than $0$ and one root is lesser than $0$

$$f(0)\le0$$
$$a\le\dfrac{15}{4}$$

Hence intersection of all conditions would give $a\in\left(-\infty,\dfrac{15}{4}\right]$

So all integral values of $a\in[1,100]=\{1,2,3,10,11\cdots\cdots100\}$, hence sum will be $6+\dfrac{91}{2}\cdot(10+100)=6+5005=6+5005=6+5005=5011$

So answer should be $4$ but actual answer is $7$. I tried to debug it a lot but didn't find any mistake. Please help me in this.

What you did was basically correct, but there are a few issues. First, one small point is that with your ranges, when a value is included on a boundary, you should use a square bracket instead of a round one. For example, with the result of checking the discriminant being non-negative, you should have stated $a\in(-\infty,4]\cup[10,\infty)$ instead of $a\in(-\infty,4)\cup(10,\infty)$. I see later you mostly assumed it excluded those values (but it should include them as the discriminant is $\ge 0$), so this seems to have been intentional on your part.

With your Case $1$, I'm not sure why repeated the calculations for the discriminant done earlier. Perhaps it was just a copy & paste mistake?

Another minor point is that, technically, you should have Case $2$ be for one root greater than $0$ and one root being less than or equal to 0. Thus, you would consider $a \le \frac{15}{4}$. However, since only integral values of $a$ are involved, it doesn't make any difference to the end results in this case.

Next, your list of integral values of $a$ is stated correctly, but on the next line with the summation, you excluded $10$. As such, the actual sum would be $5001 + 10 = 5011$ instead.

Note the sum of the digits is $7$, so undoubtedly the question had intended to ask this instead of for the number of digits. Also, since the maximum sum has $4$ digits, there's no way the actual sum could have $7$ digits. In addition, to me, at least, asking for the number of digits in this type of problem doesn't make a lot of sense. Unless you make a significant mistake, it's fairly easy to see the result is $\ge 1000$, but even if all values from $1$ to $100$ were summed, the result would be $5050$, so it's fairly clear the answer would be $4$.


Finding a parametric solution via three specific solutions. The linear equation $$x - y + 2z = 3$$ has $[3, 0, 0], [0, 1, 2]$, and $[6, 1, -1]$ as specific solutions.

The complete set of solutions corresponds to the set of points in the plane determined by the given equation. If we set $y = s$ and $z = t$, then a parametric solution is given by $[3 + s - 2t, s, t]$.

How do they found this parametric solution?

Thank you.

\begin{align}
&\space x-y+2z = 3 \\
&\Rightarrow x = 3 - 2t + s \\
&\Rightarrow [x,y,z]=[3-2t+s,s,t]
\end{align}

Note that I used $y = s, z = t$ in the last step.

$$x-y+2z=3\implies x=3+y-2z$$

They then switched $y\iff s$, and  $z\iff t$


Logic & Reasoning Question At a track meet, every group of $n$ participants shares exactly one common friend.
Suppose runner $P$ has the largest number of friends. Determine how many friends $P$ has.

Assume for this question:

$n \in N$ such that $n ≥ 3$
Friends are mutual, EX: if $X$ is friends with $Y$, then $Y$ is friends with $X$
No one is friends with themselves
a group of runners has a "common friend" $a$ iff each runner in the group is friends with $a$

Prove the answer.

Update for Clarification

$n$ is the size of the group, not the total number of attendees.
There are an arbitrary number of groups
The answer should be expressed in terms of $n$
If $X$ is friends with $Y$ then both $X$ and $Y$ are at the trackmeet
Assume there are at least $n + 1$ attendees.

Thank you very much for your effort so far @fleablood

P can't like as many as n people.

Label n of P's friends as $a_1,... a_n$.

$(P, a_1, ...., a_{n-1})$ have P as the mutual friend.  So $(a_1,..., a_{n-1})$ have no mutual friend.  So $(a_1, ..., a_n)$ have $a_n$ as a mutual friend.  So $(P, a_2, ...., a_n)$ have both P and $a_n$ as mutual friends.  Impossible.

So P has n-1 friends or fewer.

But as any group of n people have 1 mutual friend, that mutual friend has at least n-1 friends.  So P has at least n-1 friends.

So P has n-1 friends.


Showing an infinite number of irreducible polynomials in $\Bbb Z_2 [x]$ exist Show that there are infinitely many irreducible polynomials in $\Bbb Z_2[x]$.

I have a strategy about how to do this but I need some help finding a contradiction. I want to assume there are finitely many irreducible polynomials. $p_1,...p_n$. Then I want to consider the irreducible factors of $(p_1p_2...p_n)+1$. I don't really know where to go from here. Any help would be appreciated.

$x+n$ is irreducible for all $n\in \mathbb Z$. They are also different irreducibles.

If question is about finite field with two element: each of root of your polynomial will be in some extension of $\mathbb F_2$. Think bigger extensions.

EDIT 2: Thanks everyone for downvoting a legitimate answer and an honest attempt to help. Great community here.

You can use Eisenstein's criterion to note that $x^n - 2$ is irreducible in $\mathbb{Z}_2[x]$ for every integer $n\ge 2$.

Also if you don't require your irreducible polynomials to be monic, then for any one irreducible polynomial $f$, $uf$ is also irreducible for any unit $u\in\mathbb{Z}_2[x]$. If you know that every element of $\mathbb{Z}_2$ can be expressed as a power series $\sum_{k\ge 0}a_k 2^k$ where $a_k\in\{0,1\}$, then by Hensel's lemma every power series with $a_0 = 1$ corresponds to a unit in $\mathbb{Z}_2$. Thus $\mathbb{Z}_2$ has infinitely many units, so the family $\{uf : u\in\mathbb{Z}_2^\times\}$ is already an infinite family of irreducible polynomials (though not monic).

EDIT: If by $\mathbb{Z}_2$ you mean the field $\mathbb{F}_2$ with two elements, then your strategy is probably easiest. If $p_1,\ldots,p_n$ denote all the primes of $\mathbb{F}_2[x]$, then none of the $p_i$'s can divide $(p_1p_2\cdots p_n)+1$, since if say $p_1$ divides $(p_1p_2\cdots p_n)+1$, then $(p_1p_2\cdots p_n)+1 = p_1q$ for some $q\in\mathbb{F}_2[x]$, so
$$1 = p_1q - (p_1p_2\cdots p_n)$$
so
$$1 = p_1(q - (p_2p_3\cdots p_n))$$
which is to say that $p_1$ is a unit (ie, $p_1$ is invertible), which of course cannot be the case if $p_1$ is prime/irreducible (these are the same in $\mathbb{F}_2[x]$ because $\mathbb{F}_2[x]$ is a UFD)


Irrational numbers in reality I have a square tile which measures 1 metre by 1 metre, by the Pythagorean identity the diagonal from one corner to another will be $\sqrt 2$ metres. However  $\sqrt 2$ is an irrational number, could someone explain how it is possible for a non-terminating and non repeating number to represent a finite length in reality?

It's not the number $\sqrt{2}$ that's non-terminating; it's the decimal expansion of the number that's non-terminating. If you try to write down the entire decimal expansion of the number, you'll be writing forever, but the number itself is just a small number between $1.4$ and $1.5$.

A rational number can be defined as a certain type of function from integers to integers se follows. Take for example, $\frac{8}{14}$. I define it to be the function that assigns to each integer, the integer gotten by multiplying it by 8 then taking the quotient on division by 14. The following is the graph of this function which we call $\frac{8}{14}$

Now take the following two segments of a function which if you repeat them you get a rational number.

 &

Now you can get another two segments from these ones as follows. The first one is 1 copy of the first image followed by 2 copies of the second and the second one is one copy of the first image followed by 3 copies of the second to get

 &

Now again we can take two more images, one that can be gotten by taking one copy of the first followed by 2 copies of the second and 1 copy of the first followed by 3 copies of the second to get

 &

It's easy to see that this approaches a function that doesn't repeat. We've invented a new function from integers to integers. Now we can define the real numbers to be all functions of this sort and call the nonrepeating ones irrational numbers. It can in fact be shown that this particular nonrepeating function is $\sqrt3$.


Calculate Infinite Limit I'm trying to calculate the limit and when I get to the last step I plug in infinity for $\frac 8x$ and that divided by -4 I get - infinity for my answer but the book says 0. Where did I go wrong?

$$
\frac {8x^3-x^2}{7+11x-4x^4}
$$

Divide everything by $x^4$
$$
\frac {\frac{8x^3}{x^4}-\frac{x^2}{x^4}}{\frac{7}{x^4}+\frac{11x}{x^4}-\frac{4x^4}{x^4}}
$$

Results
$$
\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} = \infty
$$

Continuing from where you've reached, you can conclude that $$\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4}$$

where as $x \to \infty$, we have $\frac{8}{x} \to 0$. The same goes for $\frac{1}{x^2} \to 0$, and $\frac{7}{x^4} \to 0$. We also have $\frac{11x}{x^4} \to 0$, so you can rewrite the above approximately for large $x$ as $$\frac {\frac{8}{x}-\frac{1}{x^2}}{\frac{7}{x^4}+\frac{11x}{x^4}-4} \approx \frac{0-0}{0+0-4} = 0$$

Alternatively, (this isn't a technique you are likely to understand just yet and I wouldn't recommend using it for the moment, but I am including it any way just for an alternative way), we could use L'Hôpital's rule. Since we have $$\lim_{x\to \infty} \frac {8x^3-x^2}{7+11x-4x^4} = \frac{\infty}{\infty}$$ an indeterminate form L'Hôpital once to get $$\lim_{x\to \infty} \frac {8x^3-x^2}{7+11x-4x^4} = \lim_{x\to \infty}\frac{24x^2 - 2x}{11 - 16x^3}$$
which is still an indeterminate form, apply L'Hôpital again to get $$\lim_{x\to \infty}\frac{24x^2 - 2x}{11 - 16x^3} = \lim_{x\to \infty}\frac{48x - 2}{- 48x^2}$$ still indeterminate, so a final application of L'Hôpital gives us: $$\lim_{x\to \infty}\frac{48x - 2}{- 48x^2} = \lim_{x\to \infty} -\frac{2}{x}$$ which tends to $0$ as $x \to \infty$.

Suppose that $n,m\in \mathbb{N}$:
$$f(x)=\dfrac{a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}}{b_{m}x^m+a_{m-1}x^{m-1}+\dots+b_{1}x+b_{0}}$$
We have three case:
Case $1$:
$$n< m\Longrightarrow  \lim_{x\rightarrow\infty}f(x)=0$$
Case $2$:
$$m<n\Longrightarrow \lim_{x\rightarrow\infty}f(x)=\infty$$
Case $3$:
$$n=m\Longrightarrow \lim_{x\rightarrow\infty}f(x)=\dfrac{a_{n}}{b_{m}}$$


How do we know that the set of sets that each contain real coefficient polynomials of distinct degrees is countable? If I have the set $V=\left \{S_{0},S_{1},S_{2},...\right \}$ where $S_{0}$ is the set of real constant polynomials, $S_{1}$ is the set of real polynomials with degree 1 and so on. How do we know that $V$ is countable?

It isn't. Just the set of polynomials over $\mathbb{R}$ of degree 0 is equinumerous with $\mathbb{R}$, so isn't countable; that being a subset of $\mathbb{R}[x]$ means the later one can't be countable either.


Is this a possible multivariate generalization of pointwise Mutual Information? If so, how can it be formalized? I understand that pointwise Mutual Information (pMI) is the probability of a sequence of dependent events conditioned on the probability of its elements treated as independent events.

The definition of mutual information given in Church & Hanks (1990) and based on Fano (1961) is the following:

"[...] if two points (words), x and y, have probabilities P(x) and P(y), then their mutual information, $I(x,y)$, is defined to be:

$$I(x,y) = log_2\frac{P(x,y)}{P(x)P(y)}$$

Informally, mutual information compares the probability of observing $x$ and $y$ together (the joint probability) with the probabilities of observing $x$ and $y$ independently (chance). If there is a genuine association between $x$ and $y$, then the joint probability $P(x,y)$ will be much larger than chance $P(x)P(y)$, and consequently $I(x,y) \gg 0$. If there is no interesting relationship between $x$ and $y$, then $P(x,y) \approx P(x)P(y)$, and thus, $I(x,y) \approx 0$. If $x$ and $y$ are in complementary distribution, then $P(x,y)$ will be much less than $P(x)P(y)$, forcing $I(x,y) \ll 0$.''

Pointwise Mutual Information typically applies to sequences of two events, but it has sometimes been extended to longer sequences. For example, the pointwise mutual information of a sequence ABCD would be:

$$I(ABCD) = log_2\frac{P(ABCD)}{P(A)P(B)P(C)P(D)}$$

I would like to condition of the probability of a sequence not only on its atomic elements but also on all its possible subsequences. In the context of the above example, this would be:

$$I(ABCD) = log_2\frac{P(ABCD)}{P(A)P(B)P(C)P(D)P(AB)P(BC)P(CD)P(ABC)P(BCD)}$$

Does this also fall under mutual information? If not, does this way of conditioning the probability of a set of dependent events have ab explicit mathematical treatment in probability theory and/or information theory? Is there a way to express this way of conditioning events formally in a general fashion applicable to sets of events of arbitrary length? I tried formalizing it this way (see image below) but it seems very clunky and, more importantly, I don't know where to encode the information that there will be N-1 rows in the denominator:

mutual-info-extended

References:

Church, K. W. and P. Hanks (1990). Word association norms, mutual information and lexicography. Computational Linguistics 16(1), 22-29.

Fano, R. (1961). Transmission of Information: A Statistical Theory of Communications. Cambridge, MA: MIT Press.

The second equation in your post looks sufficient, assuming order is not important. The original definition of mutual information doesn't include order, it's just "these two things happen together more often (or not) than you'd expect them to occur based on their individual probabilities, assuming independence".  Your 2nd equation says the same thing: "these 4 things happen together more (or less) frequently than they would if they were all independent". That's exactly what I'm looking for in my application (and now that I see your 2nd equation, it's obvious); while I can't immediately say it's wrong, I'm not sure why you'd want to look at the subsets. And note that the way you've written it, they are subsets, not subsequences; if they were sequences then you would have separated P(AB) and P(BA).

Since both are plausible, but the first is simpler, I'd invoke Occam's Razor; however, another way to figure out which one you want is just to try both on some data where you think you know the "correct" answer, and see which ones gives the most sense. Certainly that's what the author did of the only published paper I can find on the subject, which is called "Two multivariate generalizations of pointwise mutual information". He just tried them both and looked at which one worked better. Link below, though if you use Google Scholar on the title above you can find several versions.

https://dl.acm.org/citation.cfm?id=2043124


Would going first in Egg Russian Roulette increase or decrease your chances of winning? Egg Russian Roulette:

  6 eggs, 5 of which are hard boiled & 1 of which is raw. Players (1v1)
  each choose eggs in a turn based system whereby the egg is smashed
  against the forehead. The first player to smash the raw egg against
  his/her forehead is the loser.

Should you have the option of choosing who goes first, would going first increase or decrease your chances of winning?

Let's list all the probabilities shall we?

If you went first there is a:
$$\frac16$$
chance of you losing on the first turn,
$$\frac56\times\frac45\times\frac14=\frac16$$
chance of you losing on the third turn,
$$\frac56\times\frac45\times\frac34\times\frac23\times\frac12=\frac16$$
chance of you losing on the fifth turn,

Therefore there is a
$$\frac16+\frac16+\frac16=\frac12$$
chance of you losing, thus your chances will stay the same.

If you went first you'd have a $1\over6$ chance of getting a raw one and the sequence of moves would look like this. $$\text{you}={1\over6}$$$$\text{them}={1\over5}$$$$\text{you}={1\over4}$$$$\text{them}={1\over3}$$$$\text{you}={1\over2}$$$$\text{them}={1\over1}$$ Which you'd have the best odds in the beginning and if it did happen to go all the way to the end your opponent would be the one to lose.

If you went second though your opponent would go twice in the time you only go once. So just within three moves you would've had a ${1\over5}$ chance of getting the raw one, but your opponent would have had a ${1\over6}+{1\over4}={5\over 12}$ which is a much higher percentage that yours.

So I could easily justify either way. That's just my logic. Don't know if it helps or not!

If you forced me to pick, I'd say go second. Just so if you happened to make it to the end and you knew you were going to get the raw egg you could use it as a weapon against your opponent.


Rank of "Characteristic Matrix": $A - \lambda I$ Matrix Analysis Horn, Johnson 1st ed. Problem: 1.2.6

  If $\lambda \in \sigma(A)$ $(\lambda$ is an eigenvalue of $A)$, for $A \in \mathbb{C}^{n \times n}$, has multiplicity $1$ as a root of the characteristic polynomial $p_{A}(t) = \det(A - \lambda I)$, show that $\text{rank}(A - \lambda I) = n - 1$.

The hint suggests this route:
In a previous problem we have shown that
$$\frac{d}{dt}p_{A}(t) = \sum\limits_{i=1}^{n}p_{A_{i}}(t)$$
where $A$ is the matrix obtained by the deleting the $i$-th row and column of $A$. Since $\lambda$ is a single root of $p_{A}$, $p_{A}'(\lambda) \neq 0$ (correct? is there another reason this is true?). This implies that at least one of the characteristic polynomials of the principal submatricies of size $n-1$ is nonzero at $\lambda$.

I'm now to conclude that one of these principle submatrices is nonsingular. I can't see why though.

Lemma: Let $A  \in \mathbb{C}^{n \times n}$ and $\lambda$ be an eigenvalue for $A$. Let $\operatorname{mult}(\lambda)$ denote the multiplicity of $\lambda$ as a root of $\chi_A$. Then $\dim\ker (A-\lambda I)\leq \operatorname{mult}(\lambda)$.

Here this implies $\dim\ker (A-\lambda I)\leq 1$ and since $\lambda$ is an eigenvalue we also have $\dim\ker (A-\lambda I)\geq 1$, hence $\operatorname{rank}(A-\lambda I)=n-1$ by the null-rank theorem.

Proof of the lemma: Let $F$ be a vector space such that $\ker (A-\lambda I) \oplus F = \mathbb C^n$. Let $p=\dim \ker (A-\lambda I).$

In this new decomposition, $A$ has the form
$$
\left[
\begin{array}{c|c}
\lambda I_p & B \\
\hline
0 & C
\end{array}
\right]
$$
that is to say there exists some invertible $P$ such that $A=P^{-1}\left[
\begin{array}{c|c}
\lambda I_p & B \\
\hline
0 & C
\end{array}
\right]P$

hence $\chi_A(t) = (\lambda-t)^p\det(C-tI_{n-p})=(\lambda-t)^pQ(t)$ where $Q$ is some polynomial. That implies $$p\leq \operatorname{mult}(\lambda)$$ie $$\dim\ker (A-\lambda I)\leq \operatorname{mult}(\lambda)$$

let $A$ be a $n\times n$ matrix, and if $c$ is an eigenvalue of the characteristic equation, then $|A-cI|$ would be $0$ and hence it won't be a full rank matrix (rank $< n$)

In fact, if it is a simple eigenvalue i.e. its algebraic multiplicity is 1, same as geometric multiplicity, then the rank would be $n-1$.
If it has algebraic multiplicity > $1$ and  $\operatorname{dim}\big(\operatorname{Ker} (A-cI)\big)=g$ (geometric multiplicity), then the rank would be $n-g$.( According to the rank nullity theorem which says that $\text{nullity} + \text{rank} = n$, here nullity is $g$, so we have rank of the characteristic equation $= n-g$)


To Prove $x'Ax=\mathrm{tr}(xAx')=\mathrm{tr}(Axx')=\mathrm{tr}(xx'A)$ To prove,

  $x'Ax=\mathrm{tr}(xAx')=\mathrm{tr}(Axx')=\mathrm{tr}(xx'A)$

where

$A$ is a square matrix.
$x'$ is the transpose of $x$.
For each $x,x'$ are column vector, row vector.

Hint: note that so long as $A$ and $B$ are compatible (that is, $A$ is $m \times n$ and $B$ is $n \times m$), we have trace$(AB) = $ trace$(BA)$, so that
$$
\mathrm{tr}(x'(Ax)) = \mathrm{tr}((Ax)x') \\
\mathrm{tr}((xx')A) = \mathrm{tr}(A(xx'))
$$
All you have to show then is that $x'Ax = \mathrm{tr}(x'Ax)$.


Show that 5 lines in a pentagon are concurrent Pentagon $ABCDE$ is inscribed in a circle. For any edge of $ABCDE$, we can draw the line perpendicular to that edge that contains the centroid of the remaining three vertices. Show that these 5 lines are concurrent.

I have drawn a basic diagram but nothing has come to my mind so far.  All help is appreciated.

Long hint:

Let the coordinates of the vertices of given pentagon be $A_i=(x_i, y_i), 1\leq i\leq 5.$ Then, let us compute the equation of the line that is passing through the centroid of triangle formed by $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and is perpendicular to the side $A_4A_5$. Observe that the centroid of the said triangle has coordinates $(a, b) = (\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}).$

Now the equation of the line that contains the side $A_4A_5$ is $$Y = \dfrac{y_5-y_4}{x_5-x_4}X+\dfrac{y_4x_5-y_5x_4}{x_5-x_4}.$$
If we let the equation of the desired line be $Y=mX+n$, then it satisfies $b = am+n$ and $m\dfrac{y_5-y_4}{x_5-x_4}+1=0.$ This is a system of linear equation, so you can easily find $m,n$.

Can you proceed from here? If not, let me know and I can finish the rest.

The pentagon is regular. Using analytical geometry, let's take the centre of the circle to be the origin ( this is also the centre of the pentagon). The angle subtended at the centre by each side of the pentagon is $72^\circ$ ( I leave it to the reader to figure this out). Without any loss of generalization, we can assume one vertex to lie on the y-axis. Hence, the vertices of the pentagon are :

$$
(r\cos\frac{\pi}{10}, r\sin\frac{\pi}{10})
\\
(r,0)
\\
(r\cos\frac{9\pi}{10}, r\sin\frac{9\pi}{10})
\\
(r\cos\frac{17\pi}{10}, r\sin\frac{17\pi}{10})
\\
(r\cos\frac{25\pi}{10}, r\sin\frac{25\pi}{10})
$$
Now... your proposition can be easily proved.


If $n$ is any positive integer, prove that $\sqrt{4n-2}$ is irrational If $n$ is any positive integer, prove that $\sqrt{4n-2}$ is irrational.

I've tried proving by contradiction but I'm stuck, here is my work so far:

Suppose that $\sqrt{4n-2}$ is rational. Then we have $\sqrt{4n-2}$ = $\frac{p}{q}$, where $ p,q \in \mathbb{Z}$ and $q \neq 0$.

From $\sqrt{4n-2}$ = $\frac{p}{q}$, I just rearrange it to:

$n=\frac{p^2+2q^2}{4q^2}$. I'm having troubles from here, $n$ is obviously positive but I need to prove that it isn't an integer.

Any corrections, advice on my progress and what I should do next?

$4n-2 = (a/b)^2$ so $b$ divides $a$.

But $\operatorname{gcd}(a,b) = 1$
so $b = 1$.

So now $2$ divides $a$
so write $a = 2k$
then by substitution, we get that
$2n-1 = 2k^2$

Left side is odd but the right side is even. Contradiction!

Suppose $n$ is a natural number. If $n$ is even, then $n^2 = (2k)^2 = 4k^2$ is divisible by 4. If $n$ is odd, then $n^2 = (2k-1)^2 = 4k^2-4k+1 = 1\ \textrm{mod}\ 4$.

Hence, if $n$ is a natural number, then $n^2$ is either 0 or 1 mod 4.

Thus, if $m$ is a natural number and is 2 or 3 mod 4, then $\sqrt{m}$ is not a natural number, and hence must be irrational, since $\sqrt{m}$ is either irrational or natural.

Hence, as a corollary, since $4n-2$ is 2 mod 4, then $\sqrt{4n-2}$ is irrational.

(This shows that $\sqrt{4n-3}$ is also irrational.)


Question with linear transformation and matrices, to try to reach conclusions. Let $T:\mathbb{R}^{2\times 2} \to \mathbb{R}^{2\times 2}$.     $T(\begin{bmatrix}x&y\\z&w\end{bmatrix})$ = $\begin{bmatrix}-2x+y&-y + z\\3w&w\end{bmatrix} $.
And there's a basis $B$ such that $[T]_B = \begin{bmatrix}a&1&0&0\\0&-2&0&0\\0&0&0&b\\0&0&2&-1\end{bmatrix}$.
Then which of these statements is true?
a) $a= 0$ and $b\ne 0$.
b) $a = 1$ and $b=0$.
c) $a=-1$ and $b=1$.
d) any $a$ and $b = 0$.
e) $a=1$ and any $b$.
My approach (The idea): I know that $ImT = Col[T]_B$, so I first found out that $dimImT=3$, so $Rank([T]_B)$ should be equal to $3$. so the answer isn't c,d,e.
Now I got stuck in deciding if it's (a) or (b), since both of the answers lead to $Rank([T]_B) = 3$, and I couldn't figure out how to find the correct answer.

I would really appreciate any hints or feedback, thanks in advance.

Consider the following  results:

for a linear operator $T$ rankT =rank of the associated matirx in some ordered basis say $B$.

Then from your calulations  you see the linear map has  rank  3  so its associated matrix in some ordered  basis has also rank 3
Now you just calculate the rank of the matrix that you given.
Try to find its row-reduceed echelon form or row reducedfrom to see its rank.
I am giving final matrix in the row reduceed echelon form in the below picture that i attached. Please find it.
From that you cansee  $b$ should  be non zero to have a  rank 3 .
Which is given in only option ( a). So it the answer.
You think why option ( c) is not the answer  ,as it will give the rank of the matrix 4.
For any other queries  you can put the values of a and b to find the rank of the matrix is 3 or 4
https://m.matrix.reshish.com/rankCalculation.php


Query about Reductio Ad Absurdum If we use the method of contradiction(i.e.Reductio Ad Absurdum), and if one of our assumptions is wrong, does that mean that all our assumptions are wrong and is the statement or hypothesis proved?

You are correct: once we obtain a contradiction, we don't know which of our assumptions was false.  Any or all of them might be.  But the claim that $a$ and $b$ have no common factor is not an assumption.

In the proof that $\sqrt2$ is irrational, we start by supposing that $\sqrt2$ is rational.  A rational number is one that can be written as a fraction, so we have $\sqrt2 = \frac pq$  for some integers $p$ and $q$, and $q\ne 0$.

Now what if $p$ and $q$ had a common factor?  Then we could take the fraction $$\sqrt2 = \frac pq$$ and cancel the common factor from the numerator and denominator and obtain an equal fraction $$\sqrt 2 = \frac ab$$ which was in lowest terms.  And then $a$ and $b$ would have no common factor greater than 1, because that is what it means to write a fraction in lowest terms.  Since every fraction can be written in lowest terms, we can do this no matter what $p$ and $q$ were.

So if $\sqrt 2$ is rational, we can find integers $a$ and $b$ with $\sqrt2=\frac ab$ and $a$ and $b$ have no common factor bigger than 1.  This is not an assumption; it's how fractions work.

And if we get nonsense later on, that they have common factors anyway, we know it must be nonsense.  It couldn't be because $a$ and $b$ had a common factor bigger than 1, because we know there must be some pair of integers  that don't have a common factor bigger than 1; we can obtain them by reducing the original fraction to lowest terms, and then call them $a$ and $b$.  It can't actually be that $a$ and $b$ really had a common factor all along, because if there had been one, we would already have canceled the factor and used the names $a$ and $b$ for the result instead.  So the problem must be something else.

I'm not sure I understand the question. However:

The only assumption made is that $p/q$ has no common factors, and then this is contradicted by showing that they must.

This means that $\sqrt{2}$ cannot be written as a reduced fraction (i.e. with no common factors in $p$ and $q$). Since every fraction can be written in reduced form, there is no fraction equal to $\sqrt{2}$ - it does not help to switch to "another" fraction $a/b$.


The contradiction method used to prove that the square root of a prime is irrational The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable?
e.g. Suppose $\sqrt{4}$ is rational,
$$\begin{align}
       \sqrt{4} &=p/q        \qquad\text{where pand q are coprimes} \\
        4 &=p^2/q^2\\
      4q^2&=p^2       \tag{1} \\
         4&\mid p^2\\
         4&\mid p\\
     \text {let }p&=4m  \qquad\text{for some natural no. m}  \\
       p^2&=16m^2\\
      4q^2&=16m^2   \qquad\text{(from (1) )}\\
       q^2&=4m^2\\
         4& \mid q^2\\
         4&\mid q
\end{align}
$$
but this contradicts our assumption that $p$ and $q$ are coprime since they have a common factor $p$. Hence  $\sqrt{4}$ is not rational. But we know that it is a rational. Why?

If $4q^2 = p^2$, $4$ is a factor of $p^2$, but it does not follow that $4$ is a factor of $p$ only that $2$ is a factor of $p$. For example, $4$ is a factor of $36$, but $4$ is not a factor of $6$ (but $2$ is).

In general, if $a$ is prime and $a$ is a factor of $pq$ then $a$ must be a factor of $p$ or a factor of $q$ (or both); in particular, if $a$ is a factor of $p^2$, then $a$ must be a factor of $p$. It then follows that if $m$ is a product of distinct primes, then if $m$ is a factor of $p^2$, $m$ must be a factor of $p$. So the method used to show that the square root of a prime number is irrational extends to numbers which are the product of distinct primes, but no further.

Let $p$ be prime and suppose that $\sqrt{p}$ is rational. Then $\sqrt{p}=\frac{r}{s}$ for some integers $r$ and
$s$ with $s$ not zero. Then $s^2 p=r^2=q_1 \ldots q_n$ for some primes $q_i$ by the fundamental theorem of
arithmetic. Note that $r^2$ and $s^2$ have an even number of prime factors by expanding squares. But then
$s^2 p$ has an odd number of prime factors. This is a contradiction. Then $\sqrt{p}$ is irrational.


Real Analysis: Function Continuity Let $A, B \subset \mathbb{R}$, let $f : A \rightarrow B$, and let $x_0 \in A$. Then $f$ is continuous at $x_0$ if and only if $\lim_{x \rightarrow x_0} f(x) = f(x_0)$.

I proved the first part

$f$ is continuous at $x_o$ if $\lim_{x \rightarrow x_0} f(x) = f(x_0)$

using "$f$ is continuous at $x_o$ in $A$ ($ dom(f)$) if, for every sequence {${x_n}$} in  $A$ ($ dom(f)$) converging to $x_0$, we have $\lim_n f(x_n) = f(x_0)$"

Moreover, I tried proving the second part, which is

$\lim_{x \rightarrow x_0} f(x) = f(x_0)$ if $f$ is continuous at $x_o$

using the other definition of continuity that involves $\delta > 0$ and $\epsilon >0$, which eventually proves that $\lim_n f(x_n) = f(x_0)$.

My question is, is it correct proving the second part of the statement using aforementioned definition? or are there any other ways to prove the second part?

To prove the second part, suppose that it is not true. Then for some sequence $\{x_n\}$ converging to $x_0$, it is not true that $f(x)$ converges to $f(x_0)$, since otherwise for every sequence $\{x_n\}$ converging to $x_0$, $f(x)$ converges to $f(x_0)$, then $\lim _{x \to x_0} f(x)=f(x_0)$. Therefore $f$ is not continuous at $x_0$, which is a contradiction.

There is nothing to prove since it is just a definition that $f(x)$ is continuous at $x=x_0 \iff \lim_{x\to x_0} f(x)=f(x_0)$.

What we need to prove, in general, is that an assigned function $f(x)$ is continuous at a point according to the given definition.

Refer for example to the related

Continuous function proof by definition


$S^2$ and a point are not homotopy equivalent Is there a way to show that $S^2$ does not have the same homotopy type of a point without using homology groups?

If $S^2$ were homotopy equivalent to a point, then every vector bundle over $S^2$ would be trivial.  But, by the hairy ball theorem, the tangent bundle of $S^2$ is nontrivial.

The hairy ball theorem can be proven via Brouwer's fixed point theorem for maps $f:D^2\rightarrow D^2$, which doesn't require the knowledge of homology groups of $S^2$.

Note that if $S^2\simeq \{*\}$, then every map $f\colon X\rightarrow S^2$ is homotopic to the constant map. This is clearly not the case though for the identity map $\mbox{id}\colon S^2\rightarrow S^2$ and so it can not be the case that the sphere $S^2$ and the point $\{*\}$ are homotopy equivalent.

I should add that the identity map on the sphere is only clearly not homotopic to the constant map when using intuition. To prove this rigorously, I would think that using the degree of the map is the simplest way, but then this is essentially a homology argument. It might be possible to prove that if $f$ is homotopic to the identity map then $f$ is surjective without using a homology argument. Whether this then fits your criterion is debatable.


How to find the probability? John invites 12 people to a dinner party, half of which are men. Exactly one man and one woman are bringing desserts. If one person from this group is selected at random,what is the probability that it is a woman, or a man who is not bringing a dessert?

There are $6+5=11$ persons of the $12$ that fall in one of the classes: 1) women 2) men that do not bring a desert. So the probability is $\frac{11}{12}$.

(If the first class is meant to be: women that do not bring a desert, then there are $5+5=10$ persons and the probability is $\frac{10}{12}$)

I could be wrong, but I would have said that 10 people didn't bring deserts so we are looking at a 1/10 probability?


Value of integral $\int \frac{(1-x)^{n+1}}{x} dx$ I was working on a series, and I came up with integral, an indefinte form
$$\int \frac{(1-x)^{n+1}}{x} dx$$
I was wondering about how to solve this indefinte integral analytically. I solved it on Mathematica and got the result

so, I was thinking how to approach the problem and solve it.After seeing the answer,I got no clue about how to approach it.
Also, Mathematica always doesn't give the neat answer. Can anyone approach the problem step wise and give a more simple answer that a undergraduate can work with.

Please correct me if I've gone wrong but here goes..
$$ \int \frac{\left(1-x\right) \left(1-x\right)^n}{x}dx = \int \frac{\left(1-x\right)^{n+1}}{x}dx$$
$$ \int\frac{\left(1-x\right)^n}{x}dx - \int\left(1-x\right)^ndx$$
$$ \int\frac{\left(1-x\right)^n}{x}dx + \frac{\left(1-x\right)^{n+1}}{n+1} + k1$$
$$ lnx=t $$
$$dx = xdt$$
$$x= e^t$$
$$ \int\frac{\left(1-x\right)^n}{x}dx= \int(1- e^t)^ndt $$
$$\int(1- e^t)^ndt = \int[ 1 - {n \choose 1}e^t+ {n \choose 2}e^{2t}- {n \choose 3}e^{3t}+....+\left(-1\right)^n{n \choose n}e^{nt}]dt$$
$$= lnx- {n \choose 1}x+ \frac{{n \choose 2}x^2}{2}-\frac{ {n \choose 3}x^3}{3}+....+\frac{\left(-1\right)^n{n \choose n}x^n}{n}+k2$$
Hence,
$$\int \frac{\left(1-x\right)^{n+1}}{x}dx= lnx- {n \choose 1}x+ \frac{{n \choose 2}x^2}{2}-\frac{ {n \choose 3}x^3}{3}+....+\frac{\left(-1\right)^n{n \choose n}x^n}{n}+ \frac{\left(1-x\right)^{n+1}}{n+1}+ k$$

The best way to do it, is to expand the formula
\begin{equation}
(1-x)^{n+1}=\sum_{k=0}^{n+1} \binom{n+1}{k} (-1)^k x^k
\end{equation}
Then
\begin{equation}
\int \frac{(1-x)^{n+1}}{x} dx = \sum_{k=0}^{n+1} \binom{n+1}{k} (-1)^k \int x^{k-1} dx
\end{equation}
Finnaly you can intergrate each term.
\begin{equation}
\int \frac{(1-x)^{n+1}}{x} dx = ln(x)-(n+1)x+\sum_{k=2}^{n+1} \binom{n+1}{k} (-1)^k \frac{x^{k}}{k-1}
\end{equation}


What is the probability that a randomly chosen positive three-digit integer is a multiple of $7$? What is the probability that a randomly chosen positive
three-digit integer is a multiple of $7$?

Is my answer right?:

$$\frac{100}{7} = 14 , \qquad \frac{999}{7} = 142$$

Then there are $142 - 14 = 128$ numbers that are multiples of $7$.

Then the probability is: $\frac{128}{900}$.

Yes, except that:

I would replace $\frac{100}{7}$ with $\frac{99}{7}$ (to see why this is important consider the analogous question asking for how many multiples of $7$ there are in $\{700, \ldots, 999\}$), and
the quotients aren't quite right as written, but we can repair them with floor notation, e.g., $\lfloor\frac{999}{7}\rfloor = 142$.

Find the lower bound and upper bound that is a multiple of 7! so for a three digit integer that is a multiple of 7, 100 is the lowest possible three digit integer and 999 is the highest, however, neither of these are a multiple of 7!
So starting at 100 find a multiple of 7! (105) is the product of (15*7), that is your lower bound. Now find your upper bound,starting at 999; you find, (994) which is a product of (142*7)! Now use a THEOREM (IF m and n are integers, and m <= n,then there are n-m +1 integers from m to n, inclusive). This means take the integer 142 from the product of 142*7 from your upper bound, and subtract that from your lower bound product which is the integer 15 from the product 15*7. You will get your event (E) so E = 142 - 15 + 1 = 128. So E = 128, that is how many three digit integers are a multiple of 7! So now take your E and put that over your sample space (S) where S = all possible outcomes in a random outcomes. so (s) = 9 * 10 * 10 where the first digit of your three digit number is a natural number between 1-9, the second digit can be any natural number from 0 to 9, and the third digit number can also be a natural number between 0 and 9. So you get (S) = 9 * 10 * 10 = 900.  So E/S gives you your ratio or probability of number of three digit integers that are a multiple of 7.  You can use this for any three digit multiple, or four digit multiple ... (N) multiple where N is a non negative  member of Z.


$1+n!=m^{2}$ for some n,m$\in\mathbb{N}$ I have no idea whether this is known or not and I couldn't find anything related on Google. While I was studying , I come up with this idea $1+n!=m^{2} $ for some $n,m\in\mathbb{N}$

$1+4!=5^{2}$

$1+5!=11^{2}$

$1+?!=?^{2}$

and the question is what is the next number? Wolfram Alpha gave me this interesting graph:

Thanks in advance for your interest.

$$4!+1=5^2\\5!+1=11^2\\7!+1=71^2$$

This problem is known as Brocard's Problem, and pairs of integers which satisfy it are known as Brown Numbers. There are no other known solutions up to $10^9!+1$.

$m^2-n!-1=0\Rightarrow m=\pm\sqrt{(n!)^2+1}$ this should help you to find $n$ since you need $(n!)^2+1$ to be a square  I.e. $(n!)^2+1=a^2$ for some $a\in(2k-1)\mathbb{N}$ (odds)


Biased random walk away from cliff A drunken probabilist stands $n$ steps from a cliff's edge. He takes random steps, either towards or away from the cliff, each step independent of the past. At any point, the probability of taking a step away is $2/3$, or a step toward, $1/3$. What are his chances of eventually falling off the cliff?

The textbook has no answer. This is how I did it:

Let $p(i)$ represent the probability of death if you start $i$ steps from the cliff. I first start with the case where $i=1$. Clearly, $$p(1)=\frac{1}{3} + \frac{2}{3}p(2)$$ and $p(2)=(p(1))^2$ so we have the equation $$2(p(1))^2 + 1 - 3(p(1))=0$$ which we can easily solve.

Then to get the case where $i=n$ simply raise $p(1)$ to the $n^\text{th}$ power.

Is this correct? It seems a bit counterintuitive, since it seems as if no matter how far away the drunkard gets, sooner or later some fluke of chance will send him walking straight and steady down the cliff, with probability $1$.

A start. Let $p_1,p_2,p_3,\dots$ be the probabilities of (ultimate) disaster if we start $1,2,3,\dots$ paces from the cliff edge. Then, as you wrote, $p(1)=\frac{1}{3}+\frac{2}{3}p_2$.

Similarly, if $k\ge 2$, then
$$p_k=\frac{1}{3}p_{k-1}+\frac{2}{3}p_{k+1}.\tag{$1$}$$
You can also fit the first equation into this pattern, by defining $p_0$ to be $1$.

Equation $(1)$ is a linear second-order recurrence. You can solve it by general techniques for such recurrences, either generating functions or by working with the characteristic equation. You can also solve the recurrence by ad-hoc techniques. For Equation $(1)$ can be rewritten as
$$\frac{2}{3}(p_{k+1}-p_k)=\frac{1}{3}(p_k-p_{k-1}).$$
So if $a_k=p_{k+1}-p_k$, then $a_k$ satisfies a very simple recurrence, for which you can find an explicit solution.

I am not quite sure what you are doing above, but here's how I did it.
The person will fall off if he takes n+1 steps forward. Let's start with looking at the cases by which the man can take n+1 steps forward.

Takes all n+1 steps forward. Probability = (1/3)^(n+1)
Takes n+2 steps forward and 1 step backward. Probability = (1/3)^(n+2)*(2/3)
Takes n+3 steps forward and 2 steps backward. Probability = (1/3)^(n+3)*(2/3)^2
Takes n+4 steps forward and 3 steps backward. Probability = (1/3)^(n+4)*(2/3)^3
...

So the probability that the man jumps off will be the sum of this infinite series.

Which comes out to (9/7)*(1/3)^(n+1)


Integrate $\int x e^{x} \sin x dx$ Evaluate:
$$\int x e^{x} \sin x dx$$

Have you ever come across such an integral? I have no idea how to start with the calculation.

Solution without complex numbers:

Let $I=\int e^x x\sin xdx$ Integrating by parts:

$$I=e^x x\sin x-\int e^x x\cos xdx-\int e^x\sin xdx$$

Then one more time by parts:

$$\int e^x x\cos xdx=e^x x\cos x + I-\int e^x\cos xdx$$

So:

$$2I=e^x x\sin x-e^x x\cos x+\int e^x(\cos x-\sin x)dx$$

Now (by parts again or by direct observation):

$$\int e^x(\cos x-\sin x)dx=e^x \cos x$$

So:

$$I=\frac{e^x x\sin x-e^x x\cos x+e^x \cos x}{2}$$

By parts,

$$\int te^tdt=(t-1)e^t.$$

Then by the change of variable $t=(1+i)x$,

$$\int xe^x\sin x\,dx=\Im\left(\frac1{(1+i)^2}((1+i)x-1)e^{(1+i)x}\right)=\Im\left(\left(\frac{1-i}2x+\frac  i2\right)e^{(1+i)x}\right).$$


Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?


Recurrence relation for Pell's equation $x^2-2y^2=1$ I am wondering how to find the recurrence relation for solutions for $x$ in the Pell's equation $x^2-2y^2=1$.

I know the formula for the general term.
It is $$\frac{(3+2\sqrt2)^n+(3-2\sqrt2)^n}{2}$$ for $x_n$, the $n^{th}$ smallest solution for $x$.
Any help would be appreciated! Thanks!

I got a feeling that the recurrence formula is $x_n=6x_{n-1}-x_{n-2}$, but I wonder how to prove this relation true/false and how to derive/generate the recurrence relation.
Note that $x_{-1}=1$ and this recurrence formula applies to all nonnegative integral $n$.

If $u_n=A\alpha^n+B\beta^n$ then $\alpha$ and $\beta$ are the roots of the quadratic equation $$(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta=0$$

Let the quadratic be $p(x)=x^2-px+q$, then consider $$0=A\alpha^np(\alpha)+B\beta^np(\beta)=$$$$=(A\alpha^{n+2}+B\beta^{n+2})-p(A\alpha^{n+1}+B\beta^{n+1})+q(A\alpha^{n}+B\beta^{n})=u_{n+2}-pu_{n+1}+qu_n$$from which $$u_{n+2}=pu_{n+1}-qu_n$$Where $p=\alpha+\beta$ and $q=\alpha\beta$

You simply need to identify $\alpha$ and $\beta$ - the surrounding constants drop out in the arithmetic.

Once two successive terms are known, the constants $A$ and $B$ are determined, and there is a unique recurrence of this kind because two consecutive terms determine the rest through the recurrence.

You can find the general recurrence relationship here:

https://crypto.stanford.edu/pbc/notes/contfrac/pell.html


Solving the integral of $sin^n(x)$ I want to delete this question. The results I had trouble with while proving them are false. Furthermore, the question was marked as duplicate by (at least) five users.

For odd $n$ you can write the integral as

$$\int \sin^{2k+1} x\; \mathrm dx = \int \sin^{2k} x \sin x\;\mathrm dx = - \int (1-t^2)^{k} \;\mathrm dt$$ and then use change of variable $t = \cos x$

For even $n=2k\geq2$ use integration by part to write
$$
I=\int\sin^{2k}x\,dx=-\sin^{2k-1}x\cos x+(2k-1)\int\sin^{2k-2}\cos^2x\,dx
$$
and the basic trig relation $\sin^2x+\cos^2x=1$ allows to write the last expression as
$$
I=\sin^{2k-1}x\cos x-(2k-1)\int\sin^{2k}x\,dx+(2k-1)\int\sin^{2k-2}x\,dx
$$
i.e.
$$
I=\frac1{2k}\left[\sin^{2k-1}x\cos x+(2k-1)\int\sin^{2k-2}x\,dx\right]
$$
Repeat the procedure $k-1$ more times.


contour integrating 1/z over unit square equal zero? After I have studied contour integration and related theorems for some time, I encounter a quite confusing problem:

integrating $\frac{1}{z}$ about a unit square centered at the origin(positive orientation).

Let $F(z)$ be the antiderivative of $\frac{1}{z}$, seperating the contour into 4 straight lines, then

$$\int_\Gamma \frac{1}{z}dz = (F(0.5+0.5i) - F(0.5-0.5i)) + (F(-0.5+0.5i) - F(0.5+0.5i)) +... = 0$$

I expect the answer to be $2\pi i$ however that results in zero. What mistake did I make?

p.s. please do not use residue theorem or any theorems to directly show that the integral equal $2\pi i$. I would like to see some sort of direct integration. Thanks.

Hint: $F(z)=log(z)$ has a branch cut. See the Wikipedia page on complex logarithm.

Observe the domain (i mean simply connected region) where the branch of logarithm fn is analytic.Now see if you take closed curve which is homotopic to a circle in that simply connected region.And then you do integral and you find it will be 2 *pi *i *n where n is a positive integer,it is the winding number,it depends no. of times of rotation on the circle or the closed curve.


Infinite dimensional intermediate subfields of an algebraic extension of an algebraic number field Let $K$ be an algebraic number field, i.e. a finite extension of $\mathbb{Q}$.
  Let $L/K$ be an infinite dimensional algebraic extension of $K$.
  Are there infinitely many infinite dimensional intermediate subfields of $L/K$?

It's easy to see that there are infinitely many finite dimensional intermediate subfileds of $L/K$, but I have no idea about the question.

Not necessarily. Clearly it is possible for there to be infinitely many subfields of infinite degree, take $\overline{\mathbf{Q}}/\mathbf{Q}$ for example. On the other hand, there exist infinite extensions of any number field $K$ with no intermediate fields of infinite degree over $K$.

For a concrete example, if $L = K_\infty/K$ is a $\mathbf{Z}_p$-extension (i.e. its Galois group is isomorphic to $\mathbf{Z}_p$, the $p$-adic integers). Such an extension always exists, e.g. as a finite index subfield of $K(\mu_{p^\infty})$, the field given by adjoining all $p$-power roots of unity. Since $\mathbf{Z}_p$ has no nontrivial closed subgroups of infinite index, the fundamental theorem of Galois theory implies that there are no intermediate fields of infinite degree over $K$.

Consider the set $P$ of prime numbers. The number of infinite subsets of $P$ it self is infinite, in fact uncountable. For each such infinite subset $A\subset P$
consider the extension of $K_A$ obtained by adjoining all square roots $\surd p,\ p\in A$. It can be shown that they are all distinct subfields. %K% itself may contain atmost a finite number of square roots of $p$, but it will not affect. (For example, to start with we can redefine $P$ to the infinite set of primes such that $\surd p$ is not in $K$).


Algebra I: Cyclic Generators The direct product $\mathbb{Z}_{45} \times \mathbb{Z}_{98}$ is cyclic and isomorphic to $\mathbb{Z}_{4410}$ because $gcd(45,98)=1$; furthermore the element $n=([1]_{45},[1]_{98})$ is a cyclic generator. Find the smallest positive integer $m$ such that $mn=([29]_{45},[17]_{98})$


Uniform random distribution on a unit disk a) A point is uniformly chosen in the unit disk $0 ≤ x^2 + y^2 ≤ 1$. Find the probability that its
  distance from the origin is less than $r$, for $0 ≤ r ≤ 1$.

  b) Compute its expected distance from the origin.

  c)Let the coordinates of the point be $(X, Y )$. Determine the marginal p.d.f. of $X$. Are $X$ and $Y$ independent?

I did the part a) using geometry that the area of the circle with a radius of $r$ is divided by the area of the unit circle, such that $$P(R\leq r)=\frac{\pi r^2}{\pi\cdot 1^2}=r^2$$

Part b) is attempted to solve by differentiating the cdf, such that $$f(r)=\frac{d}{dr}r^2=2r,\hspace{3mm} E(R)=\int_{0}^1r\times 2r\,dr=\frac{2}{3}.$$

But this result does not seem to be right... And I do not know about part c)

A simulation was done using python to visualize this distribution,
and it gives image like this.

Why are the points concentrated near the central area?
from scipy.stats import uniform
import matplotlib.pyplot as plt
import math
r = uniform.rvs(scale =1,size=5000)
pi = 3.14159265359
theta = uniform.rvs(scale =2*pi,size=5000)
x = []
y = []
for i in range (5000):
    x.append(r[i]*math.cos(theta[i]) )
    y.append(r[i]*math.sin(theta[i]) )
fig=plt.figure()
ax=fig.add_axes([0,0,2,3])
ax.scatter(x, y)```

Your results for parts a) and b) are correct.

For part c), note that $Y\le\sqrt{1-X^2}$. Thus, knowing $X$ reduces the possible range of $Y$, so the two variables can’t be independent.

To find the marginal distribution of $X$ you can use $P(x\le X\le X+\mathrm dx)=\frac1\pi\left(2\sqrt{1-x^2}\right)\mathrm dx$, since the density in the circle is uniformly $\frac1\pi$ and an infinitesimal strip of width $\mathrm dx$ at $x$ has area $\left(2\sqrt{1-x^2}\right)\mathrm dx$. Thus $f_X(x)=\frac2\pi\sqrt{1-x^2}$. (Incidentally, this tells you that $\int_{-1}^1\sqrt{1-x^2}=\frac\pi2$.)

The probability a point is closer to the origin than $r$ is the area of a disk of radius $r$ to that of the disk of radius $1$.  That is, $\frac{\pi r^2}{ \pi 1^2} = r^2$.
The expected distance is $\int\limits_{s=0}^r s (2 \pi s)\ ds$
as can be easily seen:

$p(y\mid x) = {\cal U}[-\sqrt{1-x^2}, +\sqrt{1-x^2}]$

as can be easily seen here:

And clearly $X$ and $Y$ are not independent.  If $X=.99$, then $Y$ must be small...


If $X \sim \mathsf{Unif}(0,1)$, find the CDF of $Y = \ln\left(\frac{X}{1-X}\right)$ Assume the random variable $X$ follows distribution $\mathsf{Unif}(0,1)$ and define the new random variable $Y = \ln\left(\frac{X}{1-X}\right)$.
Find CDF of $Y.$

My attempt.

$F(X \leq x) = \int_{0}^{x} \ln(\frac{t}{1-t})\,dt = \ln(1-x) + x\left(\frac{x}{1-x}\right).$

Thus,

$$
F(x) =\begin{cases}
      0 & x < 0 \\
      \ln(1-x) + x\left(\frac{x}{1-x}\right) & 0\leq x
\leq 1 \\
      1 & x > 1
\end{cases}
$$

Would this be right?

$\mathbb{P}(Y\leq y)=\mathbb{P}(\log(\frac{X}{1-X})\leq y)=\mathbb{P}(\frac{X}{1-X}\leq e^y)=\mathbb{P}(X\leq\frac{e^y}{1+e^y})=\frac{e^y}{1+e^y}$

since $0<\frac{e^y}{1+e^y}<1$

The cdf of $Y$ is $F_Y(y) = P(Y \le y)$. I know you want to integrate something, but you can't yet because you don't even know the pdf of $Y$. You have to rewrite in the form:

$$P(Y \le y) = P(X \le h(y))$$

Then you can proceed as follows

$$P(X \le h(y)) = \int_{-\infty}^{h(y)} 1_{(0,1)}(x) dx = \int_{-\infty}^{h(y)} 1_{(0,1)}(x) dx$$

$$ = \int_{\mathbb R} 1_{(-\infty,h(y))}(x) 1_{(0,1)}(x) dx = \int_{\mathbb R} 1_{(-\infty,h(y)) \cap (0,1)}(x) dx$$

And I think you know how to do the intersection based on what you tried, but I'll spell it out for you anyhoo:

$$(-\infty,h(y)) \cap (0,1) = \emptyset, h(y)\le 0$$

$$(-\infty,h(y)) \cap (0,1) = (0,h(y)), h(y) \in (0,1)$$

$$(-\infty,h(y)) \cap (0,1) = (0,1), h(y) \ge 1$$

Now for the $h(y)$:

$$F_Y(y) = {P}(Y\le y)={P}(\ln(\frac{X}{1-X})\le y)={P}(\frac{X}{1-X}\le e^y)={P}(X\le\frac{e^y}{1+e^y})$$

$$=\int_{\mathbb R} 1_{(-\infty,h(y)) \cap (0,1)}(x) dx, h(y)=\frac{e^y}{1+e^y} \in (0,1)$$

$$\therefore, F_Y(y) = {P}(Y\le y) = \int_{\mathbb R} 1_{(0,\frac{e^y}{1+e^y})}(x) dx=\frac{e^y}{1+e^y}$$

Observe that all our results hold for $y \in \mathbb R$.


I need to show that this sequence is increasing and I'm almost there but I need help on last step. Let $(1+\frac{1}{n})^n$ be a sequence and $f(x)=(1+\frac{1}{x})^x $ on $[1,inf)$.  I need to show that f is non-decreasing by showing that $f'(x)\ge0$.  So far I have:

Let $g(x)=ln(f(x))$, where $ln$ means the natural log.  So $g'(x)=ln(1+\frac{1}{x})-\frac{1}{x+1}$.  Thus $g'(x)\ge0$ iff $ln(1+\frac{1}{x})\ge\frac{1}{x+1}$.  Which is equivalent to showing $\int_1 ^{1+1/x}\frac{1}{t}dt\ge\frac{1}{x+1}$.  Then I said let P be the trivial partition $({1,1+\frac{1}{x}})$.  So by definition of Riemann integrate, $L(P,1/t)=\frac{1}{x+1}$.

This is where I get stuck.  What would be my next step.

I am not looking for a different way to show this, this is how my teacher wants us to do this problem.

The interval $I=[1,1+1/x]$ has length $1/x$. At its lowest (on $I$), the function $t\mapsto 1/t$ takes on the value $\frac{1}{1+1/x}=\frac{x}{x+1}$, hence
$$\int_{1}^{1+1/x}\frac{dt}{t}\geq \frac{1}{x}\times\frac{x}{x+1}=\frac{1}{x+1}$$


Strategies to guess choices for multiple choice questions Multiple choice questions (MCQs) are common in examinations over here in Singapore. A set of, say, $40$ questions are given to students, and each is accompanied with a list of $4$ choices of answer, $A, B, C$ or $D$.

  Suppose a student did not study for a particular test. Out of desperation, he decides to "guess" a choice for each question to secure at least some points. He considers two possible strategies:

  For each question, he picks a random choice as his answer.
  He picks a random choice, say, $B$ and uses it for all his answer.

  Assuming the correct answers are randomly distributed and his guess is completely random, which strategy would give a higher probability of securing more correct answers than the other?

Intuitively, the second strategy would give a higher probability of securing more marks. However, I am unable to come up with a mathematical proof (or disproof).

For the second strategy, we only need consider the probability that the correct answer is the choice that was picked. Under the assumption that the correct answers are randomly distributed, the probability of a particular choice being the correct answer is $\frac{1}{4}$. Hence, the student would get approximately $25\%$ of his guesses correct.

However, we can also use the same argument for the first strategy to say that the student would also get approximately $25\%$ of his guesses correct. This would imply that both strategies are equally effective, but I am pretty sure the second strategy is more effective.

EDIT: In order to prevent psychological factors from distorting reality,  I decided to write a program (in C#) that simulates the aforementioned MCQ tests. I configured the program to simulate the taking of $1000$ randomly generated MCQ tests with $40$ questions each using both strategies.

It turns out that the percentage scores for both strategies have the same average ($\approx25\%$) and the same standard deviation ($\approx 6.84$)!

Code:

static Random rng = new Random((int)DateTime.Now.Ticks);

static void GenerateAnswers(int[] answers)
{
    for (int i = 0; i < answers.Length; i++)
    {
        answers[i] = rng.Next(4);
    }
}

static int Strategy1(int[] answers)
{
    int score = 0;
    foreach (int answer in answers)
    {
        if (rng.Next(4) == answer)
        {
            score++;
        }
    }
    return score;
}

static int Strategy2(int[] answers)
{
    int choice = rng.Next(4);
    return answers.Count(x => x == choice);
}

Both will give an the same probability of securing marks. (Both will give you an expected marks of 25%.)

The key thing to note is that each question is independent of another.  This means that if the answer to the first question is B, then it does not make the answer to the second question more likely to be B (or any other choice, for that matter).

Thus, you can consider each question separately: the 25% chance of getting the first question correct does not affect the 25% chance that you'll get the second question correct, so whatever you choose for the first question isn't going to make a certain answer for the next question more correct than another.

Note: However, psychology of test setters might come into play, where test setters unknowingly avoid having consecutive answers that are both C, or other biases, and this might affect the resultant probability.

Suppose four unprepared students sit for the test, say $W,X,Y,Z$.  They decide that $Z$ will always guess $A$, $Y$ will always guess $B$, $X$ will always guess $C$, and $W$ will always guess $D$.

On any given test, it is unlikely that all four students will get the same score.  At random one of them, say $Y$, may get slightly more than 25%, while another, say $W$, will get slightly less than 25%.  Afterwards $Y$ will go on math.stackexchange with intuition that always picking $B$ is the winning strategy, because it leads to good results.  But also $W$ will go on math.stackexchange with intuition that picking at random is a better strategy, because always picking $D$ leads to bad results.

OP is $Y$.


Show that the function is not continuous using the sequence definition. Consider $f$, given by $f(x,y)$= $xy^2/(x^2+y^4)$ when $(x,y)\ne(0,0)$ and $f(0,0)=0$.

Show that $f$ is not continuous at $(0,0)$ using the sequence definition. (which is: a function is continuous if any sequence of points $(X_n,Y_n)$ in $R^2$ with $\lim X_n=0$  and  $\lim Y_n=0$  then  $\lim f(X_n,Y_n)=f(0,0)=0$.

****Sorry, I don't know how to type piecewise functions****


Equivalent definitions of equivalence between linear codes, why is it? I'm studying coding theory and we're facing the problem to tell when two given linear codes $C$ and $C'$ over $\mathbb F_q^n$ (some others call $\mathbb F_q^n$ as $GF(q)$ just in case you don't know one of the two notations) and this is our definition:
$\textbf{DEF}$: Let $C, C'\subseteq \mathbb F_q^n$ be two linear codes (i.e. vector subspaces of $\mathbb F_q^n$), then you say that $C$ and $C'$ are equivalent, and write that $C \approx C'$ if given $G_1$ and $G_2$, generator matrices of $C$ and $C'$ respectively, you can obtain $G_2$ starting from $G_1$ executing a finite number of the following operations:

Swapping two any columns in $G_1$
Replacing a column in $G_1$ with itself multiplied by a non-zero scalar $\lambda\in \mathbb F_q\setminus \{0_{\mathbb F_q}\}$

$----------------------------------$
Next we're showing (or trying to, thing that I don't understand) that this definition is equivalent to this second one:
$\textbf{DEF}$: Let $C, C'\subseteq \mathbb F_q^n$ be two linear codes, then you say that $C$ and $C'$ are equivalent, and write that $C \approx C'$ if given $G_1$ and $G_2$, generator matrices of $C$ and $C'$ respectively, $\exists\:P\in GP_N(\mathbb F_q)\;\lvert\; G_2 = G_1\cdot P$
$----------------------------------$
Here above $GP_N(\mathbb F_q)$ represents the group of generalized permutation matrices
Our teacher argumentates it this way: he says that the two aforementioned operations are the restriction to the columns of G of global operations over $\mathbb F_q^n$, i.e

$\forall \sigma\in S_N\;\exists\;\phi_\sigma\:\colon \mathbb F_q^N\mapsto \mathbb F_q^N$ mapping $(x_1, x_2, \dots, x_N)$ to $(x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, \dots, x_{\sigma^{-1}(N)})$ with $\phi_\sigma$ being a linear isomorphism. ($\leftarrow$ our teacher says this corresponds to swapping columns. Why????)
$$\mathbb T_{\mathbb F_q}\colon=\bigg\{\begin{bmatrix}
    \lambda_{1} & & \\
    & \ddots & \\
    & & \lambda_{N}
  \end{bmatrix}\;\bigg\lvert\; \lambda_i\in\mathbb F_q\setminus \{0_{\mathbb F_q}\} \bigg\}\subseteq GL_N(\mathbb F_q)$$Even on this, how is this an operation? It makes more sense if I think of getting a vector $\underline{v}$ and computing $\underline{v}\cdot T$ with $T\in\mathbb T_{\mathbb F_q}$ which will give me the vector $(\lambda_1 v_1, \lambda_2 v_2, \dots, \lambda_N v_N)$ and therefore if I do it for a matrix $G$ with "fitting" dimensions I'll get the first column of $G$ multiplied by $\lambda_1$, the second by $\lambda_2$ and so on... am I correct? And in that case, is $N=n$?

Sorry for not being too short about it, my biggest problem is with the permutations part, we never worked with it in our classes (I attend computer engineering) except for counting how many permutations exists for a set of $N$ elements and how many of them have a given amount of fixed points... this is confusing me a lot. We don't have an algebra course and we're doing it in coding theory.

Your question 1: any permutation of columns can be obtained by repeatedly swapping columns. This is because the permutation group $S_{N}$ can be generated by involutions. So your professor should not be saying that this given permutation of columns represents a swapping of columns, but that it represents a series of column swaps.
Your question about $\mathbb{T}_{\mathbb{F}_{q}}$ I don't entirely understand. Yes this maps a vector $(v_{1},\ldots, v_{N}) \mapsto (\lambda_{1}v_{1}, \ldots, \lambda_{N}v_{N})$, so each entry has been multiplied by a scalar. If you want to do this same operation to all of the vectors, this corresponds to multiplying each column of the generator matrix by the appropriate scalar.
I would say the two statements you give about the generator matrices are clearly equivalent statements, but that this definition of code equivalence is not good. For example over $\mathbb{F}_{3}$,
$$\begin{align*}G_{1} &= \begin{bmatrix} 1 & 1 & 1\\0&1&-1 \end{bmatrix}
& G_{2} &= \begin{bmatrix} -1 & -1 & -1\\0&1&-1\end{bmatrix}\end{align*}$$
define the exact same code, but the codes would not be considered equivalent under your definition because the rules don't allow for row operations (which do not change the row space).
A proper definiton of code equivalence should also allow the elementary row operations, and then the final equivalence should be $C_{1} \sim C_{2}$ if there exists $A \in \mathrm{GL}_{k}(\mathbb{F})$, $P \in \mathrm{GP}_{N}(\mathbb{F})$ with $G_{2} = AG_{1}P$.

This is better; I don't think of SPLAG as an introductory text, but he makes the corret distinction. Not the full linear group:

This is mysterious Wikipedia repeats the business about just two operations but then describes the correct things on the matrices.
I have one or two books that include codes, I will need to check. The books may or may not include arbitrary finite fields...
A few pages in Lattices and Codes by Wolgang Ebeling. I have the second edition. He quite definitely defines generator matrix in terms of a basis of the linear subspace. Thus, despite confusing wording, the third operation is permitted,  replacing a column by its sum with a different column.
Maybe someone here has taught codes...


Circle revolutions rolling around another circle I just watched this video, and I'm a bit perplexed.

Problem:

The radius of Circle A is 1/3 the radius of Circle B.

Circle A rolls around Circle B one trip back to its starting point.

How many times will Circle A revolve in total?

The intuitive answer is 3, but the correct answer is 4. I understand the trick -- that the center of Circle A must travel a distance of $2\pi(r_B + r_A)$, not $2\pi r_B$ -- but I'm still confused on one item.

At the risk of sounding very un-mathematical, how do the (infinite set of) points on the circumference of each circle map to each other to accomplish this?

Consider Circle A rolling along a straight line the length of the circumference of Circle B. Then it will revolve 3 times. It's like the universe "knows" when to apply a different point mapping when you change the arrangement of matter.

Thank you everyone for your answers. They were all informative, but I wasn't able to intuitively understand them until I saw this visual:
https://www.geogebra.org/m/v3a437ux
The key for me was to see that there are two kinds of revolution happening:

Revolutions of Circle A with respect to Circle B, and
Revolutions of Circle A with respect to the (overhead) observer.

To the observer, the revolutions of the type (2) complete before revolutions of type (1). The revolutions of type (1) happen at the $r_B/r_A$ roots of unity, and the revolutions of type (2) happen at the $(r_B/r_A + 1)$ roots of unity. It was very helpful to see precisely where these revolutions occur, because it was mentally impossible to unify the two types of revolution without the visual. It's also useful to notice that at any given moment, with respect to the observer, the points of Circle A on the far side from Circle B are moving faster than the points of Circle A on the side that is touching Circle B.
David K's answer is great for understanding that the parametric mapping of time to point pairs is the same whether Circle A is rolling along a straight line or a circle, and that we are dealing with frames of reference. I simply didn't believe the mapping was the same until I saw the visual.
zoli and Hans Lundmark's answers are great for understanding that one extra revolution must occur at some point along the rolling path. The complete answer, of course, is that this last mysterious revolution does not happen all at once, but gradually along the entire roll.
The visual was discovered in the comments on this page, which is another discussion of the same problem:
https://plus.maths.org/content/circles-rolling-circles

I don't know why people overthink these simple problems. The circumference of the larger circle is 3x the circumference of the smaller circle. Provided the small circle is not slipping, it will take 3 turns to cover the distance around the larger circle. Nothing magical here.

The reason you were perplexed by the video is because the uploader makes a mistake at 1:34 to 1:42. He counts this as a full revolution of the smaller circle, which it clearly is not. If he marked both circles with points on their circumferences, you'd see that it would take 3 revolutions. It's simply conservation of length. The rolling motion is completely irrelevant to the stated problem.

By the way, the applet you linked to was for a 4:1 scale factor, this is why in the applet it takes four revolutions.


Sum and average length of chords From a specific point A on a circle's circumference, am drawing various chords.
a) what is the sum of length of such chords and
b) what is the average length of such chords?

Assuming a radius r, I'm trying to solve a) in the below manner:

One specific chord length = 2rCos(Theta)

To get the sum, I'm integrating over -pi/2 to pi/2 ( 2rCos(Theta) dTheta )

But i don't think this is right. Please point me in the correct direction.

Thanks.

There are many possible answers, depending on the probability distribution of the parameter that defines the other endpoint of the chord. That is a standard issue in problems of geometric probability.

We give one interpretation, and derive the answer under that interpretation. Other interpretations will yield a different answer.

Imagine that we select the other endpoint $B$ by choosing $\theta$ with uniform distribution in the interval $[0,2\pi]$, and letting $B$ be the point obtained by rotating the point $A$ counterclockwise through the angle $\theta$.

Then the length $X$ of the chord $AB$ is, by basic trigonometry, $2r\sin(\theta/2)$. It follows that
$$E(X)=\int_0^{2\pi} 2r\sin(\theta/2)\frac{1}{2\pi}\,d\theta.$$
Integrate. We get
$$E(X)=\frac{4r}{\pi}.$$

If you imagine many chords of a circle arranged in a stack, with appropriate lengths, it can be made into a circle. so, the total of all the chords will give us the area of the circle. This can be easily proved and imagined. For example, take a rectangle. Have you ever wondered why the area is l x b?? It is because we are adding the length breadth number of times. Hence applying the same logic to a circle, we can understand that if we add the lengths of all the chords in a circle we would get the circle's area..pi*R^2.
The total number of chords would be equal to 2*R as the length of the circle is 2*R.
Hence the average chord length = (pi*R^2)/(2*R).
=> pi*r/2


Maximization of Mutual Information Let $X\in\{0,1\}^d$ be a Boolean vector and $Y, Z\in\{0,1\}$ are Boolean variables. Assume that there is a joint distribution $\mathcal{D}$ over $Y, Z$ and we'd like to find a joint distribution $\mathcal{D}'$ over $X, Y, Z$ such that:

The marginal of $\mathcal{D}'$ on $Y, Z$ equals $\mathcal{D}$.
$X$ are independent of $Z$ under $\mathcal{D}'$, i.e., $I(X;Z) = 0$.
$I(X; Y)$ is maximized,

where $I(\cdot;\cdot)$ denotes the mutual information. For now I don't even know what is a nontrivial upper bound of $I(X;Y)$ given that $I(X;Z) = 0$? Furthermore, is it possible we can know the optimal distribution $\mathcal{D}'$ that achieves the upper bound?

My conjecture is that the upper bound of $I(X;Y)$ should have something to do with the correlation (coupling?) between $Y$ and $Z$, so ideally it should contain something related to that.

We have the following series of inequalities:
\begin{align}
I(X;Y) & \le I(X;Y,Z) \\
& = I(X;Z) + I(X;Y|Z) \\
& = I(X;Y|Z) \\
& = I(X,Z;Y) - I(Y;Z) \\
& \le H(Y) - I(Y;Z) \\
& \le 1 \text{ bit} - I(Y;Z) ,
\end{align}
where in the third line we've used that $I(X;Z)=0$.

I start with the bound
$$1\geq H(Y)=I(X; Y)+I(Y; Z|X)$$
$$I(X; Y)\leq1-I(Y; Z|X)\leq1$$
Now it is obvious that $Z\perp Y, X$ and $Y=X_1$ uniformly distributed over $\{0, 1\}$ does the job


Is "Real Analysis with Economic Applications", by Efe A. Ok, a good book for a self-teaching beginner in real analysis? I'm interested in self teaching myself Real Analysis with the primary purpose of using it for economics, for this I decided to use Real analysis with economic applications by Efe A. Ok

Now what I was wondering was if this was a good book for somebody without a Maths background to use to self study a thing, as I'm primary interested in it for the methods used in economic theory.

Am I making a mistake reading this now or do I need more background in Maths to read this particular book? As I'm not finding it the most easy thing to read and I'm not sure if I'm understanding everything in it correctly, 50 pages in and I've already got quite a headache! Which sucks.

$(1)$ Analysis I Terence Tao

$(2)$ A Basic Course in Real Analysis (Ajit Kumar and S. Kumaresan)

Let me know if you need soft-copies of these books.


Plucker bilinear product identity In a certain textbook chapter dealing with Plucker product representations of lines in homogeneous coordinates, a certain bilinear form is introduced

$$ (L|\hat{L})=l_{12}\hat{l}_{34}+\hat{l}_{12}l_{34}+ l_{13}\hat{l}_{42}+\hat{l}_{13}l_{42}+ l_{14}\hat{l}_{23}+\hat{l}_{14}l_{23}$$

where $(l_{12}, l_{13},l_{14}, l_{23},l_{42}, l_{34},)$ are the 6 Plucker coordinates of line $L$ and similarly for $\hat{L}$.

It is then claimed, without proof, that if $L$ is the intersection two planes, represented by 4x1 homogeneous vectors P and Q, and  $\hat{L}$ is the join of two points A and B,also represented by 4x1 vectors, then

$$(L|\hat{L})=(P^TA)(Q^TB) - (Q^TA)(P^TB) $$

Can someone give or point me to a proof of the above?

The equation for the bilinear product given in the Hartley and Zisserman text (based on 4 points) was the determinant of a 4x4 matrix, where each column of the matrix represented a point.

First we'll show that the equation given in the text (based on 2 points and 2 planes) is equal to the distance from point B to the plane formed by point A and the line of intersection of planes P and Q, multiplied by the length of the normal vector to the plane formed by A / P / Q.

Second, we'll show that, if each column of a matrix is considered to be a point, the determinant of the matrix is equal to the distance of one of the points to the plane composed of the other points, multiplied by the length of the normal vector to the plane formed by the other points.

Step 1:

line L* = $PQ^T - QP^T$

plane of L* B = $PQ^TB - QP^TB$

distance from A to plane

= $(PQ^TB - QP^TB)^TA / $length of plane normal

= $(B^TQP^TA - B^TPQ^TA) / $length of plane normal

= $((Q^TB)(P^TA) - (P^TB)(Q^TA)) / $length of plane normal

$((Q^TB)(P^TA) - (P^TB)(Q^TA)) = $ (distance from A to plane) * (length of plane normal)

Step 2:

I worked through the algebra for three points in 2D space, it should be similar for four points in 3D space.  For points P0, P1, and P2:

determinant = $x0(y1 - y2) + x1(y2 - y0) + x2(y0 - y1)$

distance from P0 to line P1=>P2 = $(x0(y1 - y2) + x1(y2 - y0) + x2(y0 - y1)) / \sqrt((y2 - y1)^2 + (x2 - x1)^2)$

determinant = (distance from P0 to line P1=>P2) * $\sqrt((y2 - y1)^2 + (x2 - x1)^2)$

$\sqrt((y2 - y1)^2 + (x2 - x1)^2)$ is the length of the line normal.


Sum of two squares question Which of the following statements is true for every value of $n$?
A: If $n$ is not a sum of two squares, then neither is 69$n$
B: If $n$ is a sum of two squares, then so is 34$n$
C: If $n$ is not a sum of two squares, then neither is 15$n$
D: If $n$ is a sum of two squares, then so is 14$n$

The sum of two squares theorem tells us that an integer $n$ is representable as the sum of squares of two integers if and only if the prime decomposition of $n$ contains no term $p^k$ where $p$ is a prime of the form $p\equiv 3\pmod{4}$ with $k$ an odd integer.

A: if $n$ is not a sum of two squares, then neither is $69n$

False.  Consider $n=69$ in which case $n$ is not representable as the sum of two squares since it's prime decomposition is of the form $69=3\cdot 23$.  However, $69n=69^2=0^2+69^2$ is.
Note, it is possible however that $n$ and $69n$ are both not representable, for example when $n=7$

B: if $n$ is a sum of two squares, then so is $34n$

True.  $n$ being a sum of two squares implies that there is no term $p^k$ where $p$ is of the form $p\equiv 3\pmod{4}$ with $k$ odd in $n$'s prime decomposition.  Multiplying $n$ by $34=2\cdot 17$ does not change this since neither $2$ nor $17$ are primes of the form $p\equiv 3\pmod{4}$.  Thus, $34n$ is also representable as a sum of two squares.
Equivalently, $34 = 5^2+3^2$.  So, if $n=a^2+b^2$ you have $34n = (5^2+3^2)(a^2+b^2)=(5a+3b)^2+(5b-3a)^2$ by the Brahmagupta-Fibonacci identity

C: if $n$ is not a sum of two squares, then neither is $15n$

False.  The same counterexample as before works.  Let $n=15$.  Then $15n = 15^2+0^2$ is a sum of squares.

D: if $n$ is the sum of two squares, then so is $14n$

False.  $n$ being the sum of two squares implies that in the prime decomposition of $n$ you have either $7$ not appearing at all or $7^k$ appears with $k$ even.  $14n$ will then have $7^{k+1}$ appearing with $k$ even and thus is not representable as the sum of two squares.
In fact, we can recognize that if $n$ is the sum of two squares, then $14n$ is never the sum of two squares.

A is wrong: $69$ is not a sum of two squares but $69*69$ is.
B is true because $34= 25+9$.
C is wrong ( same as A)
D is wrong for the same reason as A.


Find the constant K in probability distribution? Find the constant K?
  I know how to find K, if there is only one "k" in the question and the rest are given number. I know that all the x value are add upto 1.

But in this question, I don't know how to start.

  One thing I know is $\Sigma(x)=1(k)+2(2k)+3(3k)+4(4k)+5(5k)$
  Does this help to find the constant k?

Please help me "Find the constant k"
Appreciate your help.

You are confused. It's pretty fine. Please note that $X=i$ means the event that X equals i. So, let me interpret the table for you.

  There's a probability of $k$ that event $X=1$ shall happen.
  There's a probability of $2k$ that event $X=2$ shall happen.
  There's a probability of $3k$ that event $X=3$ shall happen.
  There's a probability of $4k$ that event $X=4$ shall happen.
  There's a probability of $5k$ that event $X=5$ shall happen.
  No other events are possible.

Important: By Unitarity of probability, the probability of $Pr({\text{$X$ = 1,2,3,4 or 5}})=1$.

$X=1, X=2, X=3, X=4, X=5$ are five mutually exclusive events in the sample space, and there are no other events. So, $$Pr(X=1,2,3,4,5)=\sum_{i=1}^5Pr(X=i)=k+2k+3k+4k+5k = 15k$$

So, what's $k$?

$k+2k+3k+k=1$
(combine terms)
$7k=1$
(move 7 to other side)
$k=\frac{1}{7}$


Show $A$ is a Lebesgue Measurable set $\iff$ there exists $F_\sigma$ set such that $G_\delta$ set $m(G \setminus A) = m(A \setminus F) = 0$ Show $A \subseteq \mathbb{R}$ is a Lebesgue Measurable set $\iff$ there exists $F_\sigma$ set $F \subseteq \mathbb{R}$ contained in $A$ and $G_\delta$ set $G \subseteq \mathbb{R}$ containing $A$ s.t. $m(G \setminus A) = m(A \setminus F) = 0$

Note: Up to now, I have proven inner and outer regularity of $m$.  Definitions below:

Inner Regular: If $A$ is Lebesgue Measurable, then $m(A)=sup\{m(K) | K$ is an closed set contained in $A \}$

Outer Regular: If $A$ is Lebesgue Measurable, then $m(A)=inf\{m(U) | U$ is an open set containing $A \}$

Different authors define Lebesgue measurability differently. One common definition is that $A \subset \mathbb R$ is measurable if and only if for every $\epsilon > 0$ there exists an open set $G \supset A$ satisfying $m^*(G \setminus A) < \epsilon$ where $m^*$ is the Lebesgue outer measure. A simple consequence of this definition is that any set $E$ with $m^*(E) = 0$ is measurable.

Suppose $A$ is measurable. For every $k \ge 1$ there exists $G_k \supset A$ so that $m^*(G_k \setminus A) < \frac 1k$.  Define $G = \cap G_k$. Then $G$ is a $G_\delta$ set, $G \subset G_k$ for all $k$ and
$$m^*(G \setminus A) \le m^*(G_k \setminus A) < \frac 1k$$ for all $k$, forcing $m^*(G \setminus A) = 0$.

Likewise, $A$ measurable implies $\mathbb R \setminus A$ measurable, so by the above reasoning there exists a $G_\delta$ set $H \supset \mathbb R \setminus A$ with $m^*(H \setminus (\mathbb R \setminus A)) = 0$.  If you let $F = \mathbb R \setminus H$ you find that $F$ is an $F_\sigma$ set, $F \subset A$, and $$m^*(A \setminus F) = m^*(H \setminus (\mathbb R \setminus A)) = 0.$$

If $m(A)<\infty$ we have the following:
For each $j\in \mathbb{N}$ there exist by inner and outer regularity a compact set  $K_j\subset A$ and an open set $U_j\supset A$ s.t
$$
m(U_j)+1/j \leq m(A)\leq m(K_j)-1/j.
$$

Put $G=\cap_{j}A_j$ and $F=\cup_j(K_j)$. Then $G$ and $F$ are the desired $G_\delta$ and $F_{\sigma}$ sets respectively.

If $m(A)=\infty$, then decompose $A$ as $A=\cup_j (A\cap(j,j+1])$ then apply the above exposed to each $A_j=A\cap(j,j+1]$.

On the other hand if there exists  a $G$ a $G_\delta$ set satisfying $m(G\setminus A)=0$ and  $m(A\setminus F)=0$,  we have for  each test set $E$

$$
m(E)=m(E\cap G)+m(E\cap G^c)\geq m(E\cap A)+m(E\cap A^c\cap(G\setminus A)^c)\geq m(E\cap A)+m(E\cap A^c)
$$


Probability of team winning in a single-elimination tournament In a tournament of football (or whatever sport you prefer), let's say we have $2^n$ teams and for all $2^{n-1}(2^n-1)$ possible matches we know the probability of one team winning against the other. The tournament is a single-elimination tournament (therefore having $n$ rounds) and the teams are to be randomly allocated for the first round. How do we go about computing the probability of team $x$ winning the tournament?

For each team $i$ the probability of getting into the second round is
$$P_2(i)=\frac{1}{2^n-1}\sum_{j\ne i}P_v(i,j)$$
where $P_v(i,j)$ is the probability that $i$ will defeat $j$.

Similarly, the probability that $i$ will get into the round $r+1$ is
$$P_{r+1}(i)=P_r(i)\frac{1}{2^{n-r+1}-1}\sum_{j\ne i}P_v(i,j)P_r(j)$$

So, all you need to do is compute $P_n(i)$ recursively.

@sds's answer is not correct. Use an all equal skilled 4 team tournament for example. Any team has a win chance of 0.25. But his formula gives 0.75.

Should be:

$$
P_2(i) = \sum_{j!=i}\frac{1}{2^{n}-1} P_v(i,j)
$$

$$
P_{r+1}(i) = \sum_{j!=i}\frac{2^{r-1}}{2^{n}-1} P_r(i,!j)P_v(i,j)P_r(j)
$$

$P_r(i,!j)$ is the probability that $i$ reached round $r$ without meeting $j$ in the previous matches.

To calculate $P_r(i,!j)$, use recursion again:

$$
P_2(i,!j) = \frac{2^{n}-2}{2^{n}-1}\sum_{k!=i,k!=j}\frac{1}{2^{n}-2} P_v(i,k)
$$

$$
P_{r+1}(i,!j) = \frac{2^{n}-2^{r}}{2^{n}-1}\sum_{k!=i, k!=j}\frac{2^{r-1}}{2^{n}-2} P_r(i,!j)P_v(i,k)P_r(k)
$$