Real Number Continuum Why real number has number of elements equal to power set of natural number??

I can't think of an intuitive answer for this.
Please suggest some insight to this.

Intuitively speaking, you can describe a bijection as follows. Warning: some cheating is involved.

First, identify a real $x$ in $[0,1]$ with its expansion base $2$, $x = 0.x_1x_2\dots$. For instance, $0 = 0.000\dots$, $1/2 = 0.100\dots$, $1 = 0.111\dots$, and so on. (It's not quite honest what I did, because some numbers have more than one binary expansion - but trust me that it won't change much).

Next, a sequence $x_1,x_2,\dots$ of $0$s and $1$s can be identified with a set of natural numbers. Simply look at the set $A = \{ n \in \mathbb{N} \ | \ x_n = 1 \}$. You can convince yourself that this correspondence is indeed a bijection between all sequences and all sets.

Finally, combining the two steps, you find a bijection between $[0,1]$ and $\mathbb{N}$. If you want to replace $[0,1]$ by $\mathbb{R}$, just find any bijection between the two (hint: think of $\tan x$).

Because $\mathbb{R}$ contains irrational numbers (which are uncountable) and $P(\mathbb{N})$ contains an infinite number of combinations of numbers.


Simple question about the heat equation (differential equations) This is a question from my textbook.

The question is to find the temperature with the following information:
f(x) = 1
u(0,t) = 0
u(L,t) = 0
u(x,t) = f(x)

The equation itself:

The solution from chegg:

From the circled line,
How did they get An = 0 from

A simple explanation would do. Thank you.

We have that $\displaystyle \int_0^{π}\cos(x)\,dx=\int_0^{π}\cos(2x)\,dx.....=\int_0^{π}\cos(nx)\,dx=\boxed{0}$


Understanding Dirichlet Process (vs Dirichlet Distribution) I'm studying the Dirichlet process but I'm confused.

I think I can visualize it, through the Chinese Restaurant Example or The Pólya urn scheme.
But I cannot see the connection with Dirichlet Distribution.

Particularly on Wikipedia regarding it, it's stated that

  The Dirichlet process can also be seen as the infinite-dimensional
  generalization of the Dirichlet distribution.

How can it be? Can you provide me more insight regarding this sentence?

One definition of the Dirichlet process is that given a probability space, $(X, A, G)$ and an arbitrary partition of $X$ given by $A_1, \dots A_k$ a distribution $G$ is a Dirichlet process with probability measure $G_0$ and mass parameter $M$ if we have $$G(A_1, \dots, G(A_k)) \sim Dir(MG_0(A_1), \dots MG_0(A_k)$$

This can be found from an article here. This explains the definition in terms of the Dirichlet distribution and where the Dirichlet name comes from.

Now then what you do is that you verify some conditions, and then apply the Kolmogorov Extension Theorem, which gives you the desired infinite-dimensional distribution from the earlier collection of finite-dimensional distributions.

EDIT: (In simpler language)

Take all subsets of the original probability space, the Dirichlet process is a distribution where any group of subsets follow the Dirichlet distribution. Now what you have is a collection of finite-dimensional distributions. To get an infinite dimensional distributions from this, you have to use the Kolmogorov extension theorem. So you check a  few properties to check that the theorem holds, and what it states is that your definition from finite dimensional distributions determines an infinite-dimensional distribution uniquely.

Add to what has been contributed to the question:

Is Dirichlet process a Dirichlet distribution?

No. A random sample from a Dirichlet distribution of order $3$ will have the format like $(0.3, 0.2, 0.5)$ with three non-negative elements add up to $1$ and similarly a random sample from a Dirichlet distribution of order $4$ will have the format like $(0.15, 0.05, 0.6, 0.2)$.  A random sample from a Dirichlet process, say $G$ can be anything and it depends on the set $X$, $G$ is defined. A random sample can be a float like $0.4$, a tuple $(32, 3, 1, 387.232)$, "rainy" or "sunny", or "basketball" and "football". The connection is that given any (measurable) partition of $X$ noted as $B_1,\ldots,B_K$, $G$ as a random probability, $G(B_1),\ldots,G(B_K)$ follows a Dirichlet distribution. I'm not sure it's a good idea to say a Dirichlet process is a Dirichlet distribution of infinite dimension, but $K$ can be any positive number, like $1$, $12345$, $10^{23}$, or $9999999999$.

How does the "process" term come into play?

The random samples already observed from a Dirichlet process will more likely to appear again. Strictly speaking, $G_0$ is the base probability measure of Dirichlet process $G$, $G(B_1),\ldots,G(B_K) \sim$ Dirichlet distribution of order $K (MG_0(B_1),\ldots, MG_0(B_k))$ with $M$ as a scale parameter (in most literature denoted as $\alpha$ or $\mu$) and random samples/observations $x_1,x_2,\ldots,x_m$ are generated from $G$, then the conditional distribution of $G$ given $x_1,\ldots,x_m$ is a Dirichlet process with base probability $MG_0$ plus the summation of delta functions from $x_1$ to $x_m$, $G(B_1),\ldots,G(B_K) \sim$ Dirichlet distribution of order $K (MG_0(B_1) +$ # of $x_i$'s in $B_1$, $MG_0(B_2) +$ # of $x_i$'s in $B_2,\ldots, MG_0(B_K) +$ # of $x_i$'s in $B_K$). You can still get random samples other than $x_1,\ldots,x_m$, but the new random sample is likely to be a repetition of existing random samples.

How does a Dirichlet process connect to a Chinese restaurant process?

$M$ random samples $x_1,\ldots,x_m$ are generated from Dirichlet process $G$ and consider the samples having the same value are customers sitting at the same table in a Chinese restaurant. The next random sample generated from $G$ is more likely to be the more frequent observation in $x_1,\ldots,x_m,$ i.e. the next customer entering the restaurant is more likely to join the most crowded table.


Evaluate $\int_0^\infty\frac{\ln x}{1+x^2}dx$ Evaluate $$\int_0^\infty\frac{\ln x}{1+x^2}\ dx$$
I don't know where to start with this so either the full evaluation or any hints or pushes in the right direction would be appreciated.  Thanks.

Hint

Write $$\int_0^\infty\frac{\ln x}{(1+x^2)}dx=\int_0^1\frac{\ln x}{(1+x^2)}dx+\int_1^\infty\frac{\ln x}{(1+x^2)}dx$$ For the second integral make a change of variable $x=\frac{1}{y}$ and see the beauty of the result.

I am sure that you can take from here.

Another general approach :

Consider
$$
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx.\tag1
$$
Rewrite $(1)$ as
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac1{a^2}\int_0^\infty\frac{x^{b-1}}{1+\left(\frac{x}{a}\right)^2}\ dx.\tag2
\end{align}
Putting $x=ay\;\color{blue}{\Rightarrow}\;dx=a\ dy$ yields
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=a^{b-2}\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy,
\end{align}
where
\begin{align}
\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy=\frac{\pi}{2\sin\left(\frac{b\pi}{2}\right)}.
\end{align}
Hence
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac\pi2\cdot\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}.\tag3
\end{align}
Differentiating $(3)$ with respect to $b$ and setting $b=1$ yields
\begin{align}
\int_0^\infty\frac{\partial}{\partial b}\left[\frac{x^{b-1}}{a^2+x^2}\right]_{b=1}\ dx&=\frac\pi2\frac{\partial}{\partial b}\left[\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}\right]_{b=1}\\
\int_0^\infty\frac{\ln x}{x^2+a^2}\ dx&=\large\color{blue}{\frac{\pi\ln a}{2a}}.
\end{align}
Thus
$$
\int_0^\infty\frac{\ln x}{x^2+1}\ dx=\large\color{blue}{0}.
$$


Find $\lim_{x\rightarrow 0} \frac{\cos x - 1}{x}$ I'm trying to find the following limit:

$$\lim_{x\rightarrow 0} \frac{\cos x - 1}{x}$$

I tried to use squeeze theorem but it's not making much sense. I did the following:

$$\begin{align}
&\lim_{x\rightarrow 0} \frac{\cos x - 1}{x} \\
-1 \le &\lim_{x\rightarrow 0} (\cos x)(x^{-1}) \le 1 \\
\lim_{x\rightarrow 0} -x^{-1} \le &\lim_{x\rightarrow 0} \cos x \le \lim_{x\rightarrow 0} x^{-1} \quad \text{*}
\end{align}$$

The last line is where I'm confused. I don't think I'm doing squeeze theorem correctly. I'm guessing you have to manipulate $\cos x - 1$ somehow. Please provide some hints.

Thanks a bunch!

P.S You cannot use L'Hopital.

It is sufficient to use the asymptotic development of $\cos x$ around $0$, namely

$$\cos x = 1 - \frac{x^2}{2} + o(x^3).$$


Are curves in a level set continuous? Wikipedia defines a level set as

  a level set of a real-valued function of $n$ real variables $f$ is a set of the form $$L_c(f) = \left\{ (x_1, \cdots, x_n) \, \mid \, f(x_1, \cdots, x_n) = c \right\}$$

I often seen level sets drawn as curves in $\Bbb{R}^2$ like this

My Question:

Why do we assume a level set is continuous? Why couldn't it be just a collection of random points all with the same value instead of curves?

If $f$ is continuous, then the level set $\{f=c\}$ has to be closed. That's about all you can say: Given an arbitrary closed subset $E$ of $\mathbb {R}^n,$ there is $f\in C^\infty(\mathbb {R}^n)$ such that $\{f=0\} =E.$ To get nice smooth things as level sets, there is usually an added assumption hanging around, such as $\nabla f\ne 0$ on $E.$

In general, yes, the level set is just a bunch of points. But for most functions you will see, the level set is a continuous loop. It doesn't have to be a loop or line, for instance the level set of the zero function at zero is the entire domain, which could be the entire plane. The reason that the picture you gave is the usual visualization for a level set (of a function with a 2D domain) is that it commonly arises that way from a 'well-behaved" function, for instance one that is continuous. The set of level curves is then a nice way of visualizing the gradient function or topographical maps, which only work for continuous multi-variable functions.


Do $a^{1/n}$ and $n$th root of a mean the same thing? I encountered a question in which I had to find $x$ and the question had $3^{1/x}$ and I got $x=1/2$ and $1/4$ as its solution. But my textbook said that "$x$th root of $3$" is valid for $x\geq 2$ and $x$ should be a natural number.
But the question did not directly mention $x$th root 3 and instead "$3^{1/x}$". Do they actually mean the same? Do they have same restrictions on $x$?
PS: The actual question was "Solve for real values of $x$: log $4$ + ($1$+ $1/2x$) log $3$= log ($3^{1/x}$ + $27$)"

In general if $b > 0; b\ne 1$ then $b^x; x \in \mathbb R$ has a rather complicated definition based on calculus which I won't actually go into detail what it is but it is all based on the fundamental relation $b^{x+y} = b^xb^y$.
But there are some basic concepts that fall into generalizations.
$b^0 = 1$ (it has to if $b^x = b^{0+x} = b^0b^x$)
And if $n\in \mathbb N$ then $b^n = \underbrace{b\cdot ..... \cdot b}$.
And if $x = \frac 1n; n\in \mathbb N$ then, yes, $b^{\frac 1n}$ will indeed have to be then $n$th root of $b$.  They are the same thing.
(It has to be..... If $b^{x+y} = b^xb^y$ then $(b^{\frac 1n})^n = \underbrace{b^{\frac 1n } \cdot ...\cdot b^{\frac 1n}} = b^{\frac 1n + \frac 1n + ....+\frac 1n} = b^1 = b$.)
So IF we assume $x$ is a natural number then $3^{\frac 1x}$ will be the $x$th root of $3$.
But you haven't actually told us what the question was or why we should assume $x$ is a natural number.
If $x$ is a real number we do have $(b^{\frac 1x})^x = b^{\frac xx} = b^1 = b$.  So whatever $b^{\frac 1x}$ is, if we raise it to the $x$ power we will get $b$.  We can casually refer to that as "the $x$th root of $b$" but if $x$ is not a natural number that might no be clear what that means. After all $3^{\frac 1{\pi}} = 3^{0.318....} \approx 1.4186....$ and $(3^{\frac 1{\pi}})^\pi =(1.4186.....)^{3.1415....} = 3$.  So we could so $3^{\frac 1\pi}$ is the $\pi$th root of $3$ but... that's not usually what we mean by "root".  Which usually means a root is a solution to a polynomial.  And as a polynomial only has natural numbers as powers as solution to $x^\pi = 3$ isn't usually called a "root".... at least, I wouldn't call it that.
But... FWIW..... The number $3^{\frac 1x}$ will always be a solution to "$w^x = 3$ solve for $w$".

Well, in general they mean the same.
but it mostly depends on who is asking you the question, so they might have one of the following differences:

$\root 2 \of 4$ means both $-1$ and $1$ while $4^{\frac{1}{2}}$ means just the principle root, $2$.
it might hint that negative numbers are to be used as exponents as the expression $\root -1 \of {2} = \frac{1}{2}$ is practically never used.
alot less popular but some writers define $\root 0 \of n = \begin{cases}
1,  & n>0 \\
\text{undefined}, & \text{otherwise}
\end{cases}$
The best option is to ask the professor or lecturer.


How to write the negation of a biconditional? I have two statements $p$ and $q$ such that:
Statement $p$: Ravi reads Mathematics
Statement $q$: Ravi reads Chemistry
Now, I am required to write the negation of $p \iff q$, i.e., I need to write $\lnot (p \iff q)$, which, if I understand correctly, is, by definition: $(p \land \lnot  q) \lor(\lnot p \land q )$
So, according to me, it should be as follows:
Ravi reads Mathematics and not Chemistry or Ravi doesn't read Mathematics and reads Chemistry.
But, the book states the answer to be as follows:
Ravi reads neither Mathematics nor Chemistry
Are the two statements equivalent? Or am I going wrong here?

Check question 3 (e) for the question asked here

The answer for question 3 (e)

You are exactly right, and your book is wrong.
Some equivalent ways to write $\lnot (p \iff q)$ are
$$
\begin{split}
\lnot (p \iff q) &\equiv \lnot(p \implies q) \lor \lnot(q \implies p) \\
&\equiv (p \land \lnot q) \lor (\lnot p \land q) \\
&\equiv p \oplus q.
\end{split}
$$
(The last expression uses the XOR ("exclusive or") operator, $\oplus$.)
However you choose to write it, it is equivalent to what you said:
"Ravi reads Mathematics and not Chemistry or Ravi doesn't read Mathematics and reads Chemistry.
"
OP, don't be discouraged by the rude or confusing answers in this thread. Trust in your logic!

A biconditional is implication that goes both ways. An implication is false exactly when $p$ is true and $q$ is false. Therefore it does not imply $q$. Additionally. The reverse implication could also be false, that $q$ is true and $p$ is false. Either one will work and both aren't required to render the biconditional false.
Simpler is that Ravis does not read both. I think you maybe are confused about the logical variables, and possibilities of true or false, they can take on, and then the variables as well as logical connectives.
edit the negation of the logical formula is one thing, but in an actual application of it, you would also have to negate the atomic sentences too, I think?
The author may be making a direct or possible indirect example of semantic truth vs syntatic truth, but we don't know because we don't know exactly what your book said.
We got a result in the book and so consider how it can be true, is literally what you've asked, not how to negate formula of biconditional.
$p,q=1$. That is what the semantics of a language might mean ultimately. So then the negation is clearly $p,q=0$. It's either a book typo or semantic truth.


Show and prove that the sequence $u_n = \frac{n^2+n-2}{4n^2-5}$ converges to $1/4$ Show and prove that the sequence $u_n = \frac{n^2+n-2}{4n^2-5}$ converges to $1/4$

Working:

To show the sequence converges to $1/4$, note that $\frac{n^2+n-2}{4n^2-5} = \frac{1+1/n -2/n^2}{4-5/n^2}$, and using the Algebraic Limit Theorem yields a limit of $1/4$.

The problem I am having is with the proof using the formal definition:

I feel like my approach is way too long-winded, can someone please confirm whether it's correct and whether there's a quicker method? Anyhow, so I did:

We need to show for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that whenever $n \ge N$, it follows $|\frac{n^2+n-2}{4n^2-5} - 1/4| < \epsilon$

Going backwards first: note that $|\frac{n^2+n-2}{4n^2-5} - 1/4| < \epsilon \implies|\frac{4n-3}{4(4n^2-5)}| < \epsilon$. Now pick $n$ large enough so that $\frac{4n-3}{4(4n^2-5)} > 0$, hence $\frac{4n-3}{4(4n^2-5)} < \epsilon \implies 0 < n^2 - \frac{1}{4\epsilon}n + \frac{3}{16\epsilon}-\frac{5}{4}$

$\implies 0 < \left[n - \left(\frac{1+\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon} \right) \right]\left[n -  \left(\frac{1-\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon} \right)  \right]$ so,

$n > \frac{1+\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon}$ or $n < \frac{1-\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon}$.

Now pick $N > \frac{1+\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon}$, then whenever $n \ge N > \frac{1+\sqrt{80\epsilon^2-12\epsilon+1}}{8\epsilon}$, we have $(8\epsilon n -1)^2 > 80\epsilon^2 - 12\epsilon + 1 \implies \epsilon > \frac{4n-3}{4(4n^2-5)} \implies |\frac{n^2+n-2}{4n^2-5} - 1/4| < \epsilon$, as required.

I'm pretty sure I did the algebra right, just want to confirm whether all my arguments are correct and if there's a quicker method.

EDIT: Following on from @David's post, is this solution correct?

Need to show $\lim\left( \frac{n^2+n-2}{4n^2-5} \right) = 1/4$, note:

$|\frac{n^2+n-2}{4n^2-5} - \frac{1}{4} | = |\frac{4n-3}{4(4n^2-5)}|$. If $n \ge 2$, then $\frac{4n-3}{4(4n^2-5)} > 0$, so $|\frac{4n-3}{4(4n^2-5)}| = \frac{4n-3}{4(4n^2-5)}$. If $n^2 > 5$ (or more conveniently assume $n \ge 3$), then $\frac{4n-3}{4(4n^2-5)} < \frac{4n}{4(3n^2)} = \frac{1}{3n} < \epsilon$.

Now pick $N_1 \in \mathbb{N}$ such that $N_1 > \frac{1}{3\epsilon}$ and set $N = \max\{3, N_1\}$. Let $\epsilon > 0$ and suppose $n \ge N$, then $|\frac{n^2+n-2}{4n^2-5} - \frac{1}{4} | = |\frac{4n-3}{4(4n^2-5)}| = \frac{4n-3}{4(4n^2-5)} < \frac{4n}{4(3n^2)} = \frac{1}{3n} < \epsilon$, as required.

It is nearly always far easier to simplify as much as you can before choosing $N$.

When I say "simplify", you don't need to simplify by finding
$$\hbox{expression}=\hbox{simpler expression}.$$
You are trying to prove an inequality, not an equality, so it is good enough to find
$$\hbox{expression}<\hbox{simpler expression}.$$
You have to be a bit careful though, as if the "simpler expression" is lots bigger than the "expression" it could change the whole problem.  I would also recommend not saying "$<\varepsilon$" at the beginning but leaving it to the end.

There is a bit of a difficulty in your problem so first I will do a simpler one: show from the definition that
$$\lim_{n\to\infty}\frac{n^2+n-2}{4n^2+5}=\frac{1}{4}\ .$$
In this case we could work as follows: assume that $n>N$, where $N$ is yet to be chosen.  Then we have
$$\eqalign{\left|\frac{n^2+n-2}{4n^2+5}-\frac{1}{4}\right|
  &=\left|\frac{4n-13}{4(4n^2+5)}\right|\cr
  &=\frac{4n-13}{4(4n^2+5)}\cr}$$
provided $n\ge4$.  Now we can make a fraction bigger by increasing the numerator and/or decreasing the denominator.  Since $4n-13<4n$ and $4n^2+5>4n^2$, we have
$$\eqalign{\frac{4n-13}{4(4n^2+5)}
  &<\frac{4n}{4(4n^2)}\cr
  &=\frac{1}{4n}\cr
  &<\frac{1}{4N}\ ,\cr}$$
and this will be less than $\varepsilon$ provided $N>1/(4\varepsilon)$.  So the proof would look something like this.

Let $\varepsilon>0$.
Choose $\displaystyle N=\max\Bigl(4,\,\frac{1}{4\varepsilon}\Bigr)$.
Suppose $n>N$.  Then
$$\left|\frac{n^2+n-2}{4n^2+5}-\frac{1}{4}\right|=\cdots\
  \hbox{[working]}\ \cdots<\varepsilon\ .$$

I hope you will agree that this makes the algebra much easier than in your solution.

Your problem can be done in much the same way.  The extra difficulty is that where we said
$$4n^2+5>4n^2\ ,$$
you will now need to say
$$4n^2-5>4n^2\ ,$$
and of course this is not true.  Instead we can do something like this: if $n^2>5$ then
$$4n^2-5>3n^2\ ;$$
now the working will go through much as before, though the details will be slightly different.

See how you go from here.  Good luck!

so i think other answer same :
$n=\frac 1m$
$m\to  0$
$$\frac{\frac 1{m^2}+\frac 1m -2}{\frac 4{m^2}-5}=
\frac{1+m-2\cdot m^2}{4-5\cdot m^2}$$


Show the improper integral $\int_0^\infty \frac{\sin(x^3-ax)}{x}dx$ converges. This is an exercise for the Dirichlet test for the convergence of improper integrals. This exercise is from the book called “A Course of Modern Analysis”. It does not specify what $a$, but I am thinking $a$ is a nonnegative number. I am not sure if this assumption is legit. Clearly, $\frac{1}{x}$ converges to 0 as $x \to \infty$. I am struggle to show $|\int_0^L \sin(x^3-ax)dx|$ for all $L$ is bounded. I tried to use u-sub but it does not make the problem easier. Any hint will be appreciated.

Hint. Note that $$\left|\frac{\sin (x^3-ax)}{x}\right|\le \frac1x$$ for all $x>0.$


The existence of concatenation functions in Godel Numbering? I know that there are many schema of Gödel Numbering, and each has its own method of Concatenation, n★m.
But is there a general proof that shows 'For every Gödel Numbering scheme there exists a Concatenation Function'?

it is more the other way around without contentation no Godel numbering.

Godelnumbering is just a translation from formulas and proofs to numbers and from the Godelnumbers back to the same  formulas and proofs.

I think in principle any translation system that can do this can be used as godelnumbering system.

Be aware:

The numbering system  also needs to be able to translate proofs to numbers and back
all translations need to be lossless and exact
it must all be done in the axioms of arithmetic.

So it is not only translating a formula to an number , it is much more

Hope this helps


Support of the density of a transformation of a random variable $Y = \frac{1}{X}$ with $f_X(x) = \frac{1}{\pi(1+x^2)}$ Suppose we are asked to show that if $X$ is a random variable with density ($\forall x \in \mathbb{R}$) $$f_X(x) = \frac{1}{\pi(1+x^2)} ,$$
then $Y = \frac{1}{X}$ has the same distribution as $X$ ($Y \sim X$).  I am well aware of the approach to take here:  indeed, we can compute the density of $Y$ by first calculating the cumulative distribution of $Y$, $F_Y$, in terms of the cumulative distribution of $X$, $F_X$.

$$\begin{align} F_Y(y) = \mathbb{P}(Y \leq y)  &= \mathbb{P}\left(\frac{1}{X} \leq y\right) \\ &=^{[\text{if } y \neq 0]} \mathbb{P}\left(X \geq \frac{1}{y}\right) \\ &= 1 - \mathbb{P}\left(X \leq \frac{1}{y}\right) \\ &= 1 - F_X \left(\frac{1}{y}\right). \end{align}$$

Continuing with the computation, we can take the derivative with respect to $y$ to obtain the density of $Y$.  To this end, we have

$$f_Y(y) = \frac{\text{d}}{\text{d}y}F_Y(y) = -f_X \left(\frac{1}{y}\right) \cdot \left(-\frac{1}{y^2}\right) = \frac{1}{y^2} f_X \left(\frac{1}{y}\right) = \frac{1}{y^2}\frac{1}{\pi\left(1 + \frac{1}{y^2}\right)} = \frac{1}{\pi(1+y^2)},$$
for all $y \neq 0.$  My question is in regards to this last restriction which comes from the third equality in the first set of equations.  The distributions surely cannot be the same if one holds for all $x$ and the other holds for all $y \neq 0$.  Am I missing something here?  It's a small but important detail for me.  I appreciate any input, thank you.

Technically, density is defined through integration (rather than differentiation). Given a distribution function $F$, any non-negative Lebesgue measurable function $f$ such that

$$F(x) = \int_{-\infty}^x f(t) \, dt, x \in \mathbb{R} \tag{1}$$

is called a density function of $F$ (with respect to Lebesgue measure).  For a more general definition under a measure context,
see page $213$ of Probability and Measure by Patrick Billingsley.

Therefore, technically speaking, given any distribution function $F$, its density is an equivalent class of functions: as long as $f_1$ and $f_2$ agree outside a set of Lebesgue measure $0$, and both of them satisfy $(1)$ and non-negativity, both of them can be referred to as density of $F$.

Regarding to your case, what really matters is whether

$$\int_{-\infty}^z f_X(z) \, dz = \int_{-\infty}^z f_Y(z) \, dz.$$

If the above equation can be verified, then $X$ and $Y$ will have the same distribution.

$\displaystyle \dfrac 1 X \le y$ would be equivalent to $\displaystyle X \ge \dfrac 1 y$ if $X$ were always positive, but it has probability $1/2$ of being negative.  If $y>0$ then you have
$$
\Pr\left( \frac 1 X \le y \right) = \Pr\left( X <0 \right) + \Pr\left( X \ge \frac 1 y \right).
$$
If $y<0$ then you have
$$
\Pr\left( \frac 1 X \le y\right) = \Pr\left( \frac 1 y \le X < 0 \right).
$$

As to your claim that the distributions of $X$ and $1/X$ cannot be the same, do you have an argument for that?  In fact they are the same.  Here's a geometric argument for that.  A fixture rotating at a uniform rate atop a tower shines lights in straight beams in directions at right angles to each other -- thus there are four such beams.  Some distance from the tower stands a high straight wall extending to infinity in both directions (built by Donald Trump to keep aliens out).  The point on the wall nearest the tower is $x=0$.  Geometry exercise: when one light shines at the point $X$, then one at a right angle to it shines at $-1/X$.  Now choose a point in time at random.  What are the distributions of $X$ and $-1/X$?  By symmetry, they are identical.


What is difference between open set and open interval? Let $\tau$ be the Euclidean topology defined on $\mathbb R$. If we define a set $S = (2,3) \cup (5,6)$. Then is the set $S$ an open set and open interval on $\tau$?

As per definition of open set,

  A subset $A$ of $\tau$ is open set if  $\forall x \in A$, $\hspace{5pt} \exists  \hspace{3pt} a,b $ such that $x \in (a,b) \subseteq A$

As per this definition, we can find $(a,b) \in S$ such that $x \in (a,b)$. $x$ is any number in set $S$. So, set $S$ is open set.

I am not sure whether I correctly proved why set $S$ is open set. But I do not know how to prove set $S$ is open interval.

$S$ is not an interval, open, or otherwise.

An interval $I$ in any ordered set has the property:

  If $a<b<c$ and $a,c\in I$ then $b\in I$.

Your set $S$ is therefore open, but not an interval. It is the union of two open intervals.

Difference-1 if we cosider thee sequence with element of  the subset of a open interval then the sequence is monotone increasing ......but if we consider a sequene with the element of the subset of an open set then the sequence is either monotone increasing or monotone decreasing .....


How to find the Galois group of a polynomial? I've been learning about Galois theory recently on my own, and I've been trying to solve tests from my university. Even though I understand all the theorems, I  seem to be having some trouble with the technical stuff. A specific example would be how to find the Galois group of a given polynomial. I know some tricks, and I manage to solve some of those questions, but some not.

For example, one of the tests asks to find the Galois group of $x^{4}-4x+2$.

I can see that it is irreducible over $\mathbb Q$ (Eisenstein), but I have no clue as to how to find its Galois group over $\mathbb Q$. Can someone tell me how to do this? General techniques concerning this sort of problems are also welcome :).

Thanks!

This is an attempt to write a canonical answer listing techniques to compute Galois groups of explicit polynomials, primarily over $\mathbb{Q}$, as described in this meta thread. This answer is CW to encourage others to add techniques.

Computability: Over any reasonable field $K$, such as $\mathbb{Q}$, $\mathbb{F}_p$, $\mathbb{Q}(t)$, etc, Galois groups are computable. See Computable fields and Galois theory by Russel Miller for a good summary of this subject. Logicians can construct countable fields $K$ where the basic field operations $+$, $-$, $\times$, $\div$ are computable but the question "is $D$ square?" is not computable. In such a field, we cannot compute the Galois group of the splitting field of $x^2-D$. However, this is not the issue that this answer seeks to address and, as described in the linked article, there are very good criteria for showing that fields encountered in ordinary practice do not exhibit this phenomenon.

Although Galois groups are computable, computation of Galois groups, both by computer systems and by students in Galois theory courses, does not proceed along a single algorithm, but rather involves a smorgasboard of methods which are used to prove facts about the group $G$, until enough facts are accumulated to single out a particular group. The main goal of this entry is to list such techniques. I've listed the methods roughly in the order I would pull them out. Some methods are left blank for now; feel free to fill them in before I do!

Several of these methods are either primarily used to prove $G$ is large, or primarily used to prove $G$ is small. In this case, I have put "LARGE" or "SMALL" after the subject heading.

I also plan to put some useful lemmas about permutation subgroups at the end, since the more you know about subgroups of $S_n$, the fewer facts you need about a group $G$ in order to identify it.

As I assume you understand already, if $f$ is a separable polynomial of degree $n$, with roots $\theta_1$, $\theta_2$, ..., $\theta_n$, and $L$ is the splitting field of $f$ over $K$, then $Gal(L/K)$ is a subgroup of $S_n$, which can be thought of as permutations of the roots.

Noticing obvious additive and multiplicative relations between roots (SMALL) Let $f(x) = x^4+2 x^2 + 3$. Since all the exponents in $f$ are even, if $\theta$ is a root then so is $-\theta$. Let the roots be $(\theta_1, -\theta_1, \theta_2, -\theta_2)$. So any Galois symmetry must take the (unordered) pair $\{ \theta_1, - \theta_1 \}$ either to $\{ \theta_1, - \theta_1 \}$ or $\{ \theta_2, - \theta_2 \}$.

Similarly, let $f(x) = x^4 + 2 x^3 + 5 x^2 + 2 x + 1$. Since this polynomial is palindromic,
if $\theta$ is a root, then so is $\theta^{-1}$. Writing the roots as $(\theta_1, \theta_1^{-1}, \theta_2, \theta_2^{-1})$, one obtains similar restrictions on $G$. One can also see more complicated multiplicative relations: If $\theta_1$ and $\theta_2$ are two roots of $x^5-2=0$, then so is $\theta_2^2/\theta_1$; relations of the form $\theta_3 = \theta_2^2/\theta_1$ quickly restrict the Galois group to be a subgroup of a $20$ element group.

Irreducibility of polynomials and transitive group actions The polynomial $f$ is irreducible if and only if the Galois group is transitive. Wolfram alpha, Mathematica, Maple, SAGE, etc. will factor polynomials for you.

To prove polynomials are irreducible by hand, one can use Eisenstein's criterion or can try to find a prime $p$ such that $f$ does not factor modulo $p$. However, there are polynomials which are irreducible but for which neither of these tests will prove them so.

Complex conjugation (LARGE) Suppose that $K$ is a subfield of $\mathbb{R}$ (such as the rationals). Let $\theta_1$, $\theta_2$, ..., $\theta_n$ be the roots of $f$, thought of as complex numbers. Again, Wolfram alpha or any computer algebra system will find the $\theta$'s for you.

If there are $r$ real roots and $2s$ complex roots among the $\theta$'s, then $G$ contains a permutation of conjugacy class $(1\ 2) (3\ 4) \cdots (2s-1 \ 2s)$

Discriminants Define $\beta = \prod_{i<j} (\theta_i - \theta_j)$ and $\Delta = \beta^2$. Since $\Delta$ is symmetric in the $\theta$'s, it is in the ground field $K$ and is quite reasonable to compute. The number $\Delta$ is called the discriminant of $f$.
The Galois group $G$ is contained in $A_n$ if and only if $\beta$ is in $K$; in other words, if and only if $\Delta$ is square in $K$.

Here is a sample Wolfram alpha session, verifying that the Galois group of the splitting field of $-1 - 2 x + x^2 + x^3$ is contained in $A_3$.

Recognizing $S_n$ and $A_n$ (LARGE) A random polynomial over $\mathbb{Q}$ has very high probability of having Galois group $S_n$. (See this paper of Cohen for a precise bound.) So it is worth being able to recognize the Galois group $S_n$ early in the process. Keith Conrad has good notes on how to do so. Galois groups appearing on problem sets in Galois theory courses have a lower probability of being $S_n$. :-)

Reduction modulo $p$ (LARGE, but useful both ways) Suppose that $p$ does not divide the discriminant of $f$. Let $f$ factor modulo $p$ into factors of degrees $(h_1, h_2, \ldots, h_d)$. Then $G$ contains an element with cycles of length $h_1$, $h_2$, ..., $h_d$. Conversely, by the Cebatarov density theorem, if $G$ contains an element of cycle type $(h_1, h_2, \ldots, h_d)$, then there is a prime $p$ such that $f$ factors modulo $p$ with factors of that size. See David Speyer's blogpost,  Hendrik Lenstra's exposition or Keith Conrad's notes.

Here is an example:

Table[Factor[X^4 + 4 X^3 + 12 X^2 + 24 X + 24, Modulus -> Prime[n]], {n, 3, 10}]

produces

{(4 + X) (1 + 2 X + X^3), (5 + X) (2 + 3 X + 6 X^2 + X^3),
 (3 + X) (8 + 9 X + X^2 + X^3), (12 + X) (2 + 4 X + 5 X^2 + X^3),
 (16 + X + X^2) (10 + 3 X + X^2), (14 + 11 X + X^2) (18 + 12 X + X^2),
 (17 + X) (19 + 3 X + 10 X^2 + X^3), (22 + X) (9 + 2 X + 11 X^2 + X^3)}

All the polynomials have cycle type $(3,1)$ or $(2,2)$, proving that $G$ contains those cycle types, and producing strong evidence that $G$ is contained in a group which only has cycles of that type. That can't quite be right, because $G$ contains the identity with cycle type $(1,1,1,1)$; searching up to $p=71$ finds such a factorization. But even running the search that far, we only see cycle types $(1,1,1,1)$, $(3,1)$, $(2,2)$, suggesting that $G = A_4$. Indeed,

Discriminant[X^4 + 4 X^3 + 12 X^2 + 24 X + 24, X]

outputs $331776=576^2$ so we know $G \subseteq A_4$, and the above data shows that $G$ contains a $3$-cycle and contains an element of type $(2,2)$, so $G = A_4$.

Distinguishing $D_4$ and $C_4$ Let $f(x) = x^4 - 5 x^2 + 5$. Since $f(x)$ is even, if $\theta$ is a root, so is $- \theta$. This shows that $G$ is a subgroup of $D_4$, the dihedral symmetries of a square. Since the polynomial is irreducible, the action on a $4$ element set is transitive, which makes $G$ either $D_4$, the cyclic group $C_4$, or $C_2 \times C_2$ with the generators $(12)(34)$ and $(13)(24)$. Since the discriminant is $2000$, which is not square, it is not contained in $A_4$, which rules out $C_2 \times C_2$.

Factoring modulo $p$ finds no factors with cycle type $(1,1,2)$, which strongly suggests $G$ is $C_4$. But that isn't a proof. This particular case comes up enough (especially on exams!) that there are special methods for it, see Keith Conrad's notes or my answer here.

Testing for $k$-transitivity (LARGE)
More generally, let $\beta = \theta_1 + 2 \theta_2 + \cdots + k \theta_k$ and let $g(x)$ be the minimal polynomial of $\beta$. If $g(x)$ has degree $n(n-1) \cdots (n-k+1)$, then $G$ is $k$-transitive on $(\theta_1, \ldots, \theta_n)$. If there is some number of the form $\theta_{s_1} + 2 \theta_{s_2} + \cdots + k \theta_{s_k}$ which is not a root of $g$, then the group $G$ is not $k$-transitive. Since the $k$-transitive groups have been classified for $k \geq 3$ (see previous link), this can be very useful in identifying $G$.

Here is a small example. Take $f(x) = x^4-4 x^3+2 x^2+4 x-6$. We check that $f$ is irreducible:

f = x^4 - 4 x^3 + 2 x^2 + 4 x - 6;
Factor[f]

We store the roots of $f$ in variables t1, t2, t3, t4. (Mathematica can actually solve this polynomial exactly, but this is just meant as an example.)

{t1, t2, t3, t4} = x /. Solve[f == 0, x];

b = t1 + 2 t2;

g = MinimalPolynomial[b, x]
(* Output is 42 - 84 x + 50 x^2 - 12 x^3 + x^4 *)

Since $g$ has degree $4$, we conclude that the group $G$ probably has an orbit of size $4$ acting on ordered pairs of distinct $\theta_i$'s. For a careful proof, we need to make sure that we don't have accidental equalities of the form $\theta_i+2 \theta_j = \theta_k + 2 \theta_{\ell}$. In particular, we check that $g(\theta_1+2 \theta_3) \neq 0$, so $(1,2)$ and $(1,3)$ are in different orbits for the $G$-action. This example is small enough for Mathematica to do exactly, but I'll do the computation numerically because that is good practice for larger examples:

N[g /. x -> t1 + 2 t3]
(* Output is 198.996 + 220.833 I *)

Relative resultants

Installing PARI, GAP or MAGMA This section is about which computer algebra systems have implemented general algorithms to do this, and where to get them.

Useful lemmas about subgroups of $S_n$

The Galois group of a polynomial is a transitive permutation group of its roots. How many real roots does it have? What can you say about non-real roots? What are the transitive subgroups of $S_4$? For those, see http://hobbes.la.asu.edu/Groups/group-data/degree4.html from Transitive Group Data.

Consider the splitting field. Since adding a real root of the polynomial gives you an extension degree of 4 and that there are non-real roots, the degree of the splitting extension is at least 8, ie, is a multiple of 8. So, the Galois group is either $D_8$ (the dihedral group of order 8) or $S_4$.


Show that $a^p \equiv 1$ (mod $p^n$) $\Rightarrow a \equiv 1$ (mod $p^{n-1}$) I'm trying to show that $a^p \equiv 1$ (mod $p^n$) $\Rightarrow a \equiv 1$ (mod $p^{n-1}$), where $p$ is an odd prime.

Since $a^p \equiv 1$ (mod $p^n$) $\iff aa^{p-1} \equiv 1$ (mod $p^{n-1}p$), I was hoping to use Fermat's theorem, which states that $a^{p-1} \equiv 1$ (mod $p$). I therefore tried substituting $a^{p-1}$ with $pk + 1$ in the equation but I didn't manage to solve it like that.

I've been trying to solve it algebraically for some time but is no where close so I could use some help and ideas.

Let $a=1+p^iX$, where $X$ is coprime to $p$.

Then $a^p-1=\begin{pmatrix}p\\1\\\end{pmatrix}p^iX+ ... +p^{pi}X^p$, where all the coefficients on the RHS except possibly the last are divisible by $p$.

We know that $p^{n} $ divides the LHS so $i>0$. Then, on the RHS, the first term has the lowest power of $p$ and so its power, $i+1$, must be at least $n$.


Simplifying $3^{-4}$ to $1/81$ How would you solve for $3^{-4}$? I know the answer is $1/81$ but I can't work out how you get there with this one.

you can think like this let $3^{-4}=a$ where a could be any real number. and we know a property $m^{-n}=1/m^n$ therefore$$3^{-4}=a\implies\frac 1{3^4}=a $$$$1=a\times3^4$$now think of a value of $a$ which when multiplied by $3^4$ gives 1. yes it is $1/3^4$


the generalized Liouville theorem The Liouville theorem state:

Let $f$ be an entire function for which there exists a positive number M such that $|f(z)|\leq M$ for all z in $\mathbb{C}$, the $f$ must be a constant.

more general form of this theorem is :

Let $f$ be an entire function for which there exists a positive number M and a polynomial $P$ such that $|f(z)|\leq M |P(z)|$ for all z in $\mathbb{C}$ then $f(z)=k P(z)$ for $k$ a constant.

who know a good proof of this generalized form ?

Generalized Liouville's theorem.  Suppose $f(z)$ is an entire function such that $|f(z)|\le c|z|^n$ for some constant $c$ and some nonnegative integer $n$ for sufficiently large $|z|$. Then $f$ is a polynomial of degree at most $n$.

Proof. By Liouville's theorem, the statement is true for $n=0$. Suppose true for $n-1$. Consider $$g(z)=\left\{\begin{array}{ll} \frac{f(z)-f(0)}{z}, & z\neq 0\\ f'(0), & z=0\end{array}\right.$$
Then $g$ is entire, and for sufficiently large $|z|$, $$|g(z)|\le c_1|z|^{n-1}$$ for some constant $c_1$. (This is because $|f(0)|/|z|, |f'(0)|$ are bounded for large $|z|$, and $a+b|z|^{n-1}\le c_1|z|^{n-1}$ for large $|z|$.)

Hence by induction hypothesis, $g$ is a polynomial of degree at most $n-1$. Then it follows easily by the definition of $g$ that $f$ is a polynomial of degree at most $n$.

If f is dominated by P like in your statement, then it is dominated by many other polynomials. If your generalization were correct, it would imply that all these polynomials are collinear. This can not be true, as soon as you consider nonconstant polynomials P.
You might want to prove that f is a polynomial, not necessarily of the type you mention.
Maybe you can start by looking at the case when $|f(z)\leq M|z|^n$.
Then you can see that the general case boils down to the latter.

Reference: see exercises 3 and 4 p.227 in Rudin's Real and Complex Analysis.


Is there a name for this graph density measure? Let $G=(V,E)$ be an undirected graph.
We define the following procedure (randomized greedy coloring):

Fix some random ordering over the vertices (each permutation will be chosen w.p. $\frac{1}{|V|!}$).
Color the graph vertices according to the order such that each vertex gets colored by the minimal color not used already by its neighbors.

We define $\mathfrak C(G)$ to be the expected number of colors according to the procedure.

Is there a name for $\mathfrak C(G)$?

For example:

$\mathfrak C(K_n)=n$ (trivial, every vertex must get a new color regardless of the order).
$\mathfrak C(C_6)=2\frac{1}{5}$ (in a cycle on 6 nodes, if the second vertex that is colored is of distance 3 from the first colored vertex, the number of colors used will be 3).

I believe this is called the chromatic density of a finite graph. See page 8 of this reference. The notation is different, but I think it is the same concept.


Proving a subgroup of a group My question states, Let H be subgroup of a group G. Prove that HH=H and that $H^{-1}=H$  I started the 1st proof by saying  HH=H  We will show that $HH\subseteq H$  Then $x\in HH$  Then $x=ab$ where $a\in H$ and $b\in H$  I am lost right now trying to prove HH=H and $H^{-1}=H$  Anybody can help me?

I already mentioned in my comment that $HH=H$.

We know that $H<G$. We have $$H^{-1}=\{h^{-1}: h\in H\}.$$

Let $x\in H^{-1}$. Then $x=h^{-1}$ for some $h\in H$. Because $H<G$, we have $h^{-1}\in H$. Thus, $x\in H$. Hence, $H^{-1}\subset H$.

Let $y\in H$. Then $y^{-1}\in H$. Hence, $y=(y^{-1})^{-1}\in H^{-1}$. Thus, $H\subset H^{-1}$.

Done.Hope it helps.

Given   that   H   be   a   sub   group   of   G
So   we   have to   show   that   HH   =   H
For that it is enough to show that HH ⊆ H and H ⊆ HH
Now   let   h_1  h_2∈HH        where   h_1∈ H,   h_2∈ H
∴   h_1∈H ,h_2∈ H   ⇒  h_1  h_2∈H      (  ∵ closed law  )
∴   h_1  h_2 ∈ HH   ⇒   h_1  h_2 ∈ H
⇒   HH⊆ H    ------------ (1)
Again    let   h ∈ H   then   we   have to   show   that   h ∈ HH
Let   h ∈ H   and   e ∈ H   ⇒ he ∈ HH
                                             ⇒ h ∈ HH

∴      h ∈ H  ⇒   h ∈ HH
                         ⇒  H ⊆ HH   ------------------------- (2)
∴      from   (1)  and  (2)  we  get  HH =  H ( ∵  HH ⊆ H and  H ⊆ HH)
Hence  proved


Minimizing Expected Value I have a problem which wants the c value that minimizes E[(X-c)2]

I started with

E[(X-c)2] = E[X]2 -2cE[X] + c2

but couldn't continue on this. Could you please help me on finishing this problem.

Thanks in advance.

Hint:

$$\dfrac{d}{dc}E[(X-c)^2] = -2E[X]+2c$$ so find  the value of $c$ that gives the minimum in the usual way

Take derivates wrt to $c$, treating any term without $c$ in it as a constant, thus differentiates to zero.

When differentiating just treat $E[X]$ as a function of $x$ and then differentiate as normal with the chain rule.


Is Induction Independent of the Other Axioms of PA? I am trying to come up with a model of first order Peano Arithmetic (PA) where induction fails. Let $PA^{-IND}$ have the same axioms as PA except the first order induction axiom schema is replaced with its negation. I need to show there exists a predicate, $P$, that makes first order induction false. $P$ must satisfy $P(0) \land \forall x(P(x) \rightarrow P(Sx))$, yet $\forall x(P(x))$ is false. We can prove multiplication is commutative using double induction on $P(x,y)= (xy=yx)$. Why Does Induction Prove Multiplication is Commutative?

Consider the 2x2 matrices $M_2(N)$ with the standard definitions for matrix addition and multiplication. Let the zero matrix be $0$ and the identity matrix be $S0$. $\forall x(Sx=x+S0)$ is a theorem of $PA^{-IND}$. Matrix multiplication is not commutative, yet we can prove multiplication is commutative for all the successors of $0$. Would $M_2(N)$ be a model of $PA^{-IND}$?

The negation of first order induction says there exists a predicate:
$P(0) \land \forall x(P(x) \rightarrow P(Sx)) \land Ex( Not(Px))$

This looks like quantification over predicates but it isn't. The induction schema requires us to add an infinite number of axioms to the language. The negation of induction only requires the addition of a single axiom. Unlike PA, $PA^{-IND}$ has a finite number of axioms. I have simply added a new predicate to the language.

I am using the axioms of PA given by Wikipedia for First Order Arithmetic. These axioms use induction to prove commutativity. Without induction, these axioms are very weak. They don't even require addition to be commutative. I would be interested in any model of $PA^{-IND}$.

You have to be careful about what you mean by $\text{PA}^-$.  The convention I expect is that $\text{PA}^-$ is the finite set of axioms for a discretely ordered semiring - which includes the axiom that multiplication is commutative. This set of axioms is typical when we want to look at systems of arithmetic with limited induction. For example, it is the set of axioms in Kaye's book The Structure of Models of Peano Arithmetic.

The original five axioms proposed by Peano, which only mention the successor operation, but not addition, multiplication, or order, are not sufficient for first-order Peano Arithmetic, because they are not sufficient to define addition or multiplication in first-order semantics.

On the other hand, some authors use an abbreviated set of axioms for Peano arithmetic. For example, Mendelson's logic text uses a smaller set of axioms that do not mention the order relation. These work because he only considers them in conjunction with the induction scheme. It seems to me that $M_2(\mathbb{N})$ is a model of this smaller set of axioms, but I would not call them $\text{PA}^-$.

Even when we look at the full set of axioms for $\text{PA}^-$, we can give an example of a model that does not satisfy induction. Tennenbaum's theorem shows that no nonstandard model of Peano arithmetic is computable. But the set of polynomials over $\mathbb{Z}$ with positive leading coefficient are a computable nonstandard model of $\text{PA}^-$, so induction must fail in that model, although I am not sure which instance of the schema fails. I believe the proof only relies on a finite number of instances of induction, so that would give a short list of candidates.

I think a simple example of PA negating induction would be the set $\mathbb N \cup C$ where $C$ is some (any) cycle in the sense that some symbol $c \in C$, and for every $x \in C$, $Sx \in C$, and $SSSSSSSx = x$.


Is this fractional equation true for all real numbers? Is this equation true for all real values of $a,b,c,x,y,z$?
If $$\frac{a}{x} = \frac{b}{y} = \frac{c}{z} = t$$
Then $$\frac{a+b+c}{x+y+z} = t$$
I've tried experimenting with several values and they make an equality, but does this equation hold true for all real values?

(Almost always) yes: from the first set of equations, you have $a=xt, b=yt, c=zt$. Therefore,
$$\frac{a+b+c}{x+y+z} = \frac{xt+yt+zt}{x+y+z} = t\frac{x+y+z}{x+y+z} = t$$
provided, of course, that $x+y+z\neq 0$.

An example of this: imagine a course where you're graded on homework, a midterm, and an exam. If you get the same grade on each of these components, then that's your final grade, regardless of how they're weighted.

$ \text{Not true for all real numbers}  $
Comparing both equations,
$ \frac{a}{x}+\frac{b}{y}+\frac{c}{z}= \frac{a+b+c}{x+y+z} $
$ (x+y+z)\times (\frac{a}{x}+\frac{b}{y}+\frac{c}{z})=a+b+c $
$ \frac{1}{x} (ax+ay+az) + \frac{1}{y}(bx+by+bz) + \frac{1}{z}(cx+cy+cz) = a+b+c$
$ a+\frac{ay+az}{x} +b+ \frac{bx+bz}{y} + c+ \frac{cx+cy}{z}=a+b+c $
$a+b+c= a+b+c +(\frac{ay+az}{x} + \frac{bx+bz}{y} + \frac{cx+cy}{z}) $
$ \frac{ay+az}{x} + \frac{bx+bz}{y} + \frac{cx+cy}{z} = 0 $
$ \text{As} $ $ x,y,z \neq 0  $
one Solution is when  $ a=b=c=0 $, and have other solutions as well but Not true for all Real Numbers.
Thanks to


A problem related to limsup and liminf. Suppose $a_n >0$ and $b_n <1$ and lim $a_n=0$, lim $b_n =1$ and $A_n =\{x: a_n \leq x  <b_n\}$   then find limsup$ (A_n)$ and liminf$(A_n) $.

As $a_n$ goes to 0 and $b_n$ goes to 1, How can I use this condition to find  limsupA_n and liminf A_n, I got stuck.

It is generally difficult to express $\limsup_n A_n$ and $\liminf_n A_n$ in terms of $\lim_n a_n$ and $\lim_n b_n$. The reason is that $\limsup_n A_n$ and $\liminf_n A_n$ are not topological notions, that is, they do not require the existence of a topology on $[0,1]$, whereas $\lim_n a_n$ and $\lim_n b_n$ are topological notions.

Since $a_n\to 0^+$ and $b_n\to 1^-$ (we know that $0 < a_n \leq b_n < 1$), we have that for every $\epsilon \in (0,1)$, there is an $n_0=n_0(\epsilon)\in\mathbb{N}$ so that $a_n < \epsilon$ and $1-b_n<\epsilon$ for all $n\geq n_0$, therefore $[a_n, b_n] \subseteq [a_{n_0}, b_{n_0}]$.

Then,

$$
\liminf_n A_n = \bigcup_{n\geq 1}\bigcap_{j\geq n}[a_j, b_j]
\supseteq \bigcup_{n\geq n_0(\epsilon)}\bigcap_{j\geq n}[a_j, b_j]
\supseteq \bigcup_{n\geq n_0(\epsilon)}[a_{n_0}, b_{n_0}] = [a_{n_0(\epsilon)}, b_{n_0(\epsilon)}]
$$

Since this holds for all $\epsilon > 0$, we can take $\epsilon \to 0^+$ to conclude that

$$
\liminf_n A_n \supseteq (0, 1).
$$

In fact, $(0,1)$ is the largest possible such set because we can see by the definition that $0,1\notin \liminf_n A_n$. Therefore, $\liminf_n A_n = (0,1)$.

Similarly, we may show that $\limsup_n A_n = (0,1)$.

$(A_n)_n$ is an increasing sequence of sets, thus it converges to $\cup_n A_n=\liminf A_n=\limsup A_n$ and $\cup_nA_n = [0,1).$


Classic Hand shake question I am asked the following question:

  My wife and I were invited to a party attended by four other
  husband-wife couples, making a total of ten people. As people arrived,
  there was some hand shaking. No one shook their own hand, and there
  were no husband-wife hand shakes. When it was over, I asked each
  person "How many people did you shake hands with?" I asked nine people
  (not including myself) and got nine different answers. How many people
  did my wife shake hands with?

Since there are 10 people in the room should I use the formula n(n-1)2, which results in 10(10-1)/2=45 which would be the sum of all the handshakes. If I divide that by ten I would get 4.5 people shook his wife's hand.

This doesn't seem correct at all. What is the right direction to head in order to solve this question?

EDIT:

I ended up trying to solve by pairing everyone together and drawing a graph showing the relations between everyone:

But this wasn't complete apparently.

Hint:  No person shook more than eight hands, so you must have gotten answers from 0 to 8 inclusive.  Who is married to the person who shook 8?

The answer lies in the question itself.
It says, there were no hand shakes between husband-wife and the man asking everyone has not included himself in the count, so he and his wife would not be counted.
Therefore the answer is 8...His wife shook hands with 8 people(excluding her husband).


Is it OK to be picky about math you find interesting? I am a layman interested in mathematics, and I would like to hear mathematicians' views on the following: Is it normal to be picky about mathematical stuff you find interesting?

I ask because 80% of math I encounter does not seem interesting to me. I want to know if this is true for most people or if there is something wrong with me.

At any particular moment, or for stretches of time, it is entirely reasonable to find various things boring... or incomprehensible, or both.

But/and as one learns more, one's opinions will inevitably change.

And, yes, admittedly, many published papers are contrived, with the main goal of having a publication to make a good impression on dept heads, deans, and funding agencies. No real intrinsic interest... and it's hard to tell, from an amateur's or novice's viewpoint.

So I'd recommend that you (and everyone...) follow their own tastes and curiosity, BUT be open to the possibility that things that seem boring or pointless or incomprehensible at one moment may later seem quite otherwise. Don't pretend to make permanent decisions...

I would say it would be unwise and rather odd for a mathematician to dismiss any part of the subject matter of mathematics as uninteresting. The subject is very large and there are deep connections between apparently very different aspects: as a simple example, consider the many different proofs of the fundamental theorem of algebra. If 80% of the maths that you encounter seems to be uninteresting to you and you can't see why that 80% of interest to anyone else, then I don't think maths is for you. (I'm personally not very excited by the the theory of differential equations, but I know it's an important subject and I appreciate all the work that's been done on it over the centuries.)


Suppose there are $10$ horses in a race and a bookmaker quot.... Suppose there are $10$ horses in a race and a bookmaker quotes odds of $r_i$ to $1$ against horse $i$ winning. Let $p_i=\frac{1}{r_i+1}, \ i=1,\dots,10$, so each $p_i$ is between $0$ and $1$. Let the summation be $p_1+p_2+...+p_{10}$

a) do you expect that the summation is greater than, smaller than, or equal to $1$? Why?

b) suppose the summation were less than $1$. Could you take advantage of this? How? [hint:by betting on all $10$ horses in the race, a better can win a constant amount of money, regardless which horse win.]

I'm having some trouble understanding what the summation in this case even stands for. I know that a) is less than one by simply plugging in but since I don't understand what the summation represents, I can't really move on to b).

Any and all help is appreciated!

a-
without even going into an analysis of the series,
we can see that the summation will be > 1,
as the ist 3 terms themselves [0.5 + 0.33 + 0.25 ] > 1

b-
if the summation were less than 1, we could place equal bets on all the horses.

since the actual probabilities would be more on the whole
since they have to sum up to 1, we should gain


factor the following expression $25x^2 +5xy -6y^2$ How to factor
$$25x^2 +5xy -6y^2$$

I tried with $5x(5x+y)-6y^2$. I'm stuck here.

I can't continue.

Forget about the $y$ for a second.

If you want to factor $25x^2 + 5x - 6$, look for two numbers whose product is $25 \cdot -6 = -150$ and whose sum is $5$. It takes just a moment to see that the two numbers are $15$ and $-10$. Then:
\begin{align*}
25 x^2 + 5x - 6 &= 25 x^2 + (15x - 10x) - 6 \\ &= (25 x^2 + 15x) - (10x + 6) \\ &= 5x(5x + 3) - 2(5x + 3) \\& = (5x-2)(5x+3).\end{align*}

With the $y$ included the process is nearly identical:
\begin{align*}
25 x^2 + 5xy - 6 y^2&= 25 x^2 + (15xy - 10xy) - 6y^2 \\ &= (25 x^2 + 15xy) - (10xy + 6y^2) \\ &= 5x(5x + 3y) - 2y(5x + 3y) \\& = (5x-2y)(5x+3y).\end{align*}


Definite Integral of $1+\sqrt{9-x^2}$? Right, so I'm to find the definite integral (interpreting it as an area)...

$\int^0_{-3}(1+\sqrt{9-x^2})dx$

How do I go about doing this?

I am to specifically use the following theorem to work it out...

If $f$ is integrable on $[a,b]$ then...

$\int^b_a f(x)\,dx = \lim_{n\rightarrow \infty}\sum^n_{i=1}f(x_i)\Delta x$

where $\Delta x = \frac{b-a}{n}$ and $x_i = a + i\Delta x$

I keep getting halfway through and getting stuck with $\lim_{n\rightarrow\infty}\frac{3\sqrt{3}}{n^3}\sum^n_{i=1}\sqrt{4n^2-3i}$

Is this correct? How do I work on from here, if so? The answer given in the book is $3 + \frac{9}{4}\pi$.

Any ideas?

This doesn't use your theorem, but it solves it in what I consider to be a fast and efficient way. It doesn't really even use calculus! If you were to graph $y = 1+\sqrt{9-x^2}$, you would see that it is in the shape of a half circle from $(-3,3)$, and then the rest of the graph is simply 1, because the square root is imaginary. From this, it is fairly easy to see that $\int_{-3}^{0}(1 + \sqrt{9-x^2})$ is just a quarter of a circle with radius $3$, with $3$ added, because the circle is elevated one unit over a distance of $3$, and $3 \cdot 1 = 3$. A circle with radius $3$ has an area of $9 \pi$, so a quarter of that circle is $\frac{9}{4}\pi$, which is of course $\frac{9}{4}\pi +3$ when the three is added. If any of this is confusing, then try to graph the function and you'll see what I mean. I'm not sure if you wanted any way to solve the integral, or for us to simplify $\lim_{n \to \infty} \frac{3\sqrt{3}}{n^3} \sum _{i = 1}^{n}(\sqrt{4n^2-3i})$, but this is definitely a fairly quick method of solving the integral itself.

Note: This isn't a complete answer, because it doesn't use the stated theorem, but it was too long for a comment, and it seemed worth mentioning

you shouldn't try to use a Riemann sum for this.
Your instructor definitely isn't expecting you to solve that way. This is a "net area" problem, check your textbook for "net area".

The short version is this: ⺆r²=the area of a circle. In this case, r will be the same as x. So basically, you know x has to be 3 by setting the equation of the circle equal to zero, which is how you always solve for x, right? This is true for all half circle equations of this format. So you know you have a radius of 3 now. You also know that the location of your interval [-3,0] is such that your interval will cut the circle into quarters >> look at the graph on your calculator. Yes, you should always graph every line you're ever given on your calculator, lol. You can see a half circle slightly raised off the x-axis, right? So from -3 to 0 on the x-axis, you only have a quarter of the total possible circle there. so you need the area of 1/4 of your circle, since only a fourth of the total possible circle lies within your interval. so (1/4)Area of your circle = (1/4)⺆r², where r=3. This is true since the area of the total possible circle would just be ⺆r², where r=3, like we discussed. This gives you (1/4)9⺆, which multiplies out to (9/4)⺆. That's the area of a quarter of the total circle, half of which you see on your graphing calculator, with only that quarter we calculated the area for being in the necessary interval of [-3,0].

Now, you need to remember that there's space under your circle thats just open, like we were saying. Like, the edges of your circle don't touch the x-axis. yeah, you have to find the area of that also, and add it to the area of the circle part that you already found. Net area of an interval is ALWAYS from the line you see, no matter what shape it is, all the way to the x-axis. Just quickly sketch the graph and color in the entire area from the line (in this case, that half circle) down to the x-axis, and that's the total area you need to find. Don't just stop at the edges of the circle. So that shape under your circle is actually just an invisible rectangle, isn't it? Sketch that rectangle onto your hand drawn graph if necessary, so you can see it. You need to add the area of that quarter circle, with the area of the invisible rectangle underneath the circle. So the area of a rectangle is base times height, and in our case, it should be 1 x 3 = 3. So the area of the quarter circle is (9/4)⺆ and the area of the invisible rectangle underneath that quarter circle is 3. So add them together to get:
(9/4)⺆ + 3 = net area.

Yes, net area of this format is that easy. And leave your answer in the above form, don't do anything else. You dont need to, because ⺆ isn't a variable, remember? Its a constant number, so you have all constant numbers in your answer, and this form is the exact answer. That's what you want. As for a more formal looking solution, just write out:

Net area = (base x height) + (1/4)⺆r²
as the formula you created to solve and plug in your values.

So DONT try a Riemann Sum, you don't have to! It's completely unnecessary, and your instructor won't accept a Riemann Sum as a correct answer to a net area problem. Your answer should always be constant numbers when you're solving for net area. I hope this helps!


License plate combinations 4 letters 3 numbers How many license plates can be created if they have $4$ letters and $3$ digits in any order?
My thinking is $26^4 \times 10^3 \times {7 \choose 4} \times {7 \choose 3}$:
$26$ possibilities for each of 4 letters, $10$ possibilities for each of 3 digits, $4$ places out of $7$ to place the letters, and $3$ places out of $7$ to place the numbers.
Is this correct?

Reiterating what was said earlier in a comment:
Your answer is almost correct however you made the mistake of picking the positions twice rather than only once.  By choosing the four positions to be occupied by letters and then again choosing which three positions were to be occupied by digits and having chosen said positions out of the original seven, you will have run the risk of picking the same position multiple times designating it to be simultaneously used by letter and digit.
Instead, the digits' positions should have been selected out of the remaining unused positions giving a total of $\binom{3}{3}=1$ way in which this step could be accomplished, this being merely $1$ way it can be left out of the final product.  This leaves the final answer as being:
$$26^4\cdot 10^3\cdot \binom{7}{4}$$

Compare this to a slightly larger problem where you have $26$ letters to choose from, $10$ digits, and $15$ special symbols (e.g. !@#\$%^ and so on...) and we want to make a ten-character password consisting of $6$ letters, $3$ digits, and $1$ special symbol with repetition allowed.
We first pick the positions used by letters, then from those remaining positions pick the positions used by digits, letting the final remaining position be occupied by the symbol, and then picking which letters, digits, and symbol they happened to be for a total count of:
$$26^6\cdot 10^3\cdot 15\cdot \binom{10}{6}\cdot \binom{4}{3}$$

There are 28 letters and 10 digts from 0 to 9. You want to chose 7 of them when the order in which you select is important. So there are $P_{38}^7$ possibilites. ($P$ stands for permutation)


Series convergence - Gauss test How do I prove that
$$\sum_{n=1}^\infty\left(\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot(2n)}\right)^{k}$$
converges for $k>2$ using Gauss test?

Easier: log the expression to get $k(\sum_{j=1}^{n} (2j+1) - \sum_{j=1}^{n} \log (2j))$, Here we have $n$ terms in both sums. After termwise subtraction we get $kn \log 1 = 0$. Hence the limit of the original series is $e^0=1$.

Using the notations of this presentation of the test, obviously, one is interested in the sequence
$$
u_n=\left(\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot(2n)}\right)^{k},$$
hence the job is to identify
$$
\frac{u_n}{u_{n+1}}=\ldots,
$$
then to identify $r\gt1$, $h$ and the bounded function $B$ such that
$$
\frac{u_n}{u_{n+1}}=1+\frac{h}n+\frac{B(n)}{n^r},
$$
and finally, to identify the values of $k$ such that $h\gt1$ (then the series converges) and the values of $k$ such that $h\leqslant1$ (then the series diverges).


Intermediate value property with no continuity Definition: A real function f has the intermediate value property on an interval I containing [a,b] if f(a) < v < f(b) or f(b) < v < f(a); that is, if v is between f(a) and f(b), there is between a and b a c ∈ [a,b] such that f(c) = v.

What is an example function that has this property, and how do I prove that a strictly increasing function f:[a,b]→ R which has this property is continuous on [a,b]?

Presumably you wanted an example of a discontinuous function satisfying IVP. If $f$ is differentiable then the Mean Value Theorem shows that $f'$ has IVP, although $f'$ need not be continuous.

An example is the identity.

Hint (for $f$ increasing): $f$ must be surjective in $[a,b]$ for all $[a,b]\subset I$


To show that R is a division ring given some property on ring R. Let $R$ be a non-zero ring such that the equation $ax=b$ has a solution in $R$ for all $a, b\in R$ with $a\neq 0$. Show that $R$ must have unity and it is a division ring.

Now to show this we need to show that every non zero element is invertible.
If we take '$a$' as non zero element with $ax=b$ then how can I use the given fact to show the ring is a division ring. Please help.

Here is how you solve the problem if it is already assumed that $R$ has a unit:

Taking $b = 1$, a solution to $ax=1$ is a right inverse for each nonzero $a \in R$. Fix such a right inverse $ac=1$. Now $c$ also has a right inverse, $cd=1$. It follows that $a = a(cd) = (ac)d = d$, so $c$ is both a right and left inverse for $a$. This proves that every nonzero element of $R$ has a two-sided inverse.

For showing that $R$ has an identity:

What you have is, for each nonzero $a \in R$, an $e_a$ so that $a e_a = a$.

If $ab = 0$ for nonzero $a, b \in R$, then one can multiply on the right by the solution to $bx = e_a$ to get $0=abx = ae_a = a$, which is a contradiction. So $R$ has no zero divisors, which, as was noted in one of the comments, is equivalent to both left and right cancellation holding. So the $e_a$'s are unique for each fixed $a$.

As rschwieb's comment below suggests, you can use cancellation together with $a e_a e_a = a e_a$ to conclude that $e_a^2 = e_a$, making $e_a$ an idempotent. Then left canceling $e_a(e_a b - b) = 0$ makes $e_a$ a left unit for $R$, and right canceling $(b e_a - b) e_a = 0$ makes $e_a$ a right unit for $R$.

This is closely related to a standard trick for breaking down a ring given an idempotent: For $e \in R$ an idempotent in an arbitrary ring, we have $x = (x-ex) + (ex)$, which gives a direct sum decomposition of $R$ as a right module over itself.

Take $b=1$ then every $a$ has an inverse since the equation has a solution. Then take $a=b$ and see that for every $a$ there exists $ax=a$.


What's the equation for a line segment? I already know that the standard equation for a line is $y=mx+b$, but what if I want the line to have specific endpoints and not go on forever? For example, the equation for a line beginning at $(3, 1)$ and ending at $(7, 2)$. Can you help me?

What's the standard equation for this?

You can make a parametric equation:
$$x=tx_1+(1-t)x_2$$
&
$$y=ty_1+(1-t)y_2$$
where t$\in[0,1]$. It is the internal section formula of a line, where i replaced m and n with t.

You can make a line segment like this:

-sqrt(x)² gives a line with a negative slope for which if x is negative, the function is undefined.

you can limit the values of y by doing: sqrt(-sqrt(x)²)² this way you cannot have any negative y values in your graph. all you have to do now is move your graph around.

for example try: -sqrt(-sqrt(x-1)²+2)²+3

Try playing around with the function to get a feel on how it works.


Evaluate the integral in terms of areas. I understand that the first one is 4 from basically adding the squares inside the signed area, but I'm unsure on how to proceed in getting the other integrals. Any help would be appreciated, thank you.

Some tricks to help you:

a. count squares

b. calculate the area of known figures, such as triangles and rectangles.

For example, the integral in (b) can be calculate as the same of the area of a trapezoid($(1+3)\cdot 2/2$), a rectangle ($1\cdot 3$), and a triangle ($3\cdot 2/2$) to give a grand total of $10$.

That should cover it, good luck.


Algebra of limits. What NBD means? 9. If $f(x)\le g(x)$ for every $x$ in the NBD of $a$, then
  $$\lim_{x\to a}f(x)\le\lim_{x\to a}g(x).$$

NBD stands for what?

Let g=f+e where e is non-negative. lim g is a subset of lim f+lim e where lim is the set of subfunction limits (analogous to subsequential limits).


Probability of selecting four letters from ENCYCLOPAEDIA From a maths textbook: "Four letters are randomly selected from the word ENCYCLOPAEDIA. Find the probability that one letter E will occur in the selection of 4 letters."

Clarification - as I understand the question, it is asking about the probability of getting combinations with exactly one E. Samples would be with replacement - I'm picturing a bag with 13 lettered tiles in it and replacing the tiles after each sample.

In the textbook we have worked through questions involving permutations with repeated values, but not combinations. And there are a number of good examples on the web about combinations with repeated values, but this problem has limited repeated values (2 A's, 2 C's, 2 E's). I actually put the letters in alphabetical order and started making my own list of 4 letter combinations and discovered it was tricky with the repeats (and going to take a while).

It seems that $^{13}C_4$ is a good place to start for the total number of combinations (for the denominator of the probability result), but that doesn't adjust for the duplicate letters.

Since I've recently been doing Javascript in my IT class I went ahead and put together a webpage that created a list of all 715 combinations and then tried to work out how to eliminate the ones that were duplicates (more than half of the starting list).

Eventually I got the Javascript created list down to 321. While it created the list it also counted the ones with one E and that total was 100.

So problem solved: 100/321 is the probability of getting one E when randomly selecting four letters out of ENCYCLOPAEDIA. By the way, this does not match the answer in the textbook, but I suspect it is far too complicated for the Yr 11 class I teach.

I just wonder if there is a way to answer this without creating a complete list of all the possible combinations? The stuff I read about combinations with repeats is not something we do in the class (or something I have ever seen done before), so is this question possible to do "without a computer"? And what about words with different numbers of repeated objects?

Let us imagine that the letters are written on slips of paper, one letter to a slip. So we have $13$ slips of paper. It is convenient to paint letters that repeat using different colours, so we have $2$ distinct $E$'s, $E_r$ (blue) and $E_r$ (red).

It is not clear whether the question asks for the probability of exactly $1$ $E$, or the probability of at least $1$ $E$. I would lean to the exactly $1$ interpretation.

We interpret "at random" to mean that all selections of $4$ slips of paper from the $13$ are equally likely. There are $\binom{13}{4}$ ways to choose $4$ slips of paper from the $13$.

Of these, $\binom{11}{3}$ have $E_b$ but not $E_r$, and $\binom{11}{3}$ have $E_r$ but not $E_b$. Thus the required probability is
$$\frac{2\binom{11}{3}}{\binom{13}{4}}.$$

Remark: At least $1$ is easiest to do by finding first the probability of the complement, no $E$. There are $\binom{11}{4}$ ways to choose $4$ slips, none of which is $E_b$ or $E_r$.

Probability would be (No. of Selections with E) / (All Possible no. of selections)

First, lets calculate 'All Possible no. of selections'

Letters of the word ENCYCLOPAEDIA are (E,E),(C,C),(A,A),N,Y,L,O,P,D,I
i.e. 3 duplicates and 7 singles.
For selections, the duplicates and singles should be considered as independent groups.

The four letters selected could have:
(a) 4 singles
(b) 2 duplicates
(c) 1 duplicate and 2 singles

(a) can happen in 10C4 ways = 210
4 out of E,C,A,N,Y,L,O,P,D,I

(b) can happen in 3C2 ways = 3
2 out of EE,CC,AA

(c) can happen in 3C1 x 9C2 ways = 108
1 out of EE,CC,AA and then
2 out of (N,Y,L,O,P,D,I and 1 each from 2 unselected doubles)

All Possible no. of selections = 210 + 3 + 108 = 321 ways

On similar principles, lets calculate 'No. of Selections with E'

Since one 'E' is always included and the other is always excluded,
we need to select 3 letters out of (C,C),(A,A),N,Y,L,O,P,D,I
i.e. 2 duplicates and 7 singles.

The three letters selected could have:
(a) 3 singles
(b) 1 duplicate and 1 single

(a) can happen in 9C3 ways = 84
3 out of C,A,N,Y,L,O,P,D,I

(b) can happen in 2C1 x 8C1 ways = 16
1 out of CC,AA and then
2 out of (N,Y,L,O,P,D,I and 1 from the unselected double)

No. of Selections with E = 84 + 16 = 100 ways

Hence, Probability = 100/321


What is the difference between a surjective and a continuous function? What is the difference between a surjective and a continuous function?
If a function is surjective then it takes all values so it is continuous and also if a function is continuous then it takes all values then it is surjective :( ?

Definition: A function $f : X\to Y$ between sets is surjective if for every $y\in Y$, there exists $x\in X$ such that $f(x) = y$. This means that every $y$ is hit by something in $x$ under $f$. As you said, this means $f$ takes on all possible values.

Definition: A function $f : (X,\tau_X)\to (Y,\tau_Y)$ between topological spaces (not just sets!) is continuous if for every $U\in\tau_Y$, $f^{-1}(U)\in\tau_X$. If you call the elements of the topologies open sets, this says that the preimage of an open set is open. This definition generalizes the definition found in calculus classes ($f : \mathbb{R}\to\mathbb{R}$ is continuous at $a\in\mathbb{R}$ if $f(a) = \lim_{x\to a}f(x)$).

There are many differences between these definitions. For one, you can talk about a function being surjective if the domain and codomain are simply sets, but you cannot talk about a function being continuous unless the domain and codomain are topological spaces. It's easy to find examples of functions that satisfy any chosen combination of these properties:

Continuous but not surjective: Consider the constant map $f : \mathbb{R}\to\mathbb{R}$ defined by $f(x) = 0$. This is continuous, but very far from surjective (for example, no $x$ satisfies $f(x) = 1$).
Surjective but not continuous: Let $f : \mathbb{R}\to\mathbb{R}$ be defined by
$$
f(x) = \begin{cases} 1/x & \text{if }x\neq 0, \\
0 & \text{if } x=0, \end{cases}
$$
This function is surjective, but not continuous (there's "bad stuff" happening at $0$!).
Continuous and surjective: Let $f : \mathbb{R}\to\mathbb{R}$ be the identity function; i.e., $f(x) = x$. $f$ is both surjective and continuous.
Neither continuous nor surjective: Let $f : \mathbb{R}\to\mathbb{R}$ be given by
$$
f(x) \begin{cases} 1 & \text{if }x\in\mathbb{Q}, \\
0 & \text{if } x\in\mathbb{R}\setminus\mathbb{Q}, \end{cases}
$$
Then $f$ is neither continuous nor surjective (the function jumps around all over the place, and only hits the values $0$ and $1$).

Function is continous at a if $\lim\limits _{x\to a}f(x) = f(a)$ by definiton.
On the other hand function $f:A \to B$ is surjective if range of f is whole set B.Thus there is no relation between continuity and surjectivity


What is the probability that in a family of 5 children, girls outnumber boys. (Assuming probability of a boy is ${\frac{1}{2}}$) How I look at this is that the total number of outcomes is $5!$ and in the given event, either there are $3,4$ or $5$ girls in the family. Now I should calculate the total number of possible outcomes where there are 3 girls and 4 girls and 5 girls and hence find the probability. But is this approach correct? Also is there any shorter approach to this question?

If for a random chosen family with $5$ children $G$ denotes the number of girls and $B$ denotes the number of boys, then - if the probability on a boy equals the probability of a girl - $(G,B)$ and $(B,G)$ have the same distribution.

This implies that: $$P(G>B)=P(B>G)$$

Next to that we have: $$P(G>B)+P(B>G)=1$$

Now draw conclusions.

You have to create a tree diagram to get all possible sample space. then see how many space have girls outnumbered. I can't draw it here.
There are 32 possible outcomes. And 16 outcomes have no of girls 3 or more.
So probability of outnumbered girls is 16/32 = 1/2

The possible outcomes are like these
BBBBB, BBBBG, BBBGB, BBBGG, BBGBB,BBGBG,BBGGB,BBGGG, ---------------GGGGG


Weighted averaging of polynomial coefficients Suppose there are three polynomials $$y_1 = a_0 + a_1x_1, $$ $$y_2 = b_0 + b_1x_2, $$ $$y_3 = c_0 + c_1x_3,$$

and define $$ x_{\rm avg} \triangleq \dfrac{x_1(x_2-x_3)^2+x_2(x_1-x_3)^2+x_3(x_1-x_2)^2}{(x_1-x_2)^2+(x_1-x_3)^2+(x_2-x_3)^2}, $$
$$ y_{\rm avg} \triangleq \dfrac{y_1(y_2-y_3)^2+y_2(y_1-y_3)^2+y_3(y_1-y_2)^2}{(y_1-y_2)^2+(y_1-y_3)^2+(y_2-y_3)^2}. $$

I want to find $d_0 = f(a_0,b_0,c_0,a_1,b_1,c_1)$ and $d_1 = f(a_0,b_0,c_0,a_1,b_1,c_1)$ such that $$y_{\rm avg} = d_0 + d_1x_{\rm avg}.$$

Is this possible? I would guess yes, but I'm not really sure. If it is, how should I solve for $d_0,d_1$ and does the solution extend to higher order polynomials? I've tried substituting the polynomials $y_1,y_2,y_3$ into the $y_{\rm avg}$ equation, but I'm unsure how to extract meaningful coefficients that are independent of $x_1,x_2,x_3$.

Here is a counterexample. Assuming $x_{\rm avg} = M$ on the diagonal $x_1 = x_2 = x_3 = M$, then $y_{\rm avg} = d_0 + d_1 x_{\rm avg} = d_0 + d_1 M$, for some fixed $M$. For simplicity, fix $a_0 = 1, b_0 = 0, c_0 = 0$ and $a_1 = 1, b_1 = 2, c_1 = 3$.

If $M = 0$ then $y_{\rm avg} = 0 = d_0$. If $M = 1$ then $y_{\rm avg} = 2 = d_1$, but if $M = 2$ then $y_{\rm avg} \approx 3.86 \neq 2d_1$.


Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$ Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$

It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?

I think I have got close to proving this but I'm not sure if this is a valid proof - but here goes...

Using AM-GM we can show that:
$$a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\tag{1}$$
$$ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{2}$$
We can also show that:
$$a^2+b^2+c^2\ge ab+bc+ca\tag{3}$$
We can therefore infer that:
$$a^2+b^2+c^2\ge ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{4}$$
Using these facts we can say that:
$$a^2+b^2+c^2=3\sqrt[3]{a^2b^2c^2}+\delta_1\text{ where }\delta_1\ge0\tag{5}$$
$$ab+bc+ca=3\sqrt[3]{a^2b^2c^2}+\delta_2\text{ where }\delta_2\ge0\tag{6}$$
We can then use (4) to further infer that:
$$\delta_1\ge\delta_2\ge0$$
Finaly we can state that:
$$\begin{align}
a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}&=6\sqrt[3]{a^2b^2c^2}+\delta_1\\
&\ge6\sqrt[3]{a^2b^2c^2}+\delta_2\\
&\ge2\left(3\sqrt[3]{a^2b^2c^2}+\frac{\delta_2}{2}\right)\\
&\ge2\left(ab+bc+ca-\frac{\delta_2}{2}\right)\\
\end{align}$$
I am hoping this aproach triggers a thought in someones brain who can then come up with the final proof.


$4$-regular graph with exactly one perfect matching Can there be a $4$-regular graph with exactly one perfect matching? That is a graph that does have a perfect matching, but not two (not necessarily disjoint) perfect matchings.

Based on a well-know result due to Kotzig, a graph with a unique perfect matching has a cut edge (see for example the book: Matching Theory by Lovasz and Plummer). But a 4-regular graph cannot have a cut edge, so it cannot have a unique perfect matching.

EDIT 2020-04-13: this suggestion of how to give an easier proof is wrong.  I am sorry. I will leave it up as it might be interesting nonetheless. I'll strike the wrong statement. (Of course, obviously, 3-regular graphs can contain a bridge. My apologies.)

Ghodrati has already given a correct proof. It is worth pointing out

  it is also possible to deduce this from Petersen's classic theorem on that every bridgeless cubic graph has a perfect matching.

(Arguably, this is a more lightweight tool that Kotzig's theorem used by Ghodrati; that's the main reason why I make this remark).

To see this, simply note that if there were a $4$-regular graph with a unique perfect matching, then removing this matching would leave a $3$-regular graph without a perfect matching; by Petersen's theorem, it would have to have a bridge;  but a $3$-regular graph obviously cannot have any bridge, as is easily seen by summing degrees. This proves that a $4$-regular graph with a unique perfect matching is impossible.


Is it possible to create a new language that translates into words at least the basic operations of math, to make it easier to understand? I was wondering if it's possible to make maths easier to calculate and understand, and more natural for kids and even adults. And then I realized how easy it is for human beings to manipulate words and from them make whole sentences and express complex concepts easily than what we do with maths.

So my question: Is it possible to create a new language that translates into words at least the basic operations of math? and if it is, how do we make those operations with words?

I hope you guys understand my question and that I'm not trying to reinvent math. I'm trying to make it easier for the daily routine and to kids and adults that have been struggling with math as myself.
Sorry for any mistakes I might made, as English isn't my native language.

Well, math is... you know, math. There are a lot of rigorous formulas that aren't really present in the english language. You could rename numbers, operations, et cetera, but at the end, the math will still exist and stay the same.
For example, $1+1=2$. If you named $1$ as "quing" and $2$ as "quang" and $+$ as "quong", then "quing quong quing is quang" would still be $1+1=2$, just in a different form.
This is what makes language different from math. In math, $1+1=2$, no matter what. But in language, you have interpetations, connotations, and all that sort of stuff.

I think there's a fundamental misunderstanding here. Language in general is often open to manipulation, but the language of mathematics points to a set of unambiguous abstractions, so there's no leeway for much else. You can reword anything in any way you choose, but unless you're simply renaming conventions, you're bound to introduce non-mathematical prose. This is actually why there's a discussion in education about what's known as disciplinary literacy: it's important for students of mathematics at almost any level to be able to properly interpret a mathematical text as a mathematical text, understand the abstractions, and become literate in mathematical language.
I think what you're actually trying to accomplish is a systematized way of explaining concepts, and while I certainly think that's a laudable goal, I also think what you're looking for lies in the study of mathematics education rather than the language of mathematics itself, since what you're looking for has to do with how best to convey to others the meaning of mathematical language.


Prove that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers Show that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers

Let $\sqrt {n^2-1}=\alpha$

Then $(n+\alpha)(n-\alpha)=1$ and it is easy to see (prove by induction) that $(n+\alpha)^k = t+u\alpha \text{ and } (n-\alpha)^k = t-u\alpha$ with $t$ and $u$ integers so: $$1=(n+\alpha)^k(n-\alpha)^k = (t+u\alpha)(t-u\alpha)= t^2-u^2\alpha^2 $$

So $$u\alpha = \sqrt{t^2-1} \text{ and } (n+\alpha)^k=t+\sqrt{t^2-1}$$


Solving difference of cubes: what rule requires multiple solutions For my own amusement, I am working through Beginning and Intermediate Algebra by Tyler Wallace.
In Example 483 ( Quadratic In Form ) we arrive at the factors
  $x^3 =1; x^3=8$

Taking the cube root of each side gives us 1;2
Both answers are 'correct' but the answer is not complete. We need to factor the difference in cubes expressions $x^3-1 and x^3-8.
What is the rule that says we have to go on to break down the difference in cubes even though it seems we have an answer to the equation.

For example:
$$x^3=1
\implies x^3-1=0$$
$$\implies (x-1)(x^2+x+1)=0 $$
Clearly, $x=1$ is a root.
The other root lies in the solution to the equation $x^2+x+1 = 0$
Of course, it's discriminant is negative, so it has no real solutions, but you can find complex solutions using the quadratic formula:
$\frac{-1+\sqrt{3}i}{2},\frac{-1-\sqrt{3}i}{2}$

$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
We can prove it by expending.
So $$x^3-8=(x-2)(x^2+2x+4)=(x-2)((x+1)^2+3).$$
I hope it what you are looking  for.


How do you prove$(a^m)^{-1} = (a^{-1})^m$ for a being any non-zero real and m being a positive integer from the axioms of real numbers? How do you prove $(a^m)^{-1} = (a^{-1})^m$ for $a$ being any non-zero real and $m$ being a positive integer, from the axioms of real numbers only?

You should use induction. The base case is $m=1$ and the relation to be proved is
$$
(a^1)^{-1}=(a^{-1})^1
$$
which is true.

Suppose the statement holds for $m$; then
\begin{align}
(a^{m+1})^{-1}
&=(a^ma)^{-1} &&\text{definition of powers}\\
&=(a^m)^{-1}a^{-1} &&\text{because $(xy)^{-1}=x^{-1}y^{-1}$} \\
&=(a^{-1})^ma^{-1} &&\text{induction hypothesis} \\
&=(a^{-1})^{m+1} &&\text{definition of powers}
\end{align}

The proof of $(xy)^{-1}=x^{-1}y^{-1}$:
\begin{align}
(xy)(x^{-1}y^{-1})
&=(yx)(x^{-1}y^{-1}) &&\text{commutativity} \\
&=y\bigl(x(x^{-1}y^{-1})\bigr) &&\text{associativity} \\
&=y\bigl((xx^{-1})y^{-1}\bigr) &&\text{associativity} \\
&=\dots \\
&=1
\end{align}
which you can finish.


Uniqueness of coefficients in infinite sin series $a(x) = \sum\limits_{n=1}^{\infty} a_{i} \sin(n x) =0 $ if we can say

$$a(x) = \sum\limits_{n=1}^{\infty} a_{n} \sin(n x) =0, \forall x$$

does this generally imply that all constant coefficients $a_i$ should be zero, or can I construct any number of solutions that may provide the right cancellations to make it zero?

My intuition for superposition of different modes of $\sin(nx)$ is that if $a_i$ is always same-signed (I might be able to show this in my application), then this should be unique, but if each coefficient can have a different sign then it could be anything.

I was looking for someone to confirm or deny what I was thinking, and maybe point me towards more mathematical ideas about why that is so.

Let me expand the comments in an answer: there are two concepts here - trigonometric series and Fourier series:

Let $S$ be $\frac{1}{2}a_0+\sum_{n \geq 1}{(a_n\cos(nx) + b_n\sin(nx))}$, where $a_n, b_n$ are complex numbers; then S is called a trigonometric series and the Riemann theory of such series tries to associate to $S$ some meaning and deal with issues of unicity of the meaning etc.

IF $a_n, b_n$ come from an absolutely integrable (Lebesgue) periodic function $f$ on $[0, 2\pi]$ by the usual formulas $a_n = \frac{1}{\pi}\int_{0}^{2\pi} f(x)\cos(nx)dx$ and $b_n = \frac{1}{\pi}\int_{0}^{2\pi} f(x)\sin(nx)dx$, S is called a Fourier series and it satisfies some properties like $a_n, b_n$ converge to zero (Riemann-Lebesgue lemma), $S$ is Cesaro (also called Feijer in this context) and Abel (also called radial) summable to $f$ though it may not be convergent anywhere to $f$ (Kolmogorov 1920's or so), while if $f$ is continuos, a famous theorem of Carleson (1960's) shows that $S$ converges a.e to $f$ (while a classic theorem of Feijer (1910's) shows that $S$ is uniformly Cesaro (and Abel) summable to $f$.

The distinction between trigonometric and Fourier series is non-trivial even if we are given that $a_n, b_n$ converge to zero in a positive decreasing manner as the series $\sum_{n \geq 2}\frac{\sin(nx)}{\log n}$ is convergent everywhere, uniformly convergent on any compact set that doesn't contain an integral multiple of $2\pi$ including zero, so its sum is a continuous function except at $0$ and integral multiples of $2\pi$, but it is NOT a Fourier series; on the other hand its conjugate cosine sum $\sum_{n \geq 2}\frac{\cos(nx)}{\log n}$ is uniformly convergent on compact sets outside integral multiples of $2\pi$, $\infty$ at such and its sum is integrable so the cosine series IS a Fourier series.

In general, if $S$ can be associated to a function $f$ in a strong enough manner (absolute convergence, uniform convergence, $L^p$ convergence, $p \geq 1$, other more esoteric forms of convergence) then $S$ is the Fourier series of $f$, so people generally study only Fourier series for that reason but as noted pointwise convergence or a.e uniform convergence are not good enough.

As unicity goes: for Fourier series a.e convergence is enough to give it - so if we know apriori that a trigonometric series comes from an integrable function by the formulas above and the series converges a.e to $0$, all the coefficients are $0$. However the proof requires work (follows from Feijer theorem) as simple integration of $S$ against $\cos(nx), \sin(nx)$ and the use of the ortogonality relations for those to conclude is not justified unless we have strong assumptions on $S$ like uniformly bounded partial sums a.e etc

However for a general trigonometric series while convergence on a set of non-zero measure implies $a_n, b_n$ going to zero (Cantor-Lebesgue lemma), the rest of the issues are considerably more complicated and they are treated by what is called the Riemann theory of trigonometric series which was at its apogee in the 1920's when the Russian school led by N Bary obtained almost definitive results regarding unicity and the like; in essence convergence everywhere or even on a set with countable complement implies unicity (so indeed the answer to the original question is true) but the proof is complicated (though known from Riemann), while for more general sets of zero measure, convergence a.e outside them may or may not imply unicity and while a full classification is not quite known as the property  of such to be "unicity sets" seem to depend on arithmetic not topological properties, lots of results are known. However simple summability (Abel, Cesaro) etc is not very useful here as for example the Poisson kernel is Abel summable to $0$ everywhere except at $1$ but its coefficients definitely do not go to zero at infinity.

Note that $$\int_0^{2\pi}\sin(m\theta)\sin(n\theta)\,d\theta=\frac12\int_0^{2\pi}\cos((n-m)\theta) - \cos((n+m)\theta)\,d\theta = 0$$if $m\ne \pm n$.
And when $m = n$, the integral is $\pi$.

Now, what is $\int_0^{2\pi}a(x)\sin(mx)\,dx$?


Given the transfinite hierarchy of Borel sets, prove that ${\bf \Sigma}_\alpha^0 \subseteq {\bf \Sigma}_{\alpha+1}^0$ for all ordinal $\alpha < \beta$ We define the transfinite hierarchy of Borel sets $\langle {\bf \Sigma}^0_\alpha, {\bf \Pi}^0_\alpha \rangle_{\alpha \in \rm{Ord}}$ as follows:

$$\begin{aligned} &\begin{cases}
{\bf \Sigma}^0_1 &= \{B \subseteq \mathbb R \mid B\text{ is open}\}\\
{\bf \Pi}^0_1    &= \{B \subseteq \mathbb R \mid B\text{ is closed}\}\end{cases}\\
&\begin{cases}
{\bf \Sigma}^0_{\alpha + 1} &= \{\bigcup_{n \in \mathbb N}B_n \mid \forall n \in \mathbb N: B_n \in {\bf \Pi}^0_\alpha\}\\
{\bf \Pi}^0_{\alpha + 1}    &= \{\bigcap_{n \in \mathbb N}B_n \mid \forall n \in \mathbb N: B_n \in {\bf \Sigma}^0_\alpha\}\end{cases} \text{ for all ordinal } \alpha\\
&\begin{cases}
{\bf \Sigma}^0_\alpha &= \{\bigcup_{n \in \mathbb N}B_n \mid \forall n \in \mathbb N: B_n \in \bigcup^0_{\xi < \alpha}{\bf \Pi}^0_\xi\}\\
{\bf \Pi}^0_\alpha    &= \{\bigcap_{n \in \mathbb N}B_n \mid \forall n \in \mathbb N: B_n \in \bigcup^0_{\xi < \alpha}{\bf \Sigma}^0_\xi\}
\end{cases} \text{ for all limit ordinal } \alpha \end{aligned}$$

  Prove that ${\bf \Sigma}_\alpha^0 \subseteq {\bf \Sigma}_\beta^0 \quad {\bf \Sigma}_\alpha^0 \subseteq {\bf \Pi}_\beta^0 \quad {\bf \Pi}_\alpha^0 \subseteq {\bf \Pi}_\beta^0 \quad {\bf \Pi}_\alpha^0 \subseteq {\bf \Sigma}_\beta^0$ for all ordinals $\alpha < \beta$

Is my proof correct? Or have I missed something that makes an inductive argument necessary? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:

It follows directly from the construction of ${\bf \Sigma}^0_{\alpha + 1}$ and ${\bf \Pi}^0_{\alpha + 1}$ that ${\bf \Pi}^0_\alpha \subseteq {\bf \Sigma}^0_{\alpha + 1}$ and ${\bf \Sigma}^0_\alpha \subseteq {\bf \Pi}^0_{\alpha + 1}$. Hence ${\bf \Pi}_\alpha^0 \subseteq {\bf \Sigma}_\beta^0$ and ${\bf \Sigma}_\alpha^0 \subseteq {\bf \Pi}_\beta^0$ for all $\alpha < \beta$.

We know that every open set is a countable union of open intervals with rational endpoints and that every open interval is a countable union of closed intervals. Hence every open set is a countable union of closed intervals (and thus closed sets). It follows by De Morgan's laws that every closed set is a countable intersection of open set. Thus ${\bf \Sigma}_1^0 \subseteq {\bf \Sigma}_2^0$ and ${\bf \Pi}_1^0 \subseteq {\bf \Pi}_2^0$.

We proceed to prove that ${\bf \Sigma}_\alpha^0 \subseteq {\bf \Sigma}_{\alpha+1}^0$ and ${\bf \Pi}_\alpha^0 \subseteq {\bf \Pi}_{\alpha+1}^0$ for all $\alpha$ by transfinite induction. We previously prove it for $\alpha=1$. Let the assertion hold for $\alpha$, i.e. ${\bf \Sigma}_\alpha^0 \subseteq {\bf \Sigma}_{\alpha+1}^0$ and ${\bf \Pi}_\alpha^0 \subseteq {\bf \Pi}_{\alpha+1}^0$. It follows that $\{\bigcap_{n \in \mathbb N}B_n \mid \forall n \in \mathbb N: B_n \in {\bf \Sigma}^0_\alpha\} \subseteq \{\bigcap_{n \in \mathbb N}B_n \mid \forall n \in \mathbb N: B_n \in {\bf \Sigma}^0_{\alpha+1}\}$ and thus ${\bf \Pi}_{\alpha+1}^0 \subseteq {\bf \Pi}_{\alpha+2}^0$.

Similarly, $\{\bigcup_{n \in \mathbb N}B_n \mid \forall n \in \mathbb N: B_n \in {\bf \Pi}^0_\alpha\} \subseteq \{\bigcup_{n \in \mathbb N}B_n \mid \forall n \in \mathbb N: B_n \in {\bf \Pi}^0_{\alpha+1}\}$ and thus ${\bf \Sigma}_{\alpha+1}^0 \subseteq {\bf \Sigma}_{\alpha+2}^0$. This completes the proof.

You didn't prove that $\Sigma^0_\alpha$ $\subseteq$ $\Sigma^0_{\alpha + 1}$, $\Pi^0_\alpha$ $\subseteq$ $\Pi^0_{\alpha+1}$ for all ordinal $\alpha$ since what you have done is mere induction for all natural numbers, not transfinite induction. To prove this statement by transfinite induction, you need to show that if $\alpha$ is a limit ordianl and it holds for all $\xi \lt \alpha$, then it holds for $\alpha$.(and the argument you have given above cannot be applied to this case)

Instead of seperating the orignal statement into two parts and proving them independently of each oher, one can prove the whole statement at once and I personally find it easier. I give my proof below.

  we proceed by induction on $\beta$

  1) $\beta$ = 2

  trivial

  2) $\beta$ = $\xi$ + 1

  and
  $\Sigma^0_\alpha$ $\subseteq$ $\Sigma^0_\xi$,
  $\Sigma^0_\alpha$ $\subseteq$ $\Pi^0_\xi$,
  $\Pi^0_\alpha$ $\subseteq$ $\Sigma^0_\xi$,
  $\Pi^0_\alpha$ $\subseteq$ $\Pi^0_\xi$ holds for all $\alpha$ $\lt$ $\xi$

  $\alpha$ $\lt$ $\beta$ implies $\alpha$ $\le$ $\xi$

  i) $\alpha$ $\lt$ $\xi$

  $\Sigma^0_\alpha$ $\subseteq$ $\Sigma^0_\xi$ $\subseteq$ $\Pi^0_\beta$

  $\Sigma^0_\alpha$ $\subseteq$ $\Pi^0_\xi$ $\subseteq$ $\Sigma^0_\beta$

  $\Pi^0_\alpha$ $\subseteq$ $\Sigma^0_\xi$ $\subseteq$ $\Pi^0_\beta$

  $\Pi^0_\alpha$ $\subseteq$ $\Pi^0_\xi$ $\subseteq$ $\Sigma^0_\beta$

  ii) $\alpha$ = $\xi$

  $\Sigma^0_\alpha$ $\subseteq$ $\Pi^0_\beta$

  $\Pi^0_\alpha$ $\subseteq$ $\Sigma^0_\beta$

  ii-1) $\xi$ = $\gamma$ +1 (succesor ordinal)

  $\Sigma^0_\xi$ = {$\bigcup_{n\in\omega}B_n$ : $B_n \in \Pi^0_\gamma$} $\subseteq$ {$\bigcup_{n\in\omega}B_n$ : $B_n \in \Pi^0_\xi$} = $\Sigma^0_{\xi +1}$

  $\Pi^0_\xi$ = {$\bigcap_{n\in\omega}B_n$ : $B_n \in \Sigma^0_\gamma$} $\subseteq$ {$\bigcap_{n\in\omega}B_n$ : $B_n \in \Sigma^0_\xi$} = $\Pi^0_{\xi +1}$

  ii-2) $\xi$ : limit ordinal

  $\Sigma^0_\xi$ = {$\bigcup_{n\in\omega}B_n$ : $B_n \in \Pi^0_\gamma$ for some $\gamma \lt \xi$} $\subseteq$ {$\bigcup_{n\in\omega}B_n$ : $B_n \in \Pi^0_\xi$} = $\Sigma^0_{\xi +1}$

  $\Pi^0_\xi$ = {$\bigcap_{n\in\omega}B_n$ : $B_n \in \Sigma^0_\gamma$ for some $\gamma \lt \xi$} $\subseteq$ {$\bigcap_{n\in\omega}B_n$ : $B_n \in \Sigma^0_\xi$} = $\Pi^0_{\xi +1}$

  hence

  $\Sigma^0_\alpha$ $\subseteq$ $\Sigma^0_\beta$ and

  $\Pi^0_\alpha$ $\subseteq$ $\Pi^0_\beta$

  3) $\beta$ : limit ordinal (not zero)

  and
  $\Sigma^0_\alpha$ $\subseteq$ $\Sigma^0_\xi$,
  $\Sigma^0_\alpha$ $\subseteq$ $\Pi^0_\xi$,
  $\Pi^0_\alpha$ $\subseteq$ $\Sigma^0_\xi$,
  $\Pi^0_\alpha$ $\subseteq$ $\Pi^0_\xi$ holds for all $\alpha$ $\lt$ $\xi$,  $\xi$ $\lt$ $\beta$

  then

  $\alpha$ $\lt$ $\beta$ ($\alpha$ $\lt$ $\xi$ for some $\xi$ $\lt$ $\beta$) implies

  $\Sigma^0_\alpha$ $\subseteq$ $\Pi^0_\beta$

  $\Sigma^0_\alpha$ $\subseteq$ $\Pi^0_\xi$ $\subseteq$ $\Sigma^0_\beta$

  $\Pi^0_\alpha$ $\subseteq$ $\Sigma^0_\xi$ $\subseteq$ $\Pi^0_\beta$

  $\Pi^0_\alpha$ $\subseteq$ $\Sigma^0_\beta$

  the induction step is now completed.

  $\therefore$  $\Sigma^0_\alpha \subseteq \Sigma^0_\beta$, $\Sigma^0_\alpha$ $\subseteq \Pi^0_\beta$, $\Pi^0_\alpha \subseteq \Sigma^0_\beta$, $\Pi^0_\alpha \subseteq \Pi^0_\beta$ for all ordinals $\alpha \lt \beta$


On the sequence of positive integers satisfying $\sigma(n)\mid (n(\sigma_0(n))^2)$ If my calculations with my computer were rights, when I've consider the sequence of integers $n\geq 1$ satisfying  $$\sigma(n)\mid (n(\sigma_0(n))^2),$$
where $\sigma(n)=\sum_{d\mid n}d$ is the sum of divisors function and $\sigma_0(n)=\sum_{d\mid n}1$ the number of such divisors, then such sequence starts as

  1, 3, 6, 15, 28, 33, 42, 84, 91...

but I believe that such sequence isn't in The On-Line Encyclopedia of Integer Sequence.

  Question. Are right my first terms? Is known such sequence in previous encyclopedia or are there references in the literature?

I've curiosity to know what's about the number of terms in the sequence for the first $10^k$ with $k=4,5,6,\cdots, N$, thats is a plot of the counting function for previous sequence.

  Question. Can you provide us a graph of the counting function of such sequence for $10^N$, for a $N$ large? Thanks in advance.

OEIS refer to section B2 in Guy, Unsolved Problems in Number Theory, for types of "semi" perfect numbers. They do define the harmonic numbers but do not say much more. Anyway, your sequence is a superset of the harmonic numbers (named by Pomerance in 1973), also called Ore numbers.

The harmonic numbers begin
$$  1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128, 8190, 18600, 18620, 27846, 30240, 32760, 55860, 105664,   $$
and are discussed at https://oeis.org/A001599


Prove that $x$ is congruent to $y \pmod m$ if and only if $x = km + y$ Prove that $x$ is congruent to $y \pmod m$ if and only if $x = km + y$

I'm not sure how to prove this, and honestly I don't really understand why it's true in the first place.

If $x \equiv y \pmod m$, then after division of $x$ by $m$, I have a remainder of $y$, assuming $0 \leq y < m$. So, via the division algorithm, $x = mq +y$.
The other direction is trivial using the division algorithm.


Is the image of a contractible space under a contraction contractible? Is the image of a contractible space under a contraction contractible?

More specific to what I'm looking for: is the image of the unit ball $B_1(0)\subset\mathbb{R}^n$ under a contraction mapping $F:\mathbb{R}^n\to\mathbb{R}^n$ still contractible?

No. For instance, the map
\begin{align*}
F: \mathbb{R}^2 &\to \mathbb{R}^2 \\
(x,y) &\mapsto \frac{1}{4\pi}\left(\cos(2\pi x), \sin(2\pi x)\right)
\end{align*}
takes the unit ball to a circle, which is not contractible. That the map is a contraction can be seen by the derivative, or by noticing that it is the composition of the projection in the first coordinate with a walk on the circle with constant velocity equal to $1/2$. (This should make for a quick visualization that it is indeed a contraction.)

The map $H_t(x)=tF(x), t\in [0,1]$ restricted to the unit ball defines a deformation retract of the unit ball to a point.