Proving $<$ is transitive on $\mathbb{Q}$. I feel a little bit stupid asking this;

I am asked to prove that, for all rational numbers if, x < y and y < z then x < z.

I have said this;

$ x + 0 < y $

$ x - z + z < y$

$ x - z < y- z $

but $ y - z < 0$

so $ x - z < y- z $ implies $ x - z < 0 $.

I had a little search before posting this just to make sure its not a duplicate, if it is i will delete it right away, sorry and thanks in advance.

ORDERING OF THE RATIONALS:

Let x and y be rational numbers. We say that x > y iff x - y is a positive rational number, and x < y iff x - y is a negative rational number.

I think this is the information that was missing.

We have $$z-x=(z-y)+(y-x)$$
We know $z-y$ is a positive rational number, since $y<z$.  We also know that $y-x$ is a positive rational number, since $x<y$.  We now need some sort of property that the sum of two positive rational numbers is again a positive rational number.  With that property we know that $z-x$ is a positive rational number, and hence $x<z$.

As Peter Franek pointed out in his comment, your proof is flawed.

You used if $x-z<y-z$ and $y-z<0$ then $x-z<0$

This is the same as saying if $x<y$ and $y<z$ then $x<z$.

All you have to do to see that is set $x'=x-z,y'=y-z,z'=0$

As for how to prove this, use the axioms of the real number system.

By the axiom:

$x\leq y$ and $y\leq z$ imply $x\leq z$ for any real numbers $x,y,z$.

Now, say $x<y$ and $y<z$, it's the same as $x\leq y$ and $y\leq z$ and $x\neq y$ and $x\neq z$.

This implies that $x\leq z$. What we have left to prove is that $y\neq z$

Assume $y=z$, then $x\leq y$ and $y\leq x$ and $x\neq y$, that is $x=x$ and $x\neq x$ which leads to a contradiction. Thus, $x\neq z$.

This ends the proof.


Rationals in [0,1] are $F_{\sigma}$? I have the following problem that I think I have worked out.

  Let $A$ be the set of rational numbers in $[0,1]$. Is $A$ an $F_{\sigma}$ set?

My Attempt: Yes, $A$ is an $F_{\sigma}$ set. We recall that an $F_{\sigma}$ set is one which may be expressed as the countable union of closed sets. Since the set of rationals $\mathbb{Q}$ are countable, so too is the set $A$. Hence, we may enumerate it as $A=\{q_{k}\}_{k=1}^{\infty}$. Now, each singleton set $\{q_{k}\}$ is closed in $\mathbb{R}$ (with the standard topology), since its complement $\{q_{k}\}^{c}=(-\infty,q_{k})\cup(q_{k},\infty)$ is open. Thus, we have $A=\bigcup_{k=1}^{\infty}\{q_{k}\}$, so $A$ is indeed an $F_{\sigma}$ set.

My concerns: I think my above argument is okay, but I am shaky on one part. Does it matter if I consider each $\{q_{k}\}$ as closed in $\mathbb{R}$, or do I need to show explicitly that they are closed in $[0,1]$? Thanks in advance for any help!

If $S$ is closed in $\Bbb R$, then $S\cap X$ is closed in $X$ for all $X\subseteq \Bbb R$, so showing closedness in $\Bbb R$ is sufficient.

Yes it is a set. What you did is sufficient. Good job!


Finding equilibrium in a predator prey system Using predator prey system

$\frac{dR}{dt}=6R\:-2RW$

$\frac{dW}{dt}=-4W\:+5RW$

$When\:the\:system\:is\:equilibrium\:with\:W\ne 0,\:R\:\ne 0\:then\:RW\:=?$

Apologize for formatting

you are looking for the non trivial special solutions where $w = constant, r = constant.$  that is $$r \neq 0, w \neq 0, 2r(3 - w) = 0 \text{ and } w(-4 + 5r)=0 $$ that is $$ w = 3, r = \frac 45.$$

"When the derivatives are 0 and it's an equation, use Cramer and some determination!" (It's a linear equation, solve for R and W and multiply them together)

"I like the rhyme because it's sublime"


Bounding ${(2d-1)n-1\choose n-1}$ Claim: ${3n-1\choose n-1}\le 6.25^n$.

Why?
Can the proof be extended to
obtain a bound on ${(2d-1)n-1\choose
    n-1}$, with the bound being $f(d)^n$
for some function $f$?

(These numbers describe the number of some $d$-dimensional combinatorial objects; claim 1 is the case $d=2$, and is not my claim).

First, lets bound things as easily as possible.  Consider the inequality $$\binom{n}{k}=\frac{(n-k)!}{k!}\leq\frac{n^{k}}{k!}\leq e^{k}\left(\frac{n}{k}\right)^{k}.$$ The $n^k$ comes from the fact that $n$ is bigger then each factor of the product in the numerator.  Also, we know that $k!e^k>k^k$ by looking at the $k^{th}$ term in the Taylor series, as $e^k=1+k+\cdots +\frac{k^k}{k!}+\cdots $.

Now, lets look at the similar $3n$ and $n$ instead of $3n-1$ and $n-1$.  Then we see that $$\binom{3n}{n}\leq e^{n}\left(3\right)^{n}\leq\left(8.15\right)^{n}$$and then for any $k$ we would have $$\binom{kn}{n}\leq\left(ke\right)^{n}.$$

We could use Stirlings formula, and improve this more.  What is the most that this can be improved?  Apparently, according to Wolfram the best possible is $$\binom{(k+1)n}{n}\leq \left(\frac{(k+1)^{k+1}}{k^k}\right)^n.$$

(Notice that when $k=2$ we have $27/4$ which is $6.25$)

Hope that helps.

Just use the Stirling formula
$$n! \sim \sqrt{2\pi n} (n/e)^n$$
for large $n$ and neglect the $\sqrt{2\pi n}$ factor for a while. That gives us a good estimate for your expression
$$(3n-1)! / [ (n-1)!(2n)! ] \sim (2n)^{-2n}(n-1)^{1-n} (3n-1)^{3n-1} $$
Note that the powers of $e$ cancel. For large $n$, you may also approximate the bases of the powers simply by $2n,n,3n$, respectively. Then you get
$$\sim 2^{-2n} 3^{3n} = (27/4)^{n} $$
because the powers of $n$ cancel, too. Note that $27/4 = 6.75$. I have only calculated the estimate - which is actually a better result because it shows that the number $6.75$ is exact in the large $n$ limit. To prove the inequality, you have to carefully watch whether the Stirling formula underestimates or overestimates it. At any rate, it's straightforward to check that the inequality holds for any positive $n$.

It's enough to check it explicitly for a few first small values of $n$, and for larger ones, it can be shown that the $(27/4)^n$ Ansatz is being approached from the right side by computing the sign of the first subleading correction to this approximation. If you enumerate the first few binomial coefficients over the approximation (which should be smaller than one) by Mathematica

[Table[Binomial[3 n - 1, n - 1]/(27/4)^n, {n, 1, 20}]]

you will get

  {0.14814815, 0.10973937, 0.091043032, 0.079482012, 0.071435685, 0.06542258, 0.060709598, 0.056887142, 0.053706089, 0.051005081, 0.048674402, 0.046636504, 0.04483482, 0.043226987, 0.041780567, 0.040470244, 0.039275935, 0.038181474, 0.037173681, 0.036241691}

The numbers are clearly smaller than one, and from the beginning, the decrease is uniform.

For your case of $d$ dimensions, the relevant surviving terms are replaced by
$$ \sim (2d-2)^{(2-2d)(n-1)} (2d-1)^{(2d-1)(n-1)} $$
so $27/4$ gets replaced by $(2d-1)^{2d-1}/(2d-2)^{2d-2}$. For $d=2$, you get $3^3/2^2 = 27/4$. For $d=3$, you would get $5^5/4^4$, and so on.

Note that the proof only works for 27/4 = 6.75. This number couldn't be reduced further (6.25 is a typo) and any proof that replaces 6.75 by a larger number fails to prove the original assertion.


Mathematician vs. Computer: A Game A mathematician and a computer are playing a game: First, the mathematician chooses an integer from the range $2,...,1000$. Then, the computer chooses an integer uniformly at random from the same range. If the numbers chosen share a prime factor, the larger number wins. If they do not, the smaller number wins. (If the two numbers are the same, the game is a draw.)

Which number should the mathematician choose in order to maximize his chances of winning?

For fixed range:

range = 16;
a = Table[Table[FactorInteger[y][[n, 1]], {n, 1, PrimeNu[y]}], {y, 1, range}];
b = Table[Sort@DeleteDuplicates@ Flatten@Table[
Table[Position[a, a[[y, m]]][[n, 1]],
{n, 1, Length@Position[a, a[[y, m]]]}], {m, 1, PrimeNu[y]}], {y, 1, range}];
c = Table[Complement[Range[range], b[[n]]], {n, 1, range}];
d = Table[Range[n, range], {n, 1, range}];
e = Table[Range[1, n], {n, 1, range}];
w = Table[DeleteCases[DeleteCases[Join[Intersection[c[[n]], e[[n]]],
Intersection[b[[n]], d[[n]]]], 1], n], {n, 1, range}];
l = Table[DeleteCases[DeleteCases[Complement[Range[range], w[[n]]], 1],
n], {n, 1, range}];
results = Table[Length@l[[n]], {n, 1, range}];
cf = Grid[{{Join[{"n"}, Rest@(r = Range[range])] // ColumnForm,
Join[{"win against n"}, Rest@w] // ColumnForm,
Join[{"lose against n"}, Rest@l] // ColumnForm,
Join[{"probability win for n"}, (p = Drop[Table[
results[[n]]/Total@Drop[results, 1] // N,{n, 1, range}], 1])] // ColumnForm}}]
Flatten[Position[p, Max@p] + 1]

isn't great code, but fun to play with for small ranges, gives

and perhaps more illuminating

rr = 20; Grid[{{Join[{"range"}, Rest@(r = Range[rr])] // ColumnForm,
Join[{"best n"}, (t = Rest@Table[
a = Table[Table[FactorInteger[y][[n, 1]], {n, 1, PrimeNu[y]}], {y, 1, range}];
b = Table[Sort@DeleteDuplicates@Flatten@Table[Table[
Position[a, a[[y, m]]][[n, 1]], {n, 1,Length@Position[a, a[[y, m]]]}],
{m, 1,PrimeNu[y]}], {y, 1, range}];
c = Table[Complement[Range[range], b[[n]]], {n, 1, range}];
d = Table[Range[n, range], {n, 1, range}];
e = Table[Range[1, n], {n, 1, range}];
w = Table[DeleteCases[DeleteCases[Join[Intersection[c[[n]], e[[n]]],
Intersection[b[[n]], d[[n]]]], 1], n], {n, 1, range}];
l = Table[DeleteCases[DeleteCases[Complement[Range[range], w[[n]]], 1], n],
{n,1, range}];
results = Table[Length@l[[n]], {n, 1, range}];
p = Drop[Table[results[[n]]/Total@Drop[results, 1] // N,
{n, 1, range}], 1];
{Flatten[Position[p, Max@p] + 1], Max@p}, {range, 1, rr}]/.Indeterminate-> draw);
Table[t[[n, 1]], {n, 1, rr - 1}]] // ColumnForm,
Join[{"probability for win"}, Table[t[[n, 2]], {n, 1, rr - 1}]] // ColumnForm}}]

compares ranges:

Plotting mean "best $n$" against $\sqrt{\text{range}}$ gives

For range=$1000,$ "best $n$" are $29$ and $31$, which can be seen as maxima in this plot:

Update

In light of DanielV's comment that a "primes vs winchance" graph would probably be enlightening, I did a little bit of digging, and it turns out that it is. Looking at the "winchance" (just a weighting for $n$) of the primes in the range only, it is possible to give a fairly accurate prediction using

range = 1000;
a = Table[Table[FactorInteger[y][[n, 1]], {n, 1, PrimeNu[y]}], {y, 1, range}];
b = Table[Sort@DeleteDuplicates@Flatten@Table[
   Table[Position[a, a[[y, m]]][[n, 1]], {n, 1,
     Length@Position[a, a[[y, m]]]}], {m, 1, PrimeNu[y]}], {y, 1, range}];
c = Table[Complement[Range[range], b[[n]]], {n, 1, range}];
d = Table[Range[n, range], {n, 1, range}];
e = Table[Range[1, n], {n, 1, range}];
w = Table[    DeleteCases[    DeleteCases[
 Join[Intersection[c[[n]], e[[n]]], Intersection[b[[n]], d[[n]]]],
  1], n], {n, 1, range}];
l = Table[
DeleteCases[DeleteCases[Complement[Range[range], w[[n]]], 1],
n], {n, 1, range}];
results = Table[Length@l[[n]], {n, 1, range}];
p = Drop[Table[
results[[n]]/Total@Drop[results, 1] // N, {n, 1, range}], 1];
{Flatten[Position[p, Max@p] + 1], Max@p};
qq = Prime[Range[PrimePi[2], PrimePi[range]]] - 1;
Show[ListLinePlot[Table[p[[t]] range, {t, qq}],
DataRange -> {1, Length@qq}],
ListLinePlot[
Table[2 - 2/Prime[x] - 2/range (-E + Prime[x]), {x, 1, Length@qq + 0}],
PlotStyle -> Red], PlotRange -> All]

The plot above (there are $2$ plots here) show the values of "winchance" for primes against a plot of $$2+\frac{2 (e-p_n)}{\text{range}}-\frac{2}{p_n}$$

where $p_n$ is the $n$th prime, and "winchance" is  the number of possible wins for $n$ divided by total number of possible wins in range ie $$\dfrac{\text{range}}{2}\left(\text{range}-1\right)$$ eg $499500$ for range $1000$.

Show[p // ListLinePlot,  ListPlot[N[
Transpose@{Prime[Range[PrimePi[2] PrimePi[range]]],
 Table[(2 + (2*(E - Prime[x]))/range - 2/Prime[x])/range, {x, 1,
   Length@qq}]}], PlotStyle -> {Thick, Red, PointSize[Medium]},
DataRange -> {1, range}]]

Added

Bit of fun with game simulation:

games = 100; range = 30;
table = Prime[Range[PrimePi[range]]];
choice = Nearest[table, Round[Sqrt[range]]][[1]];
y = RandomChoice[Range[2, range], games];  z = Table[
Table[FactorInteger[y[[m]]][[n, 1]], {n, 1, PrimeNu[y[[m]]]}], {m, 1, games}];
Count[Table[If[Count[z, choice] == 0 && y[[m]] < choice \[Or]
Count[z, choice] > 0 && y[[m]] < choice, "lose", "win"],
{m, 1, games}], "win"]

& simulated wins against computer over variety of ranges

with

Clear[range]
highestRange = 1000;
ListLinePlot[Table[games = 100;
table = Prime[Range[PrimePi[range]]];
choice = Nearest[table, Round[Sqrt[range]]][[1]];
y = RandomChoice[Range[2, range], games];
z = Table[Table[FactorInteger[y[[m]]][[n, 1]], {n, 1, PrimeNu[y[[m]]]}], {m,
  1, games}];
Count[Table[ If[Count[z, choice] == 0 && y[[m]] < choice \[Or]
  Count[z, choice] > 0 && y[[m]] < choice, "lose", "win"], {m, 1,
 games}], "win"], {range,2, highestRange}], Filling -> Axis, PlotRange-> All]

Added 2

Plot of mean "best $n$" up to range$=1000$ with tentative conjectured error bound of $\pm\dfrac{\sqrt{\text{range}}}{\log(\text{range})}$ for range$>30$.

I could well be wrong here though. - In fact, on reflection, I think I am (related).

The mathematician should choose 990.  It has the most prime factors of 2, 3 ,5, 11 at the high end of numbers to 1000.
Since the computer is going to pick a random number, there is a high probability (large set)
1.  That the number picked will be less than 990
2.  That it will get an easily factored number containing 2,3,5 or 11 (there are many).

The mathematician does not want to be in this large set, and should win most games.


Show that the random variables $X$ and $Y$ are uncorrelated but not independent Show that the random variables $X$ and $Y$ are uncorrelated but not
  independent

  The given joint density is

  $f(x,y)=1\;\; \text{for } \; -y<x<y \; \text{and } 0<y<1$, otherwise
  $0$

My main concern here is how should we calculate $f_1(x)$

$f_1(x)=\int_y dy = \int_{-x}^{1}dy + \int_{x}^{1}dy = 1+x +1=2\; \; \forall -1 <x<1$

OR Should we do this?

$f_1(x)$=$$ \begin{cases}
      \int_{-x}^{1}dy = 1+x &&  -1<x<0  \\
      \int_{x}^{1}dy = 1-x & & 0\leq x <1 \\
   \end{cases}
$$

In the second case, how do I show they are not independent.

I can directly say that the joint distribution does not have a product space but I want to show that $f(x,y)\neq f_1(x)f_2(y)$

Also, for anyone requiring further calculations,

$f_2(y) = \int dx = \int_{-y}^{y}dx = 2y$

$\mu_2= \int y f_2(y)dy = \int_{0}^{1}2y^2 = \frac23$

$\sigma_2 ^2 = \int y^2f_2(y)dy - (\frac23) ^2 = \frac12 - \frac49 = \frac1{18}$

$E(XY)= \int_{y=0}^{y=1}\int_{x=-y}^{x=y} xy f(x,y)dxdy =\int_{y=0}^{y=1}\int_{x=-y}^{x=y} xy dxdy$ which seems to be $0$? I am not sure about this also.

$f_1(x)=1+x$ if $-1<x<0$ and $1-x$ if $0<x<1$. ( In other words $f_1(x)=1-|x|$ for $|x|<1$). As you have observed $f_2(y)=2y$ for $0<y<1$. Now it is basic fact that if the random variables are independent then we must have $f(x,y)=f_1(x)f_2(y)$ (almost everywhere). Since the equation $(1-|x|)(2y)=f(x,y)$ is not true we can conclude that $X$ and $Y$ are not independent.

$EXY=0$ is correct. Also $EX=\int_{-1}^{1}x(1-|x|)dx=0$ so $X$ and $Y$ are uncorrelated.

A slightly different approach:

Joint density of $(X,Y)$ is

\begin{align}
f(x,y)&=1_{-y<x<y\,,\,0<y<1}
\\&=\underbrace{\frac{1_{-y<x<y}}{2y}}_{f_{X\mid Y=y}(y)}\cdot\underbrace{2y\,1_{0<y<1}}_{f_Y(y)}
\end{align}

Since (conditional) distribution of $X$ 'given $Y$' depends on $Y$, clearly $X$ and $Y$ are not independent. In fact, $X$ conditioned on $Y=y$ has a uniform distribution on $(-y,y)$, which gives $E\,[X\mid Y]=0$.

Therefore, by law of total expectation,

\begin{align}
E\,[XY]&=E\left[E\left[XY\mid Y\right]\right]
\\&=E\left[YE\left[X\mid Y\right]\right]
\\&=0
\end{align}

Similarly $E\,[X]=E\left[E\,[X\mid Y]\right]=0$, so that $\operatorname{Cov}(X,Y)=E\,[XY]-E\,[X]E\,[Y]=0$.

A more intuitive way to see that two jointly distributed variables $X,Y$ are not independent is to verify that the joint support of $(X,Y)$ cannot be written as a Cartesian product of the marginal supports of $X$ and $Y$. For this, all we need to do is sketch the support of $(X,Y)$ given by $$S=\{(x,y)\in\mathbb R^2: |x|<y<1 \}$$

In fact $(X,Y)$ is uniformly distributed over $S$, which looks like

So support of $X$ is $S_1=(-1,1)$ and that of $Y$ is $S_2=(0,1)$.

But since $S\ne S_1 \times S_2$, the random variables $X$ and $Y$ are not independent.

Related:

Uncorrelated, Non Independent Random variables
The mutual density of $X,Y$ in $\{|t|+|s|<1\}$ is constant, are $X,Y$ independent?


Prove that the equation $ e^{x}+x^{3}=10+x $ has a unique solution on the open interval $(-\infty,\infty)$. Prove that the equation
$$
e^{x}+x^{3}=10+x
$$
has a unique solution on the open interval $(-\infty,\infty)$.

Letting $f(x)=e^{x}+x^{3}-10-x$, $f(0)<0$ and $f(10)>0$, so by the IVT there must be a solution in the interval $(0,10)$.

Now I must show there cannot be more than one solution.  Generally for these proofs you assume that there is at least 2 solutions and show a contradiction.  Letting the solutions be a and b ($a<b$ WLOG) I want to use Rolle's theorem to show that $f'(x)=e^{x}+3x^{2}-1$ cannot be zero (giving the contradiction).  The problem is $f'(0)=0$.  So how do I proceed?

You can observe that if $x <-1$ then $X^3-X < 0$, thus
$$f(x) <e^{-1}-10 <-9$$

Also if $x \in [-1, 1]$ we have $x^3 \leq 1$ and $-x \leq 1$ thus
$$f(x) \leq e+2-10 < -5$$

Therefore there is no solution $x \leq 1$.

Now apply your technique on $(1, \infty)$. It is easy to show tha on this interval $f'(x) >0$.

Prove that $f(x)$ is injective over positive real numbers.

$$f'(x) > 0, \forall x\geq0 \text{ appart for 0  and } f\in C^{\infty} (\mathbb{R}, \mathbb{R}) \implies f \text{ injective } $$

Then $\forall(a, b) \in \mathbb{R^+}^2, f(a) = f(b) \iff a=b$

And $\forall x<0, f(x) < 0$ as $x^3-x\leq1$ and $e^x<1$


I'm trying to describe Godel's Incompletemeness Theorem in 1 short sentence... We at the Unemployed Philosophers Guild are adding Kurt Godel to our line of illustrious finger puppets.  On the puppet tag we always have a short biography.  Below is what we have written.  Is our description of his Theorem acceptable to a mathematician/logician?

Austrian-born philosopher and logician Kurt Friedrich Gödel studied physics before publishing his famous (two) Incompleteness Theorem(s). According to Gödel, a mathematical system can’t prove or disprove every proposition within itself (it’s “incomplete”) and can’t prove itself both complete and consistent. Gödel fled Nazi Germany, renewed his friendship with fellow émigré Albert Einstein, and became a U.S. citizen. As “the most important logician since Aristotle,” Gödel influenced computer science, artificial intelligence, and philosophy of mathematics. He was devoted to operetta.

Here is an alternate description of his work:  ...According to Gödel, if a mathematical system can prove every statement that can be constructed in the system, then there must be some contradictory statements in the system: and if there are no contradictory statements, then there are statements that cannot be proved...

Thank you!

It depends on the level of precision you want to reach.

My first remark is that "According to Gödel" feels too much like "He thought so and he said so", but maybe I'm just misunderstanding this by adding a connotation that wasn't there.

More to the point, Gödel's theorem is about formal theories that can be computed. If you want to be precise this is a detail that cannot be missed because there are many theories that are complete and prove or refute every sentence.

A quick summary would be

  "Gödel proved that no effective [here the word "effective" allows you to vaguely state that it should be computable], expressive enough, consistent mathematical system can prove or refute every sentence: it will always be incomplete. In particular, he proved that such a system cannot prove its own consistency."

This allows you to be precise, while still being accessible (the words "effective", "expressive enough" are not detailed, so as to be understood by the non-logician), and is a good reflection of what Gödel actually did. My reformulation to "Gödel proved" (rather than "According to Gödel") is simply because I feel like this emphasizes more that this isn't just speculation or an idea he had but rather an actual theorem.

Austrian-born philosopher and logician Kurt Friedrich Gödel studied physics before publishing his famous (two) Incompleteness Theorem(s).

Physics has nothing to do with the incompleteness theorems.  I doubt he spent as much time or effort in physics as logic, although Wikipedia suggests he did spend a lot more time on physics later in life.

  According to Gödel, a mathematical system can’t prove or disprove every proposition within itself (it’s “incomplete”)

No, there are logics, even rather complex ones, that are complete and consistent.  Propositional logic and "the first-order theory of Euclidean geometry is complete and decidable" (from the link below).

  and can’t prove itself both complete and consistent.

No.  Sufficiently strong logics can not be both complete and consistent.  But any inconsistent logic can prove either of those.

  Gödel fled Nazi Germany, renewed his friendship with fellow émigré Albert Einstein, and became a U.S. citizen.

Wikipedia agrees with this.  Why don't you just say "immigrant"?

  As “the most important logician since Aristotle,”

I would say Hilbert, but some logicians would agree with you.

  Gödel influenced computer science, artificial intelligence, and philosophy of mathematics. He was devoted to operetta.

CS and philosophy are a given, the other two, no idea.

  Here is an alternate description of his work: ...According to Gödel, if a mathematical system can prove every statement that can be constructed in the system, then there must be some contradictory statements in the system: and if there are no contradictory statements, then there are statements that cannot be proved...

That doesn't even make sense to me.  Here is the first incompleteness theorem:

If a logic can be verified by a Turing Machine, is capable of proving all theorems of Robinson Arithmetic, and is 1-consistent; then there is a sentence $G$ such that neither $G$ is provable nor is $\lnot G$ provable in the logic.

(Small detail, but the strengthening of the statement from 1-consistent to consistent is actually due to Rosser.)

If I could say something a bit harsh, the Internet is rife with people attempting to summarize or explain Godel's Incompleteness Theorems who never bothered to really learn them.  Please don't be another.  Silence is better.  I'm not even confident about them, so I hesitated to write this.  If you wish to learn them, I find this link does a really thorough and detailed job without containing extraneous pontifications :

https://plato.stanford.edu/entries/goedel-incompleteness/

  On the puppet tag we always have a short biography.

As far as logicians with interesting lives, I find Moses Schönfinkel really compelling.  Born a poor boy in Ukraine, talented enough to eventually study even under David Hilbert, invented one of the first completely formal logics that was in some sense the grandfather of all constructive logics.  Was committed to an asylum before he was 40 and spend his later life in poverty like how he was born.  His personal papers/work were burned by his neighbors trying to stay warm because of wartime conditions.


Infinite closed subset of $S^1$ such that the squaring map is a bijection? Is there an infinite closed subset $X$ of the unit circle in $\mathbb C$ such that the squaring map induces a bijection from $X$ to itself?

Let us think of $S^1$ as $\mathbb{R}/\mathbb{Z}$, so we want an infinite closed subset $X$ on which multiplication by $2$ is a bijection.  Suppose you have such an $X$; write $T:X\to X$ for the multiplication by $2$ map.  Each element $x\in X$  determines a biinfinite binary expansion $f_x:\mathbb{Z}\to\{0,1\}$, such that $T^n(x)=\sum_{k=1}^\infty f_x(k+n)2^{-k}$ for each $n\in \mathbb{Z}$.  Say that a finite string of $0$s and $1$s is admissible if it appears as a sequence of consecutive values of some $f_x$ (i.e., if it appears as a sequence of consecutive digits in the binary expansion of some element of $X$).  For each $n$, let $A_n\subseteq\{0,1\}^n$ consist of those sequences $s$ such that both $0^\frown s$ and $1^\frown s$ are admissible (where $^\frown$ is string concatenation).  If $A_n$ is empty for some $n$, that means that given a sequence of $n$ consecutive digits in any $f_x$, all of the preceding digits are uniquely determined.  By pigeonhole, for each $x$, some sequence of $n$ digits must appear infinitely often in the restriction of $f_x$ to $\mathbb{N}$, and it follows that every $f_x$ is periodic.  Furthermore, there is a uniform bound on the periods of all the $f_x$ (because if some particular $s\in\{0,1\}^n$ appears infinitely often in $f_x|_\mathbb{N}$, that determines the period of $f_x$, and there are only finitely many different such $s$).  So there are only finitely many different $f_x$, so $X$ is finite.  This is a contradiction.

Thus each $A_n$ is nonempty.  By König's lemma, it follows that there exists an infinite string $s:\mathbb{N}\to\{0,1\}$ such that every initial segment of $s$ is in the appropriate $A_n$.  But then since $X$ is closed, the numbers $y_0$ and $y_1$ whose binary expansions are $0^\frown s$ and $1^\frown s$, respectively, are in $X$.  Since $T(y_0)=T(y_1)$, this is a contradiction.

Thus no such $X$ exists.

For a small $\theta$, close the set $\{e^{ix} : x \in (-\theta, \theta)\}$ under square roots to get an open set $U_{\theta}$. If $\theta$ is small, the measure of $U_{\theta}$ is small. So we can take $A$ to be the complement of $U_{\theta}$.

Sorry, this only gives that $A$ is closed under squaring. Anyway, if $A$ is also required to be closed under square roots, then there is no such proper subset. This is because $(1, 0)$ is not an accumulation point of $A$ and therefore for some $\theta >0$, $\{e^{ix} : x \in (-\theta, \theta)\}$ is disjoint with $A$ and closing this under squaring gives the whole circle.


What is the modern axiomatization of (Euclidean) plane geometry? I have heard anecdotally that Euclid's Elements was an unsatisfactory development of geometry, because it was not rigorous, and that this spurred other people (including Hilbert) to create their own sets of axioms.

I have two related questions:

1) What is the modern axiomatization of plane geometry? For example, when mathematicians speak of a point, a line, or a triangle, what does this mean formally?

My guess would be that one could simply put everything in terms of coordinates in R^2, but then it seems to be hard to carry out usual similarity and congruence arguments. For example, the proof of SAS congruence would be quite messy. Euclid's arguments are all "synthetic", and it seems hard to carry such arguments out in an analytic framework.

2) What problems exist with Euclid's elements? Why are the axioms unsatisfactory? Where does Euclid commit errors in his reasoning? I've read that the logical gaps in the Elements are so large one could drive a truck through them, but I cannot see such gaps myself.

I can recommend an article Old and New Results in the Foundations of Elementary Plane Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg, The American Mathematical Monthly, Volume 117, Number 3, March 2010, pages 198-219. One of the great strengths of the article is that I am in it. Marvin promotes what he calls Aristotle's axiom, which rules out planes over arbitrary non-Archimedean fields without leaving the synthetic framework. If you email me I can send you a pdf.

EDIT: Alright, Marvin won an award for the article, which can be downloaded from the award announcement page GREENBERG. The award page, by itself, gives a pretty good response to the original question about the status of Euclid in the modern world.

As far as book length, there are the fourth edition of Marvin's book, Euclidean and Non-Euclidean Geometries, also Geometry: Euclid and Beyond by Robin Hartshorne. Hartshorne, in particular, takes a synthetic approach throughout, has a separate index showing where each proposition of Euclid appears, and so on.

Hilbert's book is available in English, Foundations of Geometry. He laid out a system but left it to others to fill in the details, notably Bachmann and Pejas. The high point of Hilbert is the "field of ends" in non-Euclidean geometry, wherein a hyperbolic plane gives rise to an ordered field $F$ defined purely by the axioms, and in turn the plane is isomorphic to, say, a Poincare disk model or upper half plane model in $F^{\; 2}.$  Perhaps this will be persuasive: from Hartshorne,

  Recall that an end is an equivalence class of limiting parallel rays

Addition and multiplication of ends are defined entirely by geometric constructions; no animals are harmed and no numbers are used. In what amounts to an upper half plane model, what becomes the horizontal axis is isomorphic to the field of ends. This accords with our experience in the ordinary upper half plane, where geodesics are either vertical lines or semicircles with center on the horizontal axis. In particular, infinitely many geodesics "meet" at any given  point on the horizontal axis.

Any axiomitization of 2-dimensional Euclieadn geometry is unjustifiable. In such a system, you just assume without proof that the undefined concept of distance satisfies certain intuitive properties and prove other properties from it like the Pythagorean theorem. If you invent a space and explicitly define a lot of the relations on it like distance and show that those definitions satisfy the intuitive properties of 2-dimensional Euclidean geometry, it follows that that space can be represented by R^2 and that for any points (x1, y1) and (x2, y2), the distance from (x1, y1) to (x2, y2) is sqrt((x2 - x1)^2 + (y2 - y1)^2). We can define the distance formula to be sqrt((x2 - x1)^2 + (y2 - y1)^2) without giving a reason then show that it satisfies the intuitive properties of distance. After we show that it satisfies the intuitive properties, we can derive from that fact that that the formula is sqrt((x2 - x1)^2 + (y2 - y1)^2) but there's no need because that can be proven directly from the definition of distance. Some people might want to know why it was defined that way so for them, it might suffice to show that that formula is the unique formula that satisfies the intuitive properties of distance. Those people might also be satisfied with seeing the distance formula being proven by defining it them showing that it satisfies those intuitive properties then reproving it from those properties because from that, they can figure out how to prove that that formula is the unique formula that satisfies the intuitive properties of distance. The article http://speedydeletion.wikia.com/wiki/Distance that I wrote actually gives an incomplete proof that that formula is the unique formula that satisfies the intuitive properties of distance. A complete proof probably includes constructing the set of all real numbers with operations from the power set of the set of all natural numbers in Zermelo-Fraenkel set theory then showing that that set is a totally ordered field, similar to what I described in my answer at What is a natural number?.


Which integers have order 6 (mod 37)? Which integers have order $6\ (\text {mod}\ 37)$?
I have tried expanding $x^6-1=0\ (\text {mod}\ 37)$ equation, i tried expanding it as $x^3-1$ and $x^3+1$ and I expanded further and tried to expand as adding $37$ to $x^6-1$ does alter as $\text {mod}\ 37$ is zero , but I could lead further.

HINT:

Try to use Fermats little theorem.

EDIT: By the above named theorem $x^{36}=(x^6)^6\equiv 1 \ \ \bmod 37$. Not surprisingly $x\equiv_{37}1$ is your first solution. $2^6=64\equiv_{37} 27$ so $27$ goes on the list. $3^6=3^2\cdot 3^2\cdot 3=81\cdot 3^2\equiv_{37}7\cdot 3^2\equiv_{37}26$ so $26$ goes also on the list. $4^6=(2^6)^2\equiv_{37}=27^2\equiv_{37}26$. And so on... You will be able to use already known information as soon as $x$ is a composite.

Look for a generator, which is any number $g$ such that $g^{12}\not\equiv 1$ and $g^{18}\not\equiv 1$. Here you get lucky, because the first number you try, $2$, is a generator. So now it's just $2^6$ and $2^{30}$.


Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$ This question actually came out of a question. In some other post, I saw a reference and going through, found this, $n>0$.

Solve for n explicitly without calculator:
$$\frac{3^n}{n!}\le10^{-6}$$

And I appreciate hint rather than explicit solution.

Thank You.

Note that, for $n=3m$, $$3^{-3m}{(3m)!}=\left[m\left(m-\frac{1}{3}\right)\left(m-\frac{2}{3}\right)\right]\cdots\left[1\cdot\frac{2}{3}\cdot\frac{1}{3}\right] <\frac{2}{9}\left(m!\right)^3.$$
So you have to go at least far enough so that
$$
\frac{2}{9}\left(m!\right)^3>10^{6},
$$
or $m! > \sqrt[3]{4500000} > 150$.  So $m=5$ (corresponding to $n=15$) isn't far enough; the smallest $n$ satisfying your inequality will be at least $16$.

Similarly, for $n=3m+1$,
$$
3^{-3m-1}(3m+1)!=\left[\left(m+\frac{1}{3}\right)m\left(m-\frac{1}{3}\right)\right]\cdots \left[\frac{4}{3}\cdot1\cdot\frac{2}{3}\right]\cdot\frac{1}{3} < \frac{1}{3}(m!)^3,$$
so you need $m!>\sqrt[3]{3000000}> 140$, and $m=5$ (that is, $n=16$) is still too small.

Finally, for $n=3m+2$,
$$
3^{-3m-2}(3m+2)!=\left[\left(m+\frac{2}{3}\right)\left(m+\frac{1}{3}\right)m\right]\cdots \left[\frac{5}{3}\cdot\frac{4}{3}\cdot1\right]\cdot\frac{2}{3}\cdot\frac{1}{3} > \frac{560}{729}(m!)^3,
$$
where the coefficient comes from the last eight terms, so it is sufficient that $m! > 100\cdot\sqrt[3]{729/560}.$  To show that $m=5$ is large enough, we need to verify that $(12/10)^3=216/125 > 729/560$.  Carrying out the cross-multiplication, you can check without a calculator that $216\cdot 560 =120960$ is larger than $729\cdot 125=91125$, and conclude that $m=5$ (that is, $n=17$) is large enough.  The inequality therefore holds for exactly all $n\ge 17$.

how about make a function?$f(n)=\frac{3^n}{n!}-10^{-6}$ or maybe $f(n)=\frac{3^n}{n!10^{-6}}$


If a continuous function is strictly decreasing before a point and strictly increasing afterwards, is the point a global minimum? I'm in the middle of a proof that a point on a function is a global minimum. Usually I'd just solve an inequality to prove by contradiction that there are no points less than the minimum. But I can't in this case, since it's a transcendental equation (can't rearrange to make $x$ subject in terms of the elementary functions.) So I'm back to the drawing board. My question is:

If a continuous everywhere function is strictly decreasing before a point and strictly increasing afterwards, is the point a global minimum? How can this be proved?

Let us asume for contradiction that the given point is $x_0$, and that it is not a global minimum. Hence there is another point $x \neq x_0$ so that $f(x)<f(x_0)$. But if $x \neq x_0$ then either $x< x_0$ or $x>x_0$, but since $f$ is strictly decreasing before $x_0$ then in the first case $f(x) > f(x_0)$, in the second case similarly $f(x)>f(x_0)$. We get a contradiction, and hence such $x$ does not exist. The continuity of $f$ is not necessary.

If you have been going downhill and at one stage things improved and never went down afterwards, then we say we turned around and that turning point was the worst. This is well-known to non-mathematicians too, does not calculus  techniques to be able to see this.


Finding all the rational roots of $25x^3+25x^2-x-1$ Finding all the rational roots of $25x^3+25x^2-x-1$.

So, I saw right away that $-1$ was a root. I then used synthetic division to factor this as:

$(25x^2-1)(x+1)$.

Then I found the roots to $25x^2-1$ as $\frac{1}{5},\frac{-1}{5}$.

I've been telling my precalculus class that once you know the roots you know the factors, and if you know the factors you know the roots. However, in this case the original polynomial does not equal $(x+1)(x-\frac{1}{5})(x+\frac{1}{5})$. Can somebody tell me exactly what went wrong here? Thanks!!

The thing is, the fundamental theorem of algebra admits a leading coefficient. Let's be more explicit. Say a polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots a_1 x + a_0$ has the roots $r_1, \cdots, r_n$, which may be complex numbers and not necessarily distinct. The leading coefficient, $a_n$, is assumed nonzero.

Then we can write $f$ as a product of linear factors in terms of its roots and its leading coefficient:

$$f(x) = a_n (x-r_1)(x-r_2)\cdots (x-r_n)$$

This is what you forgot, the leading coefficient. The roots are not sufficient to uniquely determine the polynomial on their own - you can easily see that, however you vary it, the roots would remain the same since $a_n$ is just a constant factor.

With this in mind, augmenting your procedure in this example is quite simple. Find the roots of $f(x)$ as you have, and note that the leading coefficient is $25$. Then

$$f(x) = 25(x + 1)\left( x - \frac 1 5 \right)\left( x + \frac 1 5 \right)$$

Of course, that $25$ can be distributed into the rightmost two factors for a nicer presentation if you so choose.

$$f(x) = (x + 1) \cdot 5 \cdot \left( x - \frac 1 5 \right) \cdot 5 \cdot \left( x + \frac 1 5 \right) = (x+1)(5x-1)(5x+1)$$

However this is purely cosmetic and not at all necessary.


When can you take the derivative of both sides of an equation? I know in general you cannot take the derivative of an equation to solve it because the derivative at a point depends on neighboring points of a function.

However, lots of the proofs done in my probability course, for example finding the variance of a geometric random variable is done by differentiating both sides. Why is this allowed?

considering the equation
$$x^3 = x + 6$$
here $x=2$ is the solution of differential equation
but if we differentiate both sides of equation then we get
$$3x^2 = 1$$
in this case $x=2$ gives $12=1$
not satisfied
this means that two functions are just interesect not tangent to each other
If these function are actually equal then we can differenctiate both sides
For more geometric perspective the arithematic operation act pointwise but the drivative requires the knowledge of nieghborhood
This means that derivative is not operation function
if you sketch $x^3$ and $x+6$ then you would get answer of your each question....


I can't understand logical implication I just started studying logic (high school)
anyway...for the truth table of logical implication

If sentence $A$ is true and $B$ is true then $A\implies B$ is true.

does that mean if $A$ and $B$ are both true
then there is a way to prove $B$ is true from $A$,
always?

the same for if $A$ is false can you get anything either True or false proved from this $A$?

As a logical proposition, the material conditional $A \implies B$ is a very weak one: as you've noticed, it's very easy to satisfy it just by accident. In fact, this happens whenever $A$ is false, or whenever $B$ is true. Thus, merely observing that $A \implies B$, for some specific $A$ and $B$, says very little.

Instead, the usefulness of implication lies in the fact that, precisely because of its weakness, it is often possible to assert $A \implies B$ as a universal statement (either an axiom or a provable theorem) that holds for any valuation of any free variables mentioned in the propositions $A$ and $B$.

For example, consider the statement: $$x > 2 \;\land\; x \text{ is prime} \implies x \text{ is odd}.$$ Merely observing that this statement holds for some $x$ says very little — there are plenty of numbers for which it is trivially true, either because they are odd, or because they are not primes greater than 2. What makes this statement useful is that we can prove that it holds for all $x$ — there isn't a single number which would be greater than 2 and prime, but not odd.

This illustration of logical implication might help:

a) When you truly understand logical implication,
b) then you’ll be a happy person.

You may be a happy person for other reasons. But if you’re unhappy, then certainly you don’t truly understand logical implication.  :-)

Truth table:
a  b  a⇒b
F  F  T
F  T  T
T  F  F
T  T  T

I hope this somewhat intuitive example helps.


functions with floors If
$$z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$$
find $\lfloor z \rfloor$.

I don't rely know how to do this, but I was thinking about multiplying the denominator by it's conjugate, but idk.

Assuming $\{\cdot \}$ is fractional part,
$\{\sqrt{3}\} = \sqrt{3}-1$ and $\{\sqrt{2}\} = \sqrt{2}-1$, so

$$ z = \frac{(\sqrt{3}-1)^2 - 2 (\sqrt{2}-1)^2}{\sqrt{3} + 1 - 2 \sqrt{2}}$$
Expand out the numerator and you should recognize that it is a certain integer times the denominator.

I think $\{ \cdot \}$ is fractional part function.

So, we have $\{ x \} = x- \lfloor x \rfloor : x \ge 0$,  $\{ x \} = x- \lceil x \rceil : x \le 0$


Function for which it is unknown whether it is continuous Is there any function $f:\mathbb R\rightarrow \mathbb R$ for which at least some values are known but it is unknown whether $f$ is continuous or not?

Edit: I am looking for examples from actual research, not functions explicitly constructed for that purpose.

Let $f(x)=0$ for $x\neq0$ and $f(0)=$ the first even number that is not the sum of two primes.


find expectation of non-negative integer valued RV from generating function How can we find $E\left(X\right)$ and $E\left(X^{2}\right)$ if all
we have is that $G\left(s\right)$ is the generating function for
X, which takes non-negative integer values.

I know $E\left(X\right)$ = $G'\left(1\right)$ and $G\left(s\right)=$$\sum_{i=0}^{\infty}s^{i}P\left(X=i\right)=\sum_{i=0}^{\infty}s^{i}f(i)$
but how can I take the derivative?

$E\left(X\right)$ = $G'\left(1\right)$ and $G\left(s\right)=$$\sum_{i=0}^{\infty}s^{i}P\left(X=i\right)=\sum_{i=0}^{\infty}s^{i}f(i)$

$G'(s) = \sum_{i=0}^{\infty}is^{i-1}f\left(i\right)$

$G''(s) = \sum_{i=0}^{\infty}i\left(i-1\right)s^{i-2}f\left(i\right)$

$G''(1) = \mathrm{E}(X^{2}) - \mathrm{E}(X)$

Just take the derivative with respect to s and you can play with my expression for $G''(s)$ (by splitting it into two summations) if you want to see the algebra.


A variant: note that
$$\color{red}{\mathbf 1}=0+\color{red}{\mathbf 1}=0+0+\color{red}{\mathbf 1}=0+0+\cdots+0+\color{red}{\mathbf 1}=0+0+0+\cdots$$
and
$$\color{green}{\mathbf 2}=0+\color{green}{\mathbf 2}=0+0+\color{green}{\mathbf 2}=0+0+\cdots+0+\color{green}{\mathbf 2}=0+0+0+\cdots$$
"Since the right hand sides are the same", this proves that $\color{red}{\mathbf 1}=\color{green}{\mathbf 2}$.

Incidentally, nobody appeared to have resolved the fallacy of the question, so I have provided an answer.

For all $a\in \mathbf N$, it follows $$\cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots - \cfrac 12}}}$$ such that the number of times the reciprocal in the continued fraction appears is $a$.

Proof. Note the identity $$\cfrac{1}{1+a}=1-\cfrac{1}{1+\color{red}{\cfrac 1a}}.$$ By letting $a=b-1$, it follows $$\cfrac 1b = 1-\cfrac{1}{1+\cfrac{1}{b-1}}.$$ From this we can substitute for $\color{red}{\cfrac 1a}$. $$\therefore \cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{1+\cfrac{1}{a-1}}}.$$ Clearly we can now likewise substitute for $1/(a-1)$, and the pattern will continue until for some $k\in\mathbf N$, the denominator of $1/(a-k)$ reaches $a-k=1$ since it cannot pass $0$. In consequence, we deduce as desired. (And, of course, when $a=0$, we have $1/1 = 1-0$.) This completes the proof. $\;\bigcirc$

And now, since $$\lim_{a\to\infty}\frac{1}{1+a}=0$$ then $$\boxed{\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots}}}=1}$$


Given two distinct primes $p,q$ such that $(p-1)(q-1)=A$, are $p$ and $q$ uniquely determined by $A$? I believe another way I could phrase the question is:

Given $f(m,n)\equiv(m-1)(n-1)$ where $m$ and $n$ are two primes:

Is $f$ one-to-one for pairings of $(m,n)$?

Or:

Are there only two distinct primes $(p,q)$ which make $f=A$?

I conjecture the answer is yes, there is only one pairing which exists for each output, but I can still somewhat envision there being some way to get non-distinct mappings from $A\to(m,n)$


Lunch Meeting Probability for two person to meet in given 1 hour slot and none would wait more then 15 minute. Two friends who have unpredictable lunch hours agree to meet for lunch at their favorite restaurant whenever possible. Neither wishes to eat alone and each dislikes waiting for the other, so they agree that each will arrive at a random time between noon and 1 pm, and each will wait for the other for 15 minutes or until 1:00. What is the probability that the friends will meet for lunch on a given day?

Per @AppDeveloper's request changing it from a comment to an answer:

Just and idea: consider $0$ to $30$ min, the other half is the same by symmetry.

If A arrives at $0$ min, B has to arrive between $0$ and $15$ min, i.e., $p(B\leq 15|A=0)=\frac{1}{4}$.

If A arrives after $15$ min, $p(B|A)=\frac{1}{2}$.

Applying conditional probabilities and integrating, get for $t \geq 15$ min $p(B|A)p(A)=\frac{1}{8}$ and for $0 \leq t \leq 15$,  $p(B|A)p(A)=\frac{3}{32}$ Adding together and multiplying by $2$, get $\frac{7}{16}$.

No matters who arrives first.
Assume $A$ arrives first. 2 cases then to be considered:
Case 1: $A$ arrives between 12.00-12.45.
Then the chance that both will meet equals: $3/4 \cdot 1/4 = 3/16$.
Case 2: A arrives between 12.45-13.00.
Then the chance that both will meet equals: $1/4\cdot 1= 1/4$.
Chance that both will meet then equals: $3/16 + 1/4 = 7/16$.


To discuss differentiability of function at origin and my attempt P1:  $F =  |x| +  |y|  when x,y is not equal to 0,

  =   0   when x = y = 0

P2 :Discuss the differentiability at origin  of  $F = y sin(1/x)$
 :

Near good, but for $|x|+|y|$. Do not use argument, that partial derivatives are not continuous. Theorem says "if partial derivatives... are continuous... derivative exists", but not reverse. However argument with one sided partial derivatives in enough.

For the corrected version of P2 ($f(x,y)=y\sin(1/x)$ for $x\neq0$ and 0 for $x=0$):
$$
f'_y(0,0)=\lim_{y\to0}\frac{0-0}{y}=0,\quad
f'_x(0,0)=\lim_{x\to0}\frac{0\sin(1/x)-0}{x}=0,
$$
hence (in case P2) partial derivatives exist.

Function $f$ is differentiable in $(x_0,y_0)$ if
$$
\lim_{(h,k)\to(0,0)}\frac{f(x_0+h,y_0+k)-f(x_0,y_0)-f'_x(x_0,y_0)h-f'_y(x_0,y_0)k}{\sqrt{h^2+k^2}}=0.
$$
In our case we obtain
$$
\lim_{(h,k)\to(0,0)}\frac{k\sin(1/h)-0-0-0}{\sqrt{h^2+k^2}}.
$$
Let $k=h$. Then
$$
\lim_{h\to0}\frac{h\sin(1/h)}{\sqrt{2h^2}},
$$
which doesn't exist.

Edit: I can see, that there is yet another definition of $f$ ($f(0,y) = y$). Then $f'_y(0,0)=1$ and modifications are the following:
$$
\lim_{(h,k)\to(0,0)}\frac{k\sin(1/h)-0-0-k}{\sqrt{h^2+k^2}}.
$$
Let $k=h$. Then
$$
\lim_{h\to0}\frac{h(\sin(1/h)-1)}{\sqrt{2h^2}},
$$
which again doesn't exist.

Phew!

For the first question, your answer is wrong. The limits you've given do exist at the origin, the problem is that they come to different numbers depending on the direction. Thus the derivative is not well-defined.


Distance from Ellipsoid to Plane - Lagrange Multiplier Find the distance from the ellipsoid $x^2 + y^2 + 4z^2 = 4$ to the plane $x + y + z = 6$.

I'm trying to do it using Lagrange multipliers over the distance equation, but then it just gets overwhelming and I have no idea how to go on? Can someone walk me through the computation?

This is easy if you don't insist on using Lagrange multipliers.

The normal to the ellipse at the point $(x,y,z)$ is $\nabla(x^2+y^2+4z^2) = (2x, 2y, 8z)$. At minimum or maximum distance to the plane, this must be parallel to the normal to the plane, which is $(1,1,1).$ So $x = y = 4z$.

Plug this back into the equation for the ellipse to get $36z^2 = 4$, or $z = \pm \frac13$. So the nearest and farthest points are $\pm(\frac43,\frac43,\frac13)$.

I don't know how to use that maths language so please bear with me.
(I am editing his answer as this seems to be the most relevant one apart from latex. In some countries, the undergrads are unaware of this language but it shouldn't be the barrier.)
ellipsoid: $x^2 + y^2 + 4(z^2) = 4$
plane: $x + y + z = 6$
Now distance of a point $(p,q,r)$ from a plane $ax+by+cz+d=0$ is given by
$|\frac{(ap+bq+cr+d)}{\sqrt(a^2 + b^2 + c^2)}|.$
Now let the point on the ellipsoid at minimum distance from the plane be $(x,y,z)$
Therefore minimum distance $= |\frac{(x+y+z-6)}{\sqrt3}|$
Now define a function $f(x,y,z)=(x+y+z-6)/\sqrt3$
and $g(x,y,z)=x^2 + y^2 + 4(z^2) - 4 $
The minimum absolute value of f is our answer subject to the constraint $g(x,y,z)=0.$
Now, $\nabla(f) = (1/\sqrt3)i +(1/\sqrt3)j +(1/\sqrt3)k$
and $\nabla(g) = (2x)i + (2y)j + (8z)k $
Applying Lagrange's method
At minimum value of f subject to constraint g=0
$\nabla(f) = m(\nabla(g))$ ......I've used $m$ instead of $\lambda$
therefore,
$1/\sqrt3 = 2mx, 1/\sqrt3 = 2my$, $1/\sqrt3 = 8mz$
therefore, $x=y=4z$
and $x^2 + y^2 + 4(z^2) - 4 = 0$
Solving the above two equations we get two sets of points
$(4/3, 4/3, 1/3)$ and $(-4/3, -4/3, -1/3)$
Minimum and maximum distance occurs at these points on the ellipsoid
The minimum distance is given by $|f)|$
therefore, the point of minimum distance is $(4/3, 4/3, 1/3) $
and the distance is $|(f(4/3,4/3,1/3)) |=\sqrt3$
You can also solve this problem by using six variables if you don't want to use the plane point distance formula. It looks overwhelming using six variables but its easy to do but, this method is better. For the six variable method you have to solve for finding the points on the plane also.


Prove if $\lim_{x\to +\infty}(f(x)+f'(x))=0$ then $\lim_{x\to +\infty} f(x)=0$ Let $f$ be a real function  continuously differentiable at $\Bbb R$ such that
$$\lim_{x\to +\infty}(f(x)+f'(x))=0$$
prove that

$$\lim_{x\to +\infty} f(x)=0$$
I tried tu use exponential function knowing that

$$\frac{d}{dx}f(x)e^x=(f(x)+f'(x))e^x$$
but I got nothing.
thanks in advance for an answer or un idea

Could you prove it by contradiction? If $\lim_{x\rightarrow \infty} f'(x) \neq 0$ then $\lim_{x\rightarrow \infty} f(x) = -\lim_{x\rightarrow \infty} f'(x)$ but if $f'>0$ then $f$ is increasing, and if $f'<0$ then $f$ is decreasing - that seems like a contradiction.


If $f(x) = \int_{\cos x}^0 \tan(t)\mathrm dt$, what is $f'(x)$? If $f(x) = \int_{\cos x}^0 \tan(t)\mathrm dt$, what is $f'(x)$?

This is basically the First Fundamental Theorem of Calculus.

\begin{align}
f(x)&=\int^0_{\cos(x)}\tan(t)dt\\
&=-\int_0^{\cos(x)}\tan(t)dt\\
f'(x) & = -\dfrac{d}{d\cos(x)}\int_0^{\cos(x)}\tan(t)dt\ \cdot\dfrac{d\cos(x)}{dx}\\
&= \sin(x)\cdot\tan(\cos(x))
\end{align}

Use the rule $$\frac{d}{dx}(\int_a^x{f(t)dt})=f(x)$$
And the composition rule.

Let $f(x)=\int_{\cos x}^0{\tan t dt}=-h(\cos x)$ where $h(x)=\int_0^{x}{\tan t dt}$

So, $f'(x)=\frac{d}{dx}f(x)=-h'(\cos x).(\cos x)'=\sin x.\tan(\cos x)$


Two variables limit question I proved that $f(x,y)= \dfrac{xy^2}{x^2 + y^3}$ does not have limit at origin.
I used two paths test; first I followed the $x$ axis, then I followed $x = \frac{1}{2}(y^2 + (y^4 - 4y^3)^{1/2})$ for $y<0$.

However, I am STILL looking for other solutions other ideas.

Any kind of answer, help or hint is appreciated.

Think of the path $(x,-x^{2/3})$ in the open fourth quadrant as $x\to 0^+.$ The denominator of your expression equals $0$ at every point of this path.  Meanwhile the numerator is never $0$ in that open quadrant. That should give you pause. Even if you exclude this path from the domain of $f$ (which you should have done, otherwise you are dividing by $0$), you have insurmountable problems. There is no way around it: $f$ is unbounded in every neighborhood of $(0,0),$ so there is no hope of a limit.

Use polar coordinates, so your limit becomes

$$\lim_{r\to0}\frac{r^3\cos\theta\sin^2\theta}{r^2\cos^2\theta+r^3\sin^3\theta}=\lim_{r\to0}\;r\frac{\cos\theta\sin^2\theta}{\cos^2\theta+r\sin^3\theta}=0$$


Testing pythagorean triples: $333,444,555$ In this page there is a necessary and sufficient test given for testing Pythagorean triples:

  A simpler, more powerful test is, (by naming the even leg a): $(c − a)$ and $\large\frac{(c − b)}{2}$ are both perfect squares. This is both
  necessary and sufficient for the triple to be a PT.

Using this here,we can write $a = 444,b=333,c=555$,which means $111$ and $\frac{222}{2}=111$ must be perfect squares but it is not.Hence,that will not work.

Is there any necessary and sufficient condition that will work for every and any (other than summing up the squares and checking for perfect squares?

NOTE: $333,444,555$ are Pythagorean triples as $3\times111,4\times111,5\times111$ for $3,4,5$ is a Pythagorean triple.

Yes, there is a mistake in the phrasing of the condition.

Every Pythagorean triple is of the form

$$ a = 2k mn \qquad b = k(m^2 - n^2) \qquad c = k(m^2 + n^2) $$

which means that

$$ c - a = k(m^2 + n^2 - 2mn) = k (m-n)^2 \qquad \frac{c-b}{2} = k n^2 $$

So the correct statement is that

  Let $d$ be the greatest common divisor of $c-a$ and $(c-b)/2$. Then a necessary and sufficient condition for $(a,b,c)$ to be a Pythagorean triple is that $(c-a)/d$ and $(c-b)/(2d)$ are both perfect squares.

Let 333, 444, and 555 are x,y,z of a Pythagorean triple to be tested.
Rule to remember: "a given triple is Pythagorean iff there are two integers (j,k; k>j) such that square root of their product is an integer (obviously x) where k = z+y and j=z-y"
In this example: k=555+444=999, j=555-444=111; j*k = 110889; (jk)^(0.5) = 333, which is an integer and hence x.
Test is proven.


Can I prove Pythagoras' Theorem using that $\sin^2(\theta)+\cos^2(\theta)=1$? In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

The theorem can be written as an equation relating the lengths of the sides $a$, $b$, and $c$, often called the Pythagorean equation:
$$a^2 + b^2 = c^2$$

Can I prove Pythagoras' Theorem by the following way?

Actually, my question is: does it violate any rules of mathematics, or is it alright?
Sorry, it may not be a valid question for this site. But I want to know. Thanks.

The usual proof of the identity $\cos^2 t+\sin^2 t=1$ uses the Pythagorean Theorem. So a proof of the Pythagorean Theorem by using the identity is not correct.

True, we can define cosine and sine purely "analytically," by power series, or as the solutions of a certain differential equation. Then we can prove $\cos^2 t+\sin^2 t=1$ without any appeal to geometry.

But we still need geometry to link these "analytically" defined functions to sides of right-angled triangles.

Remark: The question is very reasonable. The logical interdependencies between various branches of mathematics are usually not clearly described. This is not necessarily always a bad thing. The underlying ideas of the calculus were for a while quite fuzzy, but calculus was still used effectively to solve problems, Similar remarks can be made about complex numbers.

It is not true that you must use the Pythagorean theorem to prove that $sin^2(x)+ cos^2(x)= 1$. It depends upon how you have defined sine and cosine.  It is, for example, perfectly proper to define $sin(x)= \sum_{n= 0}^\infty \frac{(-1)^n}{(2n+ 1)!}x^{2n+ 1}$ and $cos(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$ or, equivalently, to define $sin(x)$ as the function satisfying the differential equation y''+ y= 0 with initial conditions y(0)= 0, y'(0)= 1 and $cos(x)$ as the function, y, satisfying the differential equation y''+ y= 0 with initial conditions y(0)= 1, y'(0)= 0.  In either case, you can then prove that $sin^2(x)+ cos^2(x)= 1$ without reference to the Pythagorean theorem.


3-D geometry: line intersecting 2 lines and parallel to plane Find a surface generated by line intersecting lines $$y=a=z$$ and $$x+3z=a=y+z$$ and parallel to plane $$x+y=0$$

I tried to form a line equation which intersects the  given two lines i.e.
$(y-a)+k1(z-a)=0$ and $(x+3z-a)+k2(y+z-a)=0$. But don't know how to use the other (plane) condition.

Equation of the line L1 in symmetric form is $$\frac{x}{1} = \frac{y - a}{0} = \frac{z - a}{0} \tag1$$
Equation of the Line L2 in symmetric form is $$\frac{x -a}{-3} = \frac{y - a}{-1} = \frac{z}{1} \tag2$$
Let the required line 'L' intersect L1 at P and L2 at Q.
Coordinates of P are $$(r,a,a) \tag3$$
Coordinates of Q are $$(-3s+a, -s+a, s) \tag4$$
Equation of L is  $$\frac{x -r}{-3s-r+a} = \frac{y - a}{-s} = \frac{z-a}{s-a} \tag5$$
Given that this line L is parallel to the plane $x+y=0 \tag6$
Hence, its normal will be perpendicular to L
So $$(-3s-r+a)+(-s)=0 \tag7$$
or $$r=a-4s \tag8$$
From equation 5 we have $$\frac{y-a}{z-a} = \frac{-s}{s-a}$$
Solving this for s, we will have $$s=\frac{a(y-a)}{y+z-2a} \tag9$$
From equation 5, we also have $$\frac{x -r}{-3s-r+a} = \frac{y - a}{-s} \tag10$$
Substituting the values of r and s obtained from equations 8 and 9, we get the required surface as
$$y^2 + xy + yz + zx - 2ax -2az = 0$$

Reordering the equation of the line
$$\alpha: y+k_1 z-a k_1-a=0;\;\beta:x+k_2 y+(k_2+3) z -ak_2-a=0$$
The normal vector of plane $\alpha$ is $n_1=(0,1,k_1)$ and the normal vector of $\beta$ is $n_2=(1,k_2,k_2+3)$

the direction vector of the line is the cross product $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$

The line $(y-a)+k_1(z-a)=0;\;(x+3z-a)+k_2(y+z-a)=0$ is parallel to the plane $x+y=0$ if its direction vector is perpendicular to the normal vector of the plane $x+y=0$ which is $n=(1,1,0)$

$m\cdot n=0\to (3+k_2-k_1 k_2,k_1,-1)\cdot (1,1,0)=3 + k_1 + k_2 - k_1 k_2$

the condition is that $3 + k_1 + k_2 - k_1 k_2=0$

$k_2=\dfrac{k_1+3}{k_1-1}$

$\alpha:y+k_1 z-a-a k_1=0$

$\beta:(-1+k_1) x+(3+k_1) y+(3+3 (-1+k_1)+k_1) z-a (-1+k_1)-a (3+k_1)=0$

The direction of the line is $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$ where $k_2=\dfrac{k_1+3}{k_1-1}$

$m=(k_1^2-k_1,k_1-k_1^2,k_1-1)$

a point on the line is $P\left(-6 a,0,\dfrac{3 a}{2}\right)$

a parametric equation of the line is

$(x,y,x)=P+tm$ that is

$x=-k_1 t+k_1^2 t-6a,y=k_1 t-k_1^2 t,z=\dfrac{3 a}{2}-t+k_1 t$

Note that from the first two equations $x=-y-6a$

Furthermore from the three equation we get

$k_1= -\dfrac{4 y}{x+y+4 z},\;t= -\dfrac{(x+y+4 z)^2}{4 (x+5 y+4 z)}$

substitute in the third equation $z=\dfrac{3 a}{2}-t+k_1 t$

$z=\dfrac{3 a}{2}+\dfrac{(x+y+4 z)^2}{4 (x+5 y+4 z)}+\dfrac{y (x+y+4 z)}{x+5 y+4 z}$

simplify and get the result.

The requested surface is the plane $x + y+6a=0$


Suppose A and B are sets. Prove that $A\subseteq B$ if and only if $A \cap B = A$. Suppose $A$ and $B$ are sets. Prove that $A \subseteq B$ if and only if $A \cap B = A$.

Here's how I see it being proved.
If $A$ and $B$ are sets,and the intersection of $A$ and $B$ is equal to $A$, then the elements in $A$ are in both the set $A$ and $B$. Therefore, the set of $A$ is a subset of $B$ since all the elements are contained in the interesection of sets $A$ and $B$ are equal to $A$.
Can I prove it that way?

Your proof is almost perfect and let me rectify it a bit:

Let  $A$ and $B$ be two sets. The intersection of $A$ and $B$ is equal to $A$, is equivalent to  the elements in $A$ are in both the set $A$ and $B$ which's also equivalent to the set of $A$ is a subset of $B$ since all the elements of $A$ are contained in the intersection of sets $A$ and $B$ are equal to $A$.

Suppose A is subset of B. Let X belongs to A then by hypothesis, X will belong to B. Hence X belong to A  and
X  belong to B  implies that X belongs to A intersection B.  Accordingly A is subset of A intersection B. But we know that A intersection B is always subset of A. Hence A intersection B is equal to A.
On the other hand, suppose A intersection B is equal to A. Then in particular, A is subset  of A intersection B. We know that, A intersection B is subset of B. Hence A is subset  of B.


How find this ODE $(1-x^2)y''+2xy'-2y=-2$ Question:

  Find the ODE
  $$(1-x^2)y''+2xy'-2y=-2$$

I think we can find
$$(1-x^2)y''+2xy'-2y=0$$
I have find a solution $y=x$,But I can't find all solution.

Thank you

By setting $y=1+f(x)$, we have that $f$ satisfies:
$$(1-x^2) f'' + 2x f' - 2f = 0, $$
that can be solved using Frobenius' method. By setting:
$$ f(z) = \sum_{j=0}^{+\infty} a_j z^j $$
we have:
$$ z\,f'(z) = \sum_{j=1}^{+\infty} j a_j\, z^j,\qquad f''(z)=\sum_{j=2}^{+\infty}j(j-1)a_j z^{j-2} $$
$$(1-z^2)f''(z) = \sum_{j=0}^{+\infty}\left((j+2)(j+1)a_{j+2}-j(j-1)a_j\right)z^j$$
so:
$$(j+2)(j+1)a_{j+2}-j(j-1)a_j+2ja_j-2a_j = 0 $$
or:
$$(j+2)(j+1)a_{j+2} = (j-1)(j-2) a_{j}\tag{1} $$
so the relation $a_{j+2}=\frac{j-1}{j+1}\cdot\frac{j-2}{j+2}a_j$ gives that the solution associated with the initial conditions $a_0=0,a_1=1$ is given by $f(x)=x$ while the solution associated with the initial conditions $a_0=1,a_1=0$ is $f(x)=x^2+1$.

That is a Legendre equation. Deriving a solution on your own will be very difficult. Here is some information about it http://en.wikipedia.org/wiki/Legendre_polynomials


The 3rd raw moment of a binomial distribution What is the 3rd raw moment (that is, $ E\{X^3\} $) of a Binomial distribution with parameters $n$ and $p$?

I am getting $n(n-1)(n-2)p^3 + 3n(n-1)p^2 + np$. Is it correct?

For such a distribution, it is best to compute the probability generating function (PGF) rather than MGF.  That is to say, $$P_X(t) = \operatorname{E}[t^X] = \sum_{x=0}^n t^x \binom{n}{x} p^x (1-p)^{n-x} = \sum_{x=0}^n \binom{n}{x} (pt)^x (1-p)^{n-x} = (pt+1-p)^n,$$ the last equality being a consequence of the binomial theorem.  From here, we observe/recall that $$\left[\frac{dP}{dt}\right]_{t=1} = \operatorname{E}\left[X t^{X-1}\right]_{t=1} = \operatorname{E}[X],$$ and similarly $$\left[\frac{d^2P}{dt^2}\right]_{t=1} = \operatorname{E}[X(X-1)],$$ and in general, $$\left[\frac{d^k P}{dt^k}\right]_{t=1} = \operatorname{E}[X(X-1)\cdots(X-k+1)].$$  Therefore, for $k = 1, 2, 3$, we get $$\operatorname{E}[X] = \frac{d}{dt}\left[(pt+1-p)^n\right]_{t=1} = np,$$ $$\operatorname{E}[X(X-1)] = \frac{d^2}{dt^2}\left[(pt+1-p)^n\right]_{t=1} = n(n-1)p^2,$$ $$\operatorname{E}[X(X-1)(X-2)] = n(n-1)(n-2)p^3,$$ and indeed in the general case, $$\operatorname{E}[X(X-1)\ldots(X-k+1)] = n(n-1)\ldots(n-k+1)p^k.$$  Therefore, $$\operatorname{E}[X^3] = (n(n-1)(n-2)p^3) + 3(n(n-1)p^2) + np.$$

Yes, it is correct.use the expectation values of first and second powers of the random variable, binomial to derive the third moment.Thus it will be reduced to a matter of relating some finite telescopic sums.


Solution to $x^\alpha + p x = q$? I was wondering if there was any tricks, similar in spirit to the Vieta's substitution, that would apply the equation
$$
x^\alpha + p x = q,
$$
where $p,q$ and $\alpha$ are real constants. In particular $\alpha$ is not necessarily an integer. The goal is to solve for $x$.

Thanks for your help!

There is no known closed form general solution for a random integer $\alpha$. However, rewriting the

equation as $\color{blue}x=\sqrt[\large\alpha]{q-p\color{blue}x}~$ yields the following formula: $x=\sqrt[\large\alpha]{q-p~\sqrt[\large\alpha]{q-p~\sqrt[\large\alpha]{q-\ldots}}}$

Let me choose a for alpha to make notation easier

x^a = -px+q

(a)LN( x) = LN(-px+q)

x = e^{LN(-px+q)}/a

hope that helps


How to show that $A(A+B)^{-1}B = B(A+B)^{-1}A$ I am trying this problem. Could not proceed . Please give a hint.
The exercise, if the above image is not showing clearly, is the following: Let $A$ and $B$ be two square $m \times m$ matrices for some integer $m \ge 2$ such that $A+B$ is invertible. Then show that
the equation $A(A+B)^{-1}B = B(A+B)^{-1}A$. [Note that either $A$ or $B$ or each of both may not be invertible, and that $A$ and $B$ may not commute.]

First note the following equation:
$$A(A+B)^{-1}(A+B) = (A+B)(A+B)^{-1}A.$$
[Indeed, both sides of this equation just above are $A$.] That hint may or may not be enough. If you want more help, see below:

 Distributing on each side gives.....

 $$A(A+B)^{-1}A +A(A+B)^{-1}B = A(A+B)^{-1}A + B(A+B)^{-1}A.$$

 But then subtracting both sides by $A(A+B)^{-1}A$ gives

 $$A(A+B)^{-1}B = B(A+B)^{-1}A,$$

 which is what you want.


Real-world applications of prime numbers? I am going through the problems from Project Euler and I notice a strong insistence on Primes and efficient algorithms to compute large primes efficiently.

The problems are interesting per se, but I am still wondering what the real-world applications of primes would be.

What real tasks require the use of prime numbers?

Edit: A bit more context to the question:
I am trying to improve myself as a programmer, and having learned a few good algorithms for calculating primes, I am trying to figure out where I could apply them.

The explanations concerning cryptography are great, but is there nothing else that primes can be used for?

Here is a hypothesized real-world application, but it's not by humans...it's by cicadas.

Cicadas are insects which hibernate underground and emerge every 13 or 17 years to mate and die (while the newborn cicadas head underground to repeat the process). Some people have speculated that the 13/17-year hibernation is the result of evolutionary pressures. If cicadas hibernated for X years and had a predator which underwent similar multi-year hibernations, say for Y years, then the cicadas would get eaten if Y divided X. So by "choosing" prime numbers, they made their predators much less likely to wake up at the right time.

(It doesn't matter much anyway, because as I understand it, all of the local bug-eating animals absolutely gorge themselves whenever the cicadas come out!)

EDIT: I should have refreshed my memory before posting. I just re-read the article, and the cicadas do not hibernate underground. They apparently "suckle on tree roots". The article has a few other mild corrections to my answer, as well.

Prime numbers are used in public key cryptography. It is used because you generally don't think of the really big prime numbers, so it is great for codes and to keep things safe.


The diameter of Voronoi cells in Euclidean spaces Let $A \subset \mathbb{R}^d$ and let $(x_n)_{n \in \mathbb{N}} \subset \mathbb{R}^d$ be a sequence dense in $A$.
For each $n \in \mathbb{N}$, let $V_{1,n},\dots,V_{n,n}$ the sequence of Voronoi cells associated to the points $x_1, \dots, x_n$, where the ties are broken lexicographically, i.e.:
$$V_{1,n} =\{x \in \mathbb{R}^d \mid \forall k\in\{2,\dots,n\}, |x-x_1|\le |x-x_k|\} \\
V_{2,n} =\{x \in \mathbb{R}^d \mid |x-x_2|<|x-x_1| \land \forall k\in\{3,\dots,n\}, |x-x_2|\le |x-x_k|\} \\
\vdots \\
V_{n,n} =\{x \in \mathbb{R}^d \mid \forall k\in\{1,\dots,n-1\}, |x-x_n| < |x-x_k|\}
$$
If $x \in \mathbb{R}^d$ and $n \in \mathbb{N}$, define $V_n(x)$ as the unique element in $V_{1,n},\dots, V_{n,n}$ that contains $x$.

Is it true that
$$\forall x \in A, \operatorname{diam}\big(A \cap V_n(x)\big) \to 0, n \to \infty?$$

I suspect that this result should hold due to the finite dimensionality of $\mathbb{R}^d$, since at least in this case we can obtain bounded sets using a finite number of intersections of half-spaces.
However, it seems quite involved from a geometric point of view to obtain this claim. Has anyone any idea?
EDIT: note that we can WLOG assume that $A$ is the closure of the set whose points are those of the sequence $(x_n)_{n \in \mathbb{N}}$.
Some context: I'm trying to prove the aforementioned result to obtain that if $g \colon A \to \mathbb{R}$ is a continuous function, then
$$\forall x \in A, \sup_{y \in A \cap V_n(x)} |g(x) - g(y)| \to 0, n \to \infty.$$

For additional insight, one key intuition might be that, by definition of density, any part of the set will eventually be dotted by an arbitrarily fine cloud of points.
Note that the definition of $ V_n $ is to choose among $ V_{1,n} \cdots V_{n,n} $ with $ n $ points present from the start. Sufficiently far down the sequence, there should be points arbitrarily close to $ x $ “all around” in order to “force” $ V_n $ into an arbitrarily small corner.
Otherwise, some angle would remain open, in which an open ball could exist, the interior of which must intersect a region of $ A $ where the sequence could then not be dense.
Edit: I see now that this is already plain for the particular case $ A = \mathbb{R}^d $, but does not help understanding what happens when the cells are unbounded. Would it be fair to say you are asking why it is that the unboundedness must resolve when capping by $ A $?
It should be in principle easier to show that the diameter at least does not diverge to infinity. Any point in the sequence is eventually “surrounded” by other points “in the local directions of $ A $” as you so evocatively put it.
Consider any $ a \in A $ and its path connected component of $ A $, then any path emanating from it must encounter points of the sequence within an arbitrarily thin thickness and arbitrarily close to it. Therefore no “local direction” is ever free to go on forever.
This is far from rigorous, I hope it nevertheless contributes positively to the discussion.


Every metric space with a countable base is separable I know that every separable metric space has a countable base.

I was wondering if we can get a countable dense subset from every metric space that has a countable base.

Thank you very much!!

You don't even need a metric space for that direction, in holds in general. If your topology has a countable base, just pick one element out of every base set and you will get a countable dense subset.

In general in any topological space  let $A=\{a_i : a_i \in O_i\}$ you  can prove that $A$ is a dense countable subset, such that $B=\{O_i : i \in\mathbb N\}$ is the Base of topological space


When does linearity of definite Riemann integrals hold? My Calculus text book says if functions $f$ and $g$ are continuous on a closed interval $[a,b]$, then
$$
\int_a^b (f(x)+g(x)) \, dx=\int_a^b f(x) \, dx+\int_a^b g(x) \, dx
$$
where the integrals are in the Riemann sense.
However, there are many important applications for functions with discontinuities.
Does this identity also apply in all cases where all three integrals exist? If not what other constraints are needed to include functions with discontinuities?
*** Update ***
I think I have an example where excluding discontinuities is relevant.
$
\int_0^{\infty } \left(\sin (x) \cos \left(\frac{1}{x}\right)-\sin (x)\right) \, dx
$ is well defined but
$
\int_0^{\infty } \sin (x) \cos \left(\frac{1}{x}\right) \, dx-\int_0^{\infty } \sin (x) \, dx
$ involves two integrals that don't exist. It seems the sufficient conditions used in my text book are not met in this example because sin(x) is not continuous at $\infty$.

The correct theorem in the setting of Riemann integration is the following:

If $f,g:[a,b]\to\Bbb{R}$ are Riemann-integrable over $[a,b]$, then $f+g$ is also Riemann-integrable over $[a,b]$, and in this case we have
\begin{align}
\int_a^b(f+g)&= \int_a^bf+\int_a^bg.
\end{align}

A proof should be available in any good textbook (for example it's in Spivak's Calculus book). Of course, continuous functions are Riemann-integrable so you can apply this result to continuous functions. There are of course also many Riemann-integrable functions which are not continuous; the thereom holds for these functions as well. As you can see from the theorem statement, there is no mention of continuity at all!

If we integrate via Lebesgue, then the same equality holds for discontinous functions, provided $f$ and $g$ are discontinous only on a dense numerable subsetset of their domain. To be fair, the condition of being integrable doesn't depend on continuity: there exists non continous functions which are integrable. As an example you can take the floor function, which is clearly integrable but it has discontinuities on every number in $\mathbb{Z}$, which is not dense in $\mathbb{R}$.


What is the mark of the winning team? $10$ teams have participated a competition with five contestants.  According to the results, we give each person a grade. The grades are between $1$ and $50$ and we cannot use a grade twice.  The winner is the team that gets the minimum score.  What scores are possible to get by the winning team.

My attempt:  There is a lower bound $1+2+3+4+5=15$ and also an upper bound $\big\lfloor{\frac{1+2+\dots 50}{10}}\big\rfloor =127$, but I am not sure if we can reach the numbers between them.

Proposition: If the winning team can get a score of $n$, and $n>15$, then they can get a score of $n-1$.

Proof: If the score $n$ of the winning team isn't the minimum of $15$, then there must be some player on the winning team who got exactly one point more than some player not on the winning team. Swap the grades of the two, and the winning team gets one point less, while some other team gets one point more, and the other right teams are unchanged. Thus the winning team is still the winning team, with $n-1$ points.

This proves that the winning team can "reach the numbers between them". What's left for you is finding the actual maximal score they can get. In other words, can they get $127$ and still be the winning team?

Every number from $15$ up to $127$ (in fact, up to $240$) can be written as sum of five distinct numbers $\in\{1,2,\ldots,50\}$. Indeed, this is evidently possible for $15=1+2+3+4+5$. Assume it is possible for $n$ with $15\le n<240$, say $n=a_1+a_2+a_3+a_4+a_5$ with $1\le a_1<a_2<a_3<a_4<a_5\le 50$.
If $a_i+1<a_{i+1}$ for some $i$, $1\le i<5$, then we are allowed to replace $a_i$ with $a_i+1$ and obtain a representation for $n+1$. The same holds if $a_5<50$. Thus we are left only with the case that $a_{i+1}=a_i+1$ for $1\le i<5$ and $a_5=50$, but then $n=46+47+48+49+50=240$, contradicting our assumption. We conclude that $n+1$ can also be written that way.


If the rank of $A$ is equal to the number of non-zero eigenvalues, do $A$ and $A^2$ have the same rank? Let $A$ be an $n$-by-$n$ matrix over some field. If it happens that $\operatorname{rank}(A)=$ number of non-zero eigenvalues of $A$, can we say that $\operatorname{rank}(A^2)=\operatorname{rank}(A)$?

I believe we can say this (thinking about idempotent matrices) but I am not sure about the proof.  Please give some hints to get started and the main idea.

Let $J= \operatorname{diag}(J_1,...,J_k)$ be the Jordan normal form.

Note that $J^2= \operatorname{diag}(J_1^2,...,J_k^2)$

The rank of $J$ is given by the sum of the
ranks of the blocks that is, $\operatorname{rk} J = \sum_k \operatorname{rk} J_k$.
Similarly,
 $\operatorname{rk} J^2 = \sum_k \operatorname{rk} J_k^2$.

It follows from the hypothesis that the rank of the Jordan block
 corresponding to the zero eigenvalue is zero. That is, the Jordan
block is identically zero (and so is the square of the Jordan block).

For the blocks $J_k$ corresponding to non zero eigenvalues, we have
$\operatorname{rk} J_k = \operatorname{rk} J_k^2$.

It follows that $\operatorname{rk} J = \operatorname{rk} J^2$.

Alternative:

Let $N = \ker A$. We see that $z=\dim N$ is the number of zero eigenvalues of $A$. Let $b_1,..,b_z$ be a basis for $N$ and complete the basis with $b_{z+1},...,b_n$. Note that $N$ is $A$ invariant and so in this basis, $A$ has the form
$\begin{bmatrix} 0 & A_{12} \\ 0 & A_{22}\end{bmatrix}$. Furthermore, we
must have $\det A_{22} \neq 0$ otherwise $A$ would have more that $ z$ zero eigenvalues.

Then we want to show that $\ker A^2 = N$. Note that $A^2$ has the form
$\begin{bmatrix} 0 & A_{12}A_{22} \\ 0 & A_{22}^2\end{bmatrix}$ and it follows from
this that if $x \in \ker A^2$ then $x \in \ker A$.

No. Let for example $A$ the nilpotent matrix

$$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
then we have $\operatorname{rank}(A)=1$ and $\operatorname{rank}(A^2)=\operatorname{rank}(0)=0$.

Edit I used in my above answer the classic definition of the rank of matrix which is the dimension of the image of the matrix. For the definition of the rank given by the OP, the answer is yes in the field $\Bbb C$ and to see this we use that every matrix is similar to a triangular matrix.


number of squares in a rectangle. Given a rectangle of length a and width b(as shown in the figure).How many different squares of edge grater than 1 can be formed from using the cells inside .

For example if a=2,b=2 ,then the number of such squares is just 1.

In general, given an $n \times k$ grid of squares, to find the number of rectangles you can form, you would turn your grid into an $n \times k$ multiplication table, put the values into each square (i.e., the $i$th row and $j$th column would contain $i \cdot j$), and then sum them all up. Proving that this holds is a nice exercise.

For your particular question, you are asked to find the number of squares in a $3 \times 3$ grid where each square has its sides greater than $1$. This is straightforward to figure out directly from your picture (how many $3 \times 3$ squares are there? how many $2 \times 2$ squares are there?) but the solution also becomes clear to anyone who proves the statement of the previous paragraph.

The number of $3 \times 3$ squares is $1$, which is $1 \cdot 1$; the number of $2 \times 2$ squares is $4$, which is $2 \cdot 2$. Thus, the total is $1 + 4 = 5$.

Incidentally, the number of $1 \times 1$ squares is $9$, which is $3 \cdot 3$. Note that $1, 4, 9$ are the entries of the diagonal in a $3 \times 3$ multiplication table. This is no coincidence!

Given an $n \times n$ multiplication table, to find the number of squares, just add up all the elements of the diagonal. The formula for the sum of the first $n$ squares, by the way, is $n(n+1)(2n+1)/6$, which you could look up online or prove by induction.

Since you want to exclude $1 \times 1$ squares, you would subtract $n^2$ from this sum, giving a final answer of
$$S(n) = \frac{n(n+1)(2n+1)}{6} - n^2 = \frac{(n-1)n(2n-1)}{6}$$

Indeed, for $n = 3$, we find $S(3) = 5$ as desired.

In an $n\times p$ rectangle, the number of rectangles that can be formed is $\frac{np}{4(n+1)(p+1)}$ and the number of squares that can be formed is $\sum_{r=1}^n (n+1-r)(p+1-r)$.


a question about a canonical form of a quadratic form using Gauss theorem Get the following quadratic form:

$$Q(x)=x_1^{2}+x_3^2+4x_1x_2-4x_1x_3 $$

to obtain the canonical form I tried the following:

$$Q(x)=x_1^{2}+x_3^{2}+4x_1x_2-4x_1x_3=4(x_1^{2})+(x_3^{2})-4x_1x_3-3x_1x_1+4x_1x_2-(4/3)x_2x_2+(4/3)x_2x_2=(2x_1-x_3)^2-... $$

There I stopped because I rememebered that I shouldn't change the coefficient of $x_1^2$.

From here I don't know how to continue.

I thank you in anticipation for your understanding and I wait forward your answer!

There is a method I asked about at
reference for linear algebra books that teach reverse Hermite method for symmetric matrices
The main advantage is that it is a recipe with matrices, no need to carry variable names. The main disadvantage is the need to invert one matrix at the end; however, the matrix has determinant $\pm 1$ and may very well be upper triangular (it is this time).

It leads to $$ x^2 + z^2 + 4 xy - 4 zx = (x +2y-2z)^2 - 4 (y-z)^2 + z^2  $$

and comes from this matrix stuff; I did it first by hand, it did work, but I thought i would check everything. The Pari code is not quite as readable as Latex but is not too bad.

parisize = 4000000, primelimit = 500509

? m = [ 1,2,-2; 2,0,0; -2,0,1]
%2 =
[1 2 -2]

[2 0 0]

[-2 0 1]

? m - mattranspose(m)
%3 =
[0 0 0]

[0 0 0]

[0 0 0]

? p1 = [1,-2,2; 0,1,0; 0,0,1]
%4 =
[1 -2 2]

[0 1 0]

[0 0 1]

? m1 = mattranspose(p1) * m * p1
%5 =
[1 0 0]

[0 -4 4]

[0 4 -3]

? p2 = [ 1,0,0; 0,1,1; 0,0,1]
%6 =
[1 0 0]

[0 1 1]

[0 0 1]

? d = mattranspose(p2) * m1 * p2
%7 =
[1 0 0]

[0 -4 0]

[0 0 1]

? p = p1 * p2
%8 =
[1 -2 0]

[0 1 1]

[0 0 1]

? matdet(p)
%9 = 1
? q = matadjoint(p)
%10 =
[1 2 -2]

[0 1 -1]

[0 0 1]

? confirm =  mattranspose(q) * d * q
%12 =
[1 2 -2]

[2 0 0]

[-2 0 1]

? m
%13 =
[1 2 -2]

[2 0 0]

[-2 0 1]

? m - confirm
%14 =
[0 0 0]

[0 0 0]

[0 0 0]

?

? ( x + 2 * y - 2 * z)^2 - 4 * (y - z)^2 + z^2
%1 = x^2 + (4*y - 4*z)*x + z^2

=========================================================

Places on this site I put this, several typeset:

reference for linear algebra books that teach reverse Hermite method for symmetric matrices

Bilinear Form Diagonalisation

Given a $4\times 4$ symmetric matrix, is there an efficient way to find its eigenvalues and diagonalize it?

Find the transitional matrix that would transform this form to a diagonal form.

Writing an expression as a sum of squares

Determining matrix $A$ and $B$, rectangular matrix

Method of completing squares with 3 variables


Proof of triangle inequality I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)

From your definition of the absolute value, establish first $|x| = \max\{x,-x\}$ and $\pm x ≤ |x|$.
Then you can use
\begin{align*}
a + b &≤ |a| + b ≤ |a| + |b|,\quad\text{and}\\
-a - b &≤ |a| -b ≤ |a| + |b|.
\end{align*}


Is this conception of countable vs. uncountable infinity adequate? I am not a mathematician, so please point out any mistakes I am making here - I am trying to grasp the concept of countable vs. uncountable infinity in a somewhat informal way and would like to know whether that conception makes sense.

We can imagine the set of natural numbers as an axis that goes from some fixed point ($0$, or $1$ if you want) to infinity:

Clearly, the points on this axis are countable, because we know exactly how the axis goes on and can therefore make a precise calculation about how many points the rest of the axis will contain.

For the integers, we no longer have a fixed starting point, but the axis grows infinitely in two directions:

However, we still have only one axis of fixed points and can make a precise calculation of how many points the axis as a whole will contain.
The first thing that bugs me: Since the axis is calculably exactly twice as long, this should be a "larger infinity" than the for the natural numbers, right? But still, we would say that $\mathbb{Z}$ has the same cardinality as $\mathbb{N}$?

For the rational numbers, things get a little more difficult, but we can still handle it: Any rational number can be displayed as the fraction between two integers - if I understood it correctly, this is what the Cantor pairing function does? - so we can just add a second axis to account for the combinatoric possibilities yielding $\mathbb{Q}$:

The amount of points now doesn't simply add up, i.e. the axis doesn't just get longer (as in the step from $\mathbb{N}$ to $\mathbb{Z}$), but it multiplies, i.e. there are more axes now, so that's even a "larger increase of infinity". Is this correct?
But we still have finitely many axes with countably many points, so the whole amount of points is countable too.

Now for the case of real numbers, things look a bit differently.
Clearly, a one-dimensional system doesn't suffice because we need to account for the digits behind the comma, so we need at least two axes, in order to create $0.0, 0.1, 0.2, ..., 1.0, 1.1, ...$:

Now that doesn't suffice either, because from $1.1$, we can decide to either stay at $1.1$, which would be $1.10$ (Is it true that $1.1$ is in fact $1.10$ which is in fact $1.1000000...$, so that rational numbers are actually never really finite, or is this idea false and $1.1$ is really just $1.1$?) or go further to $1.11$, so we need another axis:

We are now three-dimensional and can thus account for all the numbers with two digits behind the comma, but that still doesn't suffice, because between $1.10$ and $1.11$, we also have the numbers $1.101, 1.102, 1.103, ... $, and from any point we are, we are recursively stepping one dimension deeper, because for any digit we add, we again have all the possibilities to go on from that point, so we never reach a point where we can stop adding axes:

(At this point I'm running out of imagination on how to draw a 7D diagram, sorry)

Now we are at the point where we run into an infinite number of axes - and this corresponds to the set of real numbers $\mathbb{R}$ no longer being countably infinite.

My question is:
Is it adequate to say that countably infinite corresponds to finitely many dimensions, while uncountably infinite corresponds to infinitely many dimensions, or did I go anywhere wrong in my conception?

The notion of "countably infinite" is well named.  Another word you can use is "enumerable," which is even more descriptive in my opinion.

I understand your intuition on the subject and I see where you're coming from.  Let me try to give you some insight into the agreements about infinity that have been reached over the years by the mathematicians who've worked on this problem.  (This is what I wish someone had explained to me.)

"Countably infinite" just means that you can define a sequence (an order) in which the elements can be listed.  (Such that every element is listed exactly once.)

The natural numbers are the most naturally "countable"—they're even called "counting numbers"—because they are the most basic, natural sequence.  The word "sequence" itself is wrapped up in what we mean by natural numbers, which is just one thing following another, and the next one coming after that, and the next one after that, and so on in sequence.

But the integers are countable as well.  In other words, they are enumerable.  You can specify an order which lists every integer exactly once and doesn't miss any of them.  (Actually it doesn't matter, for the meaning of "countable," if a particular element gets listed more than once, because you could always just skip it any time it appears after the first time.)

Here is an example of how you can enumerate (count, list out) every single integer without missing any:

$0, 1, -1, 2, -2, 3, -3...$

The rational numbers are also countable, again because you can define a sequence which lists every rational number.  The fact that they are listed means (at the same time) that they are listed in a sequence, which means that you can assign a counting number to each one.

The real numbers are not countable.  This is because it is impossible to define a list or method or sequence that will list every single real number.  It's not just difficult; it's actually impossible.  See "Cantor's diagonal argument."

This will hopefully give you a solid starting point to understanding anything else about infinite sets which you care to examine.  :)

Your understanding of the whole thing is a little off the road of regular Mathematical sense. I am not a mathematician or even a math major, but I know some math and let me tell you some of my understanding. The major difference btw set R and N is that N is listable but R is not, and listable is just another saying of countable. N is somewhat discrete while R is "continous". Btw every 2 reals, there will always be a real, no matter rational or irrational while between 2 consecutive integer, no integer exists. There is no "next" for a real. As to the dimension thing, you have to know linear combination and dimension is defined as the minimal number of elements such that each element of the space can be represented as a linear combination of those elements. Those elements must also be linear independent. In finite dimensional space, the situation is always simpler than infinite dim space. In finite dim, people tend to find element whose span is dense instead of being the whole space. For R, it is one dimensional because a single real can represent all others by linear combination.


Finding the total number of proper subfields of $F$? I was thinking about the following problem:

  Let $F$ be a field with $5^{12}$ elements.Then how can I find the total number of proper subfields of $F$?

Can someone point me in the right direction? Thanks in advance for your time.

I take it you mean, proper subfield.

Can you show that any subfield of $F$ contains the field of $5$ elements?

Can you show that any subfield must contain $5^r$ elements, for some $r$?

Can you show that the degree of such a subfield (over the field of $5$ elements) must be $r$? and must be a divisor of the degree of the field of $5^{12}$ elements?

Can you show that a finite field has at most one subfield of any given number of elements?

If you can do all those, you have your answer.

The number of divisors of 12 are 1,2,3,4,6,12
Therefore 6=subfields exists.
Proper subfield means except 12 then number of proper subfield of 5^12 is 5.


Group of order $|G|=6545$ has either a normal Sylow 5-subgroup or a normal Sylow 17-subgroup. Let $G$ be a group of order $|G|=6545$. Show that $G$ has either a normal Sylow 5-subgroup or a normal Sylow 17-subgroup.

My attempt (supposedly wrong):

Given $|G|=6545=5\cdot7\cdot11\cdot17$, by Sylow's theorem, we have: $n_{5}=1\text{ mod }5\text{ and }n_{5}\mid7\cdot11\cdot17=1309\Rightarrow n_{5}=1\text{ or }11\\n_{7}=1\text{ mod }7\text{ and }n_{7}\mid5\cdot11\cdot17=935\Rightarrow n_{7}=1\text{ or }85\\n_{11}=1\text{ mod }11\text{ and }n_{11}\mid5\cdot7\cdot17=595\Rightarrow n_{11}=1\text{ or }595\\n_{17}=1\text{ mod }17\text{ and }n_{17}\mid5\cdot7\cdot11=385\Rightarrow n_{17}=1\text{ or }35$

Suppose $G$ has no normal Sylow 5-subgroup or normal Sylow 17-subgroup. Then, we have $n_{5}=11$ and $n_{17}=35$. and there exist $11\cdot(5-1)+35\cdot(17-1)=604$ non-identity elements of order $5$ and $17$. Consequently we have $6545-604=5941$ elements of order 7 or 11. But this is a impossible because the only possible outcomes of $n_{7}$ and $n_{11}$ are:$1\cdot(7-1)+1\cdot(11-1)+1=17\neq5941\\1\cdot(7-1)+595\cdot(11-1)+1=5956\neq5941\\85\cdot(7-1)+1\cdot(11-1)+1=521\neq5941\\85(7-1)+595\cdot(11-1)+1=6461\neq5941$

Thus, we must have $n_{5}=1$ or $n_{17}=1$. In other words, $G$ has either a normal Sylow 5-subgroup or a normal Sylow 17-subgroup.

I know my solution is completely wrong and I'm not supposed to solve it this way. Can someone help me?

Here is one way to proceed.

First case: $n_{5} = 1$.
Then $G$ has a normal Sylow $5$-subgroup.
Second case: $n_{5} = 11$.
Let's consider the action of $G$ on the set of its Sylow $5$-subgroups by conjugation i.e the morphism $$\pi : \begin{array}[t]{rcl} G & \longrightarrow & \mathfrak{S}_{\left\{ \text{Sylow } 5\text{-subgroups of } G \right\}}\\ g & \longmapsto & \left(S \mapsto g S g^{-1}\right) \end{array}$$
Then $\frac{|G|}{|\text{Ker}(\pi)|} = |\text{Im}(\pi)| \geq 11$ since this action is transitive - all the Sylow $5$-subgroups are conjugate to each other - and divides both $|G| = 5 \cdot 7 \cdot 11 \cdot 17$ and $|\mathfrak{S}_{\text{Sylow } 5\text{-subgroups}}| = 11!$
Therefore, $$\frac{|G|}{|\text{Ker}(\pi)|} = 11, \ 5 \cdot 7, \ 5 \cdot 11, \ 7 \cdot 11 \text{ or } 5 \cdot 7 \cdot 11$$ and hence, $$|\text{Ker}(\pi)| = 17, \ 5 \cdot 17, \ 7 \cdot 17, \ 11 \cdot 17 \text{ or } 5 \cdot 7 \cdot 17$$
Furthermore, $|\text{Ker}(\pi)| \ne 5 \cdot 17 \text{ and } 5 \cdot 7 \cdot 17$.
Indeed, let's suppose that $|\text{Ker}(\pi)| = 5 \cdot 17 \text{ or } 5 \cdot 7 \cdot 17$.
Then, since $v_{5}(|G|) = v_{5}(|\text{Ker}(\pi)|)$ and $\text{Ker}(\pi) \vartriangleleft G$, the Sylow $5$-subgroups of $G$ are exactly the Sylow $5$-subgroups of $\text{Ker}(\pi)$.
Hence, $n_{5} = 11 | 7 \cdot 17$. Contradiction.
Thus, $$|\text{Ker}(\pi)| = 17, \ 7 \cdot 17 \text{ or } \ 11 \cdot 17 $$
Now, since $v_{17}(|G|) = v_{17}(|\text{Ker}(\pi)|)$ and $\text{Ker}(\pi) \vartriangleleft G$, the Sylow $17$-subgroups of $G$ are exactly the Sylow $17$-subgroups of $\text{Ker}(\pi)$.
Hence, $n_{17} \equiv 1 \pmod{17}$ and $n_{17} | 7 \cdot 11$ which implies that $n_{17} = 1$.
Thus, $G$ has a normal Sylow $17$-subgroup.

The 17-Sylow subgroup of a group of this order is ALWAYS normal.

Simple solution: First, a group of this order is solvable.  (You don't need the odd order theorem...groups of square-free order can be shown to be solvable by the Burnside transfer theorem, and by other means..there is a non-transfer proof in Marshall Hall's Theory of Groups that a square-free order group is metacyclic...i.e., a a normal cyclic subgroup with cyclic factor group.)  For solvable groups, Philip Hall's theorem (see, e.g., Th 9.3.1 in Marshall Hall's book) says that the number of Sylow-17 subgroups is a product of factors each of which is congruent to $1\pmod{17}$ and divides a chief factor.  Since the chief factors of a (necessarily solvable) square-free group are the primes dividing it, the number of 17-Sylow subgroups must be a product of ones..i.e., 1.

Thus, in fact a much more general result holds: ANY group of square-free order has a normal Sylow subgroup corresponding to its largest prime.

In fact, a group of order 6545, by the above result, must have normal subgroups of orders 7, 11, and 17, and thus has a cyclic normal subgroup of index 5, which has automorphism group a direct product of cyclic groups of orders 6, 10, and 16, and that has a unique subgroup of order 5.  Thus, a Sylow-5 element has two ways to act on the index 5 subgroup...trivially or faithfully.  This means that, up to isomorphism, there are only two groups of order 6545--the cyclic group, and a semidirect product with a 5-element acting non-trivially on an index 5 normal cyclic subgroup.  Note that another way to characterize the non-abelian group of order 6545 is as the direct product of a cyclic group of order 119 with a Frobenius group of order 55.

I should add that if you have a reference that does not contain the counting statement of Philip Hall's theorem but only the other parts of it (e.g., Isaacs's book), I posted a proof of the counting statement as it applies to Sylow subgroups here: Sylow Counting Generalization of Hall Theorem .  Note that even though this statement for Sylow subgroups applies to all finite groups you still need to show independently that the group is solvable, since that's how you know that the chief factors are all 17 or less.


Is $\mathbb Q^n$ disconnected? there is a problem on topology.

Let $n > 1$ and let $X = \{(p_1,p_2, \ldots , p_n)\mid p_i\text{ is rational}\}$. Show that $X$ is disconnected.

how to solve this problem.i am completely stuck out.

HINT: Consider the set $\{\langle p_1,p_2,\dots,p_n\rangle\in X:p_1<\sqrt2\}$. Is it open? Closed?

If we take the line $Y=(\sqrt 2)Z + (\sqrt 3,\sqrt 3,\sqrt 3, \cdots  \text{n times})$ where $Y,Z\in  R^n$. then the line cut the space into two disjoint open sets.hence it is disconnected


What is the probability that when you place 8 towers on a chess-board, none of them can beat the other. What is the probability that when you place 8 towers on a chess-board, none of them can beat the other.

Attempt: ${64 \choose 8}^{-1} \approx1$ in $4\ 400\ 000\ 000$

Correct answer: ${64 \choose 8}^{-1} \cdot 8! \approx 1$ in $9\ 000\ 000$.

I disagree with the $8!$. If there's combinations (binomial coefficient) in the denominator, why would there be permutations i.e. the order counts, in the numerator?

There are $\binom{64}{8}$ ways to place the eight rooks on the board. Out of these, there are $8!$ ways for the rooks not to be able to beat each other.

Why? There must be one rook at each row, one at each column. So the placement will define a one-to-one map from the eight columns to the eight rows. There are $8!$ such maps.

(The “towers” are called rooks in English.)

That there is combination but not permutation in the denominator is supported by the fact that we need to choose 8 position out of the total 64 to place the towers, and since the towers are identical there is no need to consider the 8! combinations of every choosed 8 positions.

Now, as to understand why there is permutation in the numerator, let' look from a different point of view. Whenever you place a tower, you cannot place another in its own row or column. Try visualizing the row and column corresponding to the tower getting shaded as you move the tower across the board. After having placed all the 8 towers, if you look from the side of the chess board, they all appear in a line. There are 8! combinations possible in this line, and here we do not avoid the 8! permutation because though we are visualizing all 8 in a line, on every next combination they will not exchange places. Rather they will exchange their row number but remain in the same column. This way we will cover every row position in all 8 column thus covering the entire chess board. This gives us 8! combinations for the towers to be placed such that none of them can beat the other.


Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$ Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$

(1) When $n=0$ we have that $3^2+1 = 10$ which is divisible by $5$ clearly.
(2) Assuming that the condition hold for $n=k.$
(3) Proving that it holds for $n=k+1$
$$3^{4(k+1) + 2} + 1 = 3^{4k + 6} + 1 = 3^4 \cdot 3^{4k+2} + 1$$
Since we assumed that $5 \mid 3^{4k+2} + 1$ we have that
$$3^4 \cdot 3^{4k+2} + 1 = 3^4 \cdot 5t, \text{ where $t \in \Bbb Z$}.$$
Thus $5 \mid 3^{4(k+1) + 2} + 1$.
Is this a valid proof? I'm not entirely sure I'm correct with this...

An alternative strategy for this kind of problem is to consider the difference between consecutive terms.
Let $f(n)=3^{4n+2}+1$. Then $f(n+1)-f(n)=80 \cdot 3^{4 n + 2}$ is a multiple of $5$.
The claim follows by induction since $f(0)=10$ is a multiple of $5$.
(Actually, this proves that  $f(n)$ is always a multiple of $10$.)


Changing streams in PhD I've a masters degree from a reputed Indian university in pure mathematics, with a specialization in Algebraic Number Theory. However, I'd like to apply for a PhD in computational math/theoretical computer science in US universities next year. How should I go about it? Considering that I've little or no formal background in the applied branches or computer science, do I stand a chance of getting accepted in a good PhD program?

Also, what should I tell them when I write SOPs? How should I make myself appear to be a good candidate for a CS PhD without having any formal background in the area?

Speaking as someone who has a masters degree in Mathematics from India and is now doing a Ph.D. in CS in Europe:

I would expect you to just be honest in SOPs about:

Why do you want to do a Ph.D.? Maybe because you read about some topic and became interested, you tried a machine learning model and realized that you would like to study more on this
Why should the university choose you? Because you have already shown the grit to understand highly rigorous and technical mathematical texts. A lot of theoretical computer science is based on pure mathematics.

Regarding the showcase of your commitment, you could do online courses to have a deeper understanding of which field to explore and which sub-field to specialize in. This would even help your CV.

You're way ahead of most U. S.-born college students. One time I asked a chemical engineering student what $\sqrt{-25}$ is. Wanna know what his answer was? 5. Let's just hope he doesn't decide to switch to electrical engineering.

So if I was you, I'd be more worried about having too many programs to choose from. If you've got some family already in America, you'll probably be able to choose a program that is prestigious and close to your family.