complex analysis - What proof uses both the Riemann Hypothesis and its negation?

Some time ago I happened to see a proof that was remarkable in that it used both the Riemann Hypothesis and its negation. That is, it considered the two cases: RH is true, and RH is false, obtaining, after a non-trivial chain of reasoning in each case, the theorem in question. I failed to bookmark it well enough that I can readily retrieve it, so I’m asking the community’s help.

What reminded me of this result was this question on MO about famous vacuously true statements. Obviously, one of the cases in the proof is vacuous and, since it involves the Riemann Hypothesis, conceivably qualifies as being famous.

Answer

There is a small list on Wikipedia: https://en.wikipedia.org/wiki/Riemann_hypothesis#Excluded_middle

summation - Sum of binomial coefficients when lower suffices is same in the series: ${m choose m}+{m+1 choose m}+{m+2 choose m}+...+{n choose m}$

I want to find out sum of the following series:
$${m \choose m}+{m+1 \choose m}+{m+2 \choose m}+...+{n \choose m}$$
My try:
${m \choose m}+{m+1 \choose m}+{m+2 \choose m}+...+{n \choose m}$ = Coefficient of $x^m$ in the expansion of $(1+x)^m + (1+x)^{m+1} + ... + (1+x)^n$

Or, Coefficient of $x^m$
$$\frac{(1+x)^{m}((1+x)^{n}-1)}{1+x-1}$$
$$=\frac{(1+x)^{m+n}-(1+x)^{m}}{x}$$

But, how to proceed further?

Note: $m≤n$

Answer

Other way
$$\left( \begin{matrix}
k \\
m \\
\end{matrix} \right)=\left( \begin{matrix}
k+1 \\

m+1 \\
\end{matrix} \right)-\left( \begin{matrix}
k \\
m+1 \\
\end{matrix} \right)
$$
we have
$$\sum\limits_{k=m}^{n}{\left( \begin{matrix}
k \\
m \\

\end{matrix} \right)}=\sum\limits_{k=m}^{n}\left[{\left( \begin{matrix}
k+1 \\
m+1 \\
\end{matrix} \right)-\left( \begin{matrix}
k \\
m+1 \\
\end{matrix} \right)}\right]=\left( \begin{matrix}
n+1 \\
m+1 \\
\end{matrix} \right)

$$
Also

Let $x_i\in \mathbb{N}$ and
$$x_1+x_2+x_3+\cdots+x_{k+2}=n+2$$

real analysis - Direct proof that if p is prime, then $ sqrt{p} $ is irrational.

Does anyone know of a simple direct proof that if p is prime, then $\sqrt{p}$ is irrational?

I have always seen this proved by contradiction and have been trying unsuccessfully to prove it constructively. I searched this site and could not find the question answered without using contradiction.

Answer

Maybe an elementary proof that I gave in a more general context, but I can't find it on the site, so I'll adapt it to this case.

Set $n=\lfloor \sqrt p\rfloor$. Suppose $\sqrt p$ is rational and let $m$ be the smallest positive integer such that $m\sqrt p$ is an integer. Consider $m'=m(\sqrt p-n)$; it is an integer, and

$$ m'\sqrt p=m(\sqrt p-n)\sqrt p=mp-nm\sqrt p $$
is an integer too.

However, since $0\le \sqrt p-n <1$, we have $0\le m' smallest positive integer such that $ m\sqrt p$ is an integer, it implies $m'=0$, which means $\sqrt p=n$, hence $p=n^2$, which contradicts $p$ being prime.

real analysis - Show that the following function is continuous

Let $ f : \mathbb{R} \rightarrow \mathbb{R} $ be a function which takes a convergent sequence and gives us a convergent sequence.

Show that $f$ is continuous.

So I saw a proof for this but I don't get it.

proof

We show that for all convergent sequences $a_n$ with limit $a$ it holds that $ \lim_{ n \rightarrow \infty } f(a_n) = f(a)$. Let $a_n$ be convergent with limit $a$. define $ b_n :=a_{\frac{n}{2}} $ if $n$ even and $ b_n := a $ if $n$ odd.

Now the unclear part

The following first equation is unclear:

And the last part with "it follows that" is unclear:

$ \lim_{ n \rightarrow \infty } f(b_n) = \lim_{ n \rightarrow \infty } f(b_{2n+1} ) = \lim_{ n \rightarrow \infty } f(a) = f(a) $. It follow that $ \lim_{ n \rightarrow \infty } f(a_n) = a. $

Answer

The sequence $\bigl(f(b_{2n+1})\bigr)_{n\in\mathbb N}$ is a subsequence of the sequence $\bigl(f(b_n)\bigr)_{n\in\mathbb N}$ and therefore, since the limit $\lim_{n\in\mathbb N}f(b_n)$ exists, you have$$\lim_{n\in\mathbb N}f(b_n)=\lim_{n\in\mathbb N}f(b_{2n+1}).$$So, the limit $\lim_{n\in\mathbb N}f(b_{2n})$ also exists, but $b_{2n}=a_n$ and therefore $\lim_{n\to\infty}f(a_n)$ exists (and it is equal to $f(a)$).

number theory division of power for the case $(n^r −1)$ divides $(n^m −1)$ if and only if $r$ divides $m$.

Let $n > 1$ and $m$ and $r$ be positive integers. Prove that $(n^r −1)$ divides $(n^m −1)$ if and only if $r$ divides $m$.

calculus - how to prove that $ln(1+x)< x$

I want to prove that: $\ln(x+1)< x$.

My idea is to define: $f(x) = \ln(x+1) - x$, so:

$f'(x) = \dfrac1{1+x} - 1 = \dfrac{-x}{1+x} < 0, \text{ for }x >0$.

Which leads to $f(x)

Is that a valid proof?
Any other ideas?

Thanks.

Answer

I think your approach is correct but you need to add some more details. Based on your approach let $f(x) = \log(1 + x) - x$ so that $f(0) = 0$. Clearly $$f'(x) = -\frac{x}{1 + x}$$ and hence $f'(x) > 0$ if $-1 < x < 0$ and $f'(x) < 0$ if $x > 0$. It follows that that $f(x)$ in increasing in $(-1, 0]$ and decreasing in $[0, \infty)$. Thus we have $f(x) < f(0)$ if $-1 < x < 0$ and $f(x) < f(0)$ if $x > 0$. It thus follows that $f(x) \leq f(0) = 0$ for all $x > -1$ and there is equality only when $x = 0$. So we can write $$\log(1 + x) \leq x$$ for all $x > -1$ and there is equality only when $x = 0$.

Note: We have considered $x > -1$ because $\log(1 + x)$ is not defined if $x \leq -1$.

calculus - Show that $e+pi$ is not integer.

It was suggested Taylor series for that.

$e^1 = \sum_{k=0}^\infty \frac{1}{k!}$

I don't know how to prove the convergence of this series, so I tried to set the upper limit to 5 (I'm doing all this with a very simple calculator, it's basically by hand). Then $e \approx 2.71$

Since $4\arctan(1)= \pi$

$4\arctan(1)=4 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$

Again doing a aproximation setting the upper limit to 5, $\pi\approx2.96$ (which I think it's pretty bad, but with my calculator it's the best I could do).

Then $e+\pi \approx 5.67$. But this only proves the approximation that I did is not integer, not the exactly value of $e+\pi$. Is there a way to prove that $e+\pi$ is not integer without relaying on approximations?

Answer

With your approximations and using Interval arithmetics, you just have to show that $5