The curious mathmo talks to David Roberts.
@christianp had a fascinating chat with David about his work on class forcing and topos theory. They discussed the continuum hypothesis, algebraic geometry, and all sorts of nontraditional mathematical universes.
aperiodical.com/2017/06/the-cu mathstodon.xyz/media/HzcgL03yP

Let $$〈\mathbf a,\mathbf b〉$$ be an inner product on a real vector space $$V$$ with $$|\mathbf a|^2=〈\mathbf a,\mathbf a〉$$. Then $$〈\mathbf a,\mathbf b〉^2\leq|\mathbf a|^2|\mathbf b|^2$$. Consider $$|x\mathbf a+\mathbf b|^2$$. We may assume $$\mathbf a\neq 0$$. If $$\mathbf b=λ\mathbf a$$, then $$〈\mathbf a,\mathbf b〉=|\mathbf a|^2|\mathbf b|^2$$. If not, then $$|x\mathbf a+\mathbf b|^2\gt 0$$ for all $$x$$, and thus $$〈\mathbf a,\mathbf b〉^2-|\mathbf a|^2|\mathbf b|^2 \lt 0$$.

that monotone bounded sequences converge:
Let $$(a_n)$$ be an increasing bounded sequence; let $$A$$ be the set of terms. Then $$A$$ is bounded, so $$a=\sup A$$ exists. For $$\epsilon > 0$$, since $$a$$ is the least upper bound, there exists $$N \in \mathbb{N}$$ so that $$a_N > a-\epsilon$$. Since $$(a_n)$$ is increasing, this means $$a_n > a-\epsilon$$ for all $$n>N$$; equivalently, $$|a_n - a|<\epsilon$$, so $$\lim a_n = a$$.
A similar argument applies to decreasing sequences.

$$(u(x)v(x))'=\lim_{\delta x\to 0}\frac{u(x+\delta x)v(x+\delta x)-u(x)v(x)}{\delta x}.$$
Add and subtract $$u(x+\delta x)v(x)$$ in the numerator. Then
$$(u(x)v(x))'=\lim_{\delta x\to 0}\frac{u(x+\delta x)v(x+\delta x)-u(x+\delta x)v(x)+u(x+\delta x)v(x)-u(x)v(x)}{\delta x}$$
$$=\lim_{\delta x\to 0}u(x+\delta x)\lim_{\delta x\to 0}\frac{v(x+\delta x)-v(x)}{\delta x}+\lim_{\delta x\to 0}\frac{u(x+\delta x)-u(x)}{\delta x}\lim_{\delta x\to 0}v(x)$$
$$=u(x)v'(x)+u'(x)v(x).$$

that $$\sqrt{2}$$ is irrational. Suppose that $$\sqrt{2}\in\mathbb Q$$. So there exists an integer $$k>0$$ such that $$k\sqrt{2}\in\mathbb Z$$. Let $$n$$ be the smallest one with this property and take $$m=n(\sqrt{2}-1)$$. We observe that $$m$$ has the following properties:
$1. \ m∈Z \\ 2. \ m>0 \\ 3. \ m\sqrt{2}\in\mathbb Z \\ 4. \ m<n$
and so we come to a contradiction! Q.E.D.

Take any list of reals - for example, the algebraics: $$\{a_i\}$$.
Cover each $$a_i$$ with an interval of size $$2^{-i}$$.
Total length is 1, but every number in the list is covered.
So every list is incomplete, hence the reals are uncountable.

Corollary: The algebraics are countable, hence there are transcendentals.

$$\forall x>0$$ we have $$e^x > 1+x$$. Now set $$x=\pi/e-1$$. Then $$e^{\pi/e-1} > \pi/e$$ and so $$e^{\pi/e} > \pi$$, thus $$e^{\pi} > \pi^e$$

that the equation $$f(x)=ax^2+bx+c$$ is always tangent to $$g(x)=-ax^2+bx+c$$ with $$(a,c,b) \in \mathbb{R}^3$$.
We get:
$$ax^2+bx+c=-ax^2+bx+c$$
$$2ax^2=0$$

Since b and c are canceled, the discriminant equals 0. There is only one solution, i.e. only one point I of intersection.

$$x_I = 0; y_I = f(0) = c$$.

Pythagorean Theorem

0. Subdivide a right triangle into two smaller right triangles.

1. $$\alpha+\tau = 90 = \sigma+\tau$$ so $$\alpha = \sigma$$. Likewise $$\beta = \tau$$.
Hence the three triangles are similar.

2. By similarity,
$\frac{s}{a}=\frac{a}{c}, \quad \frac{t}{b} = \frac{b}{c}.$

3. By construction $$c = s + t$$.

4. Combining these, we have
$c = \frac{a^2}{c} + \frac{b^2}{c},$
therefore
$c^2 = a^2 + b^2.$

Take $$n>2 \in \mathbb{N}$$. Then $$2^{1/n}$$ is irrational. Suppose $$2^{1/n} = a/b$$. Then $$2b^n=a^n$$, so $$b^n+b^n=a^n$$ giving a counter-example to Fermat's Last Theorem.

Thus every $$n^{th}$$ root of 2 is irrational from 3 onwards.

that there exist irrationals $$a$$ and $$b$$ such that $$a^{b}$$ is rational:
$$\sqrt{2}$$ is irrational. Let $$z=\sqrt{2}^{\sqrt{2}}$$. If $$z$$ is rational, the claim is proved. If not, then $$z^{\sqrt{2}}=2$$ proves the claim.

that the perpendicular to the line of equation $$y = mx$$ is $$y = -\frac{1}{m}x$$.

The angle between the abcsissa axis is $$\theta = \arctan(m)$$.
So we are looking at a slope $$m'$$ (where $$\alpha = \arctan(m')$$), such that $$\theta - \alpha = \frac{\pi}{2}$$. Indeed, we will use radians.

By rearranging, we are getting,
$m' = \tan(\arctan(m) - \frac{\pi}{2})$
$\iff m' = -\frac{1}{m}$
because $$\tan = \frac{sin}{cos}$$.

Finally, the slope $$m'$$ is $$-\frac{1}{m}$$.

that $$e$$ is irrational:
From Taylor's theorem, $$\exists\xi\in(0,1)$$ such that $$e=\sum_{n=1}^{N}\frac{1}{n!} + R_{N}$$ where $$0<R_{N}=\frac{e^{\xi}}{(N+1)!}<\frac{3}{(N+1)!}$$. Assume $$e$$ is rational: $$e=\frac{a}{b}, a,b\in\mathbb{Z}, b\neq 0$$ & let $$N>\max(b,3)$$: $$\frac{N!a}{b}-\sum_{n=1}^{N}\frac{N!}{n!} = N!R_{N}$$. Since $$N>b$$, $$N!$$ must contain $$b$$ and $$N!R_{N}\in\mathbb{Z}$$ . Since $$N>3$$: $$0<N!R_{N}<\frac{3}{N+1}<\frac{3}{4}$$ which is a contradiction.

that there are infinitely many solutions to the equation $$xy = x + y$$.
i.e. $1 = \frac{1}{x} + \frac{1}{y}$
Let $$x = c/a$$ and $$y = c/b$$. We have $$1 = \frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$$
$$\iff c = a + b$$.
So $$xy = x + y$$ is satisfied when $$x = c/a$$, $$y = c/b$$ where $$c = a + b$$, for any value of a and b.

e.g.: $$\frac{37}{15} + \frac{37}{22} = \frac{37}{15} \times \frac{37}{22} = \frac{1369}{330}$$.

that the sum of the $$n^{th}$$ first odd numbers is equal to $$n^2$$.
A odd number $$x$$ can be noted $$2k-1$$. We are looking at $$\sum_{i=1}^{n}(2i - 1)$$.
And we have $$\sum_{i=1}^{n}(-1) = -n$$.
So $$\sum_{i=1}^{n}(2i - 1) = 2\left(\sum_{i=1}^{n}i\right) - n$$
$$= 2 \times \frac{n^2 + n}{2} - n = n^2$$.

that every natural number minus its palindrome is divisble by 9. Let $$\sum_{i=1}^{n}10^{i-1}a_i = a, a_i \in {0,...,9}$$.
Its palindrome is $$\sum_{i=1}^{n}10^{n-(i-1)}a_i = \overline{a}$$.
Thus $$a - \overline{a} = \sum_{i=1}^{n}a_i(10^{i-1} - 10^{n-(i-1)})$$, or we know that $$\forall a,b \in \mathbb{N}, 10^a - 10^b \equiv 0\pmod 9$$.
Consequently, $$a - \overline{a} \equiv 0\pmod 9$$.

that there are infinitely-many Pythagorean triples (that is, integers $$x,y,z$$ such that $$x^{2}+y^{2}=z^{2}$$):

Let $$a,b$$ be any integers, then:

$$(a^{2}+b^{2})^{2}=(a^{2}-b^{2})^{2} + (2ab)^{2}$$ $$\blacksquare$$

that $$\sqrt{2}$$ is irrational:
Suppose not, i.e. $$\sqrt{2} = \frac{a}{b}$$, $$a,b \in \mathbb{Z}$$ and $$a,b$$ are coprime.
Then $$\left(\frac{a}{b}\right)^2 = 2$$. Then $$a^2 = 2b^2$$, so $$a^2$$ is even. Let $$a = 2c$$, so we have $$4c^2 = 2b^2 \implies b^2 = 2c^2$$. So $$b$$ is even.
Then both $$a$$ and $$b$$ are divisible by 2, contradicting the earlier proposition.
So $$\sqrt{2}$$ can't be expressed as the ratio of two integers, which means it's irrational.

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