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.

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:

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
\[c = \frac{a^2}{c} + \frac{b^2}{c},\]
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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