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.

http://aperiodical.com/2017/06/the-curious-mathmo-talks-to-david-roberts/ https://mathstodon.xyz/media/HzcgL03yPRK02ePv1E4

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 \(〈\mathbf a,\mathbf b〉^2-|\mathbf a|^2|\mathbf b|^2 \lt 0\). #proofinatoot

#proofinatoot 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

\( (u(x)v(x))'=\lim_{\delta x\to 0}\frac{u(x+\delta x)v(x+\delta x)-u(x+\delta x)v(x)+u(x+\delta x)v(x)-u(x)v(x)}{\delta x} \)

\( =\lim_{\delta x\to 0}u(x+\delta x)\lim_{\delta x\to 0}\frac{v(x+\delta x)-v(x)}{\delta x}+\lim_{\delta x\to 0}\frac{u(x+\delta x)-u(x)}{\delta x}\lim_{\delta x\to 0}v(x) \)

\( =u(x)v'(x)+u'(x)v(x). \) #proofinatoot

#proofinatoot 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.

#ProofInAToot

\( \forall x>0 \) 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 \(e^{\pi} > \pi^e\)

#proofinatoot 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 #proofinatoot

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.

Contradiction.

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

#proofinatoot 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.

#proofinatoot 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 \(-\frac{1}{m}\).

#ProofInAToot 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.

#proofinatoot 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.: \(\frac{37}{15} + \frac{37}{22} = \frac{37}{15} \times \frac{37}{22} = \frac{1369}{330}\).

#proofinatoot 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 \(\sum_{i=1}^{n}(2i - 1) = 2\left(\sum_{i=1}^{n}i\right) - n\)

\(= 2 \times \frac{n^2 + n}{2} - n = n^2\).

#proofinatoot 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, \(a - \overline{a} \equiv 0\pmod 9\).

#proofinatoot 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:

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

#proofinatoot 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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Joined May 2017