Let's prove that if \(k\) is an integer, then \(\gcd(k, k+1) = 1\). That is, any two consecutive integers are coprime.

Twitter proof:

— Mathspp (@mathsppblog) November 14, 2020

Let k be an integer and let d be the greatest common divisor of k and k+1. We have that (k+1)/d=k/d+1/d and both (k+1)/d and k/d are integers, so 1/d must be an integer and we can only have d=1.https://t.co/pItsAnueib

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]]>Let's prove that if you want to maximise \(ab\) with \(a + b\) equal to a constant value \(k\), then you want \(a = b = \frac{k}{2}\).

Twitter proof:

— Mathspp (@mathsppblog) November 5, 2020

Take s = k/2. If a = s+h then b = s-h, from which we get that ab = (s+h)(s-h) = s^2 - h^2. Because we know h^2 >= 0, ab is maximised when h = 0, that is a = b = s = k/2.https://t.co/9ypegSMF2F

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]]>Let's prove that, if a set has size \(n\), then that same set has exactly \(2^n\) subsets.

Twitter proof:

— Mathspp (@mathsppblog) August 6, 2020

Take a set of size n. For any of its subsets, we can label items with 0/1 depending on whether or not the item is in the subset or not, and to any such labelling corresponds a single subset. There are 2^n such labellings, hence 2^n subsets.https://t.co/5uxKAi7D9T

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]]>Let's prove that there are two irrational numbers, call them \(a\) and \(b\), such that \(a^b\) is a rational number! And let's do it in a tweet.

Twitter proof:

— Mathspp (@mathsppblog) July 23, 2020

Let a=b=√2. If a^b is rational then there is nothing to be done. Assume a^b is irrational. Redefine a=√2^√2. Notice how a^b = (√2^√2)^√2 = (√2)^2 = 2 which is clearly rational. QED.https://t.co/3sFinyzJwu

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