Wanting to write mathematics
Sep. 12th, 2023 10:43 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
alan_croweYears ago I registered a Wordpress blog, because Wordpress had support for LaTex. I've just started trying to use it. But I've created a page https://cawtech.wordpress.com/partial-differentiation-puzzle-from-an-old-book/ when I intended to create a post. I'll fix that sometime. At least the mathematics shows up.
Meanwhile, I've noticed that there are lots more HTML entities than I remember. So I can write a ∈ {a,b,c} and {a} ⊂ {a,b,c} while ℘({a,b}) = {∅,{a},{b},{a,b}}. Since there is a square root sign I can write my favourite proof of the irrationality of √2
Suppose that √n∈ℚ. Write √n = a/b where a and b are coprime. Then b√n=a. Also, by Bezout's Theorem there are r,s∈ℤ such that ra+sb=1. Multiply by √n
ra√n + sb√n = √n
We already have b√n=a so sb√n is an integer. Does anything similar happen for ra√n ? Yes, just multiply b√n=a by √n obtaining bn=a√n, from which we deduce that ra√n must also be an integer. Thus √n is the sum of two integers and itself an integer.
That is awkward; a rational square root always turns out to be a whole number! Since 1<2<4 we know that 1<√2<2. But there is no integer between one and two so √2 is irrational.
Meanwhile, I've noticed that there are lots more HTML entities than I remember. So I can write a ∈ {a,b,c} and {a} ⊂ {a,b,c} while ℘({a,b}) = {∅,{a},{b},{a,b}}. Since there is a square root sign I can write my favourite proof of the irrationality of √2
Suppose that √n∈ℚ. Write √n = a/b where a and b are coprime. Then b√n=a. Also, by Bezout's Theorem there are r,s∈ℤ such that ra+sb=1. Multiply by √n
ra√n + sb√n = √n
We already have b√n=a so sb√n is an integer. Does anything similar happen for ra√n ? Yes, just multiply b√n=a by √n obtaining bn=a√n, from which we deduce that ra√n must also be an integer. Thus √n is the sum of two integers and itself an integer.
That is awkward; a rational square root always turns out to be a whole number! Since 1<2<4 we know that 1<√2<2. But there is no integer between one and two so √2 is irrational.
