Definition: Two magnitudes are commensurable if and only if their respective lengths can be compared without remainder.Lemma: If the square of a positive whole number is even then the whole number itself is even.Assume the square root of two is commensurable with some positive whole number magnitude i.e. it is equivalent to the ratio of two commensurable magnitudes with positive whole number lengths that share no common factor that we will call: Length One and Length Two.This implies that twice the square of Length Two is equal to the square of Length One, implying that Length One is even which means that twice the square of the Length Two is equal to the square of an even number.But since twice the square of Length Two is the square of an even number, then the square of Length Two is an even number, and hence Length Two is an even number.But then Length One and Length Two share a common factor, which contradicts our assumption that Length One and Length Two share no common factors.Thus by reductio ad absurdum we must conclude that the square root of two cannot be expressed by the ratio of two commensurable magnitudes and thus there does not exist any positive whole number magnitude which can be evenly compared with the square root of two.
>>17053193>compared without remaininggibberish
>>17053240You cant visualize it?
>>17053306>visualizeyour imagination does not constitute a formal system
>>17053309Greeks did math entierly with it.
>>17053193Let us strip this down to a more common parlance, just to make sure we all agree on the meaning.It is safe to assume we can translate magnitude to be a number, specifically the reals or a subset, likely with only positive numbers in mind. So I will assume everything to be strictly greater 0.Because it is easier I will denote "~" to mean "is commensurable with". "Comparing" is very ambiguous but given he speaks of remainders it is clear that he means quotients. I suppose the most intuitive exegesis of it would be to assume either length can be used as a yard stick for the other such that some integer multiple of both winds up equal. [eqn]x \sim y \iff \exists a,b\in\mathbb{N}^*: x n = y m[/eqn]The lemma reads, slightly generalized:[eqn]x^2 \in 2\mathbb{Z} \Rightarrow x \in 2\mathbb{Z}[/eqn]So far all of this looks good.
>>17053336pardon, of course this was meant to be[eqn]x \sim y \iff \exists a,b\in\mathbb{N}^*: x a = y b[/eqn]which is strictly equivalent to[eqn]x \sim y \iff \frac{x}{y}\in\mathb{Q}[/eqn]Now, on to the proof.
