Couldn’t you stop the proof at the statement q^2 must equal 2m^2 since it’s obvious there’s no solutions to q^2 = 2m^2.
To explain why it’s obvious, squares always have an even number if factors of two (an even multiple of any prime factor since it’s a square but just focus in on 2 here for now).
A square times two always has an odd number of factors of 2 since it’s the above (an even number of factors of two) plus one more factor.
An odd number of prime factors on one side can’t be equal to an even number of factors on the other side. q^2 can never equal 2m^2 for any integer value of a or m. Therefore it’s irrational.
Yep, that does work, although it needs a bunch of extra machinery (the fundamental theorem of arithmetic, which guarantees existence and uniqueness of prime factorisation). If you're happy to take that machinery as having already been proved - it's not entirely trivial, and it's definitely not obvious! - then you can indeed stop there.
Why do I claim that it's not obvious? Consider the ring of integers with sqrt(-5): that is, all complex numbers of the form `a + b sqrt(-5)` with a, b integers. This is a ring - it has all the nice additive and multiplicative properties that the integers do - but it doesn't have unique factorisation, because 6 has two distinct factorisations.
That’s reasonable. I’ve been taught unique prime factorization is a thing since primary school but never considered the history behind that knowledge. I feel anyone with that basis could reasonably stop at the third line here. In fact they could quickly create a generalization since a^2 could clearly never equal b(c^2) unless b was also a square for integer a and c but that obviousness is based on a lot of other knowledge.
To explain why it’s obvious, squares always have an even number if factors of two (an even multiple of any prime factor since it’s a square but just focus in on 2 here for now).
A square times two always has an odd number of factors of 2 since it’s the above (an even number of factors of two) plus one more factor.
An odd number of prime factors on one side can’t be equal to an even number of factors on the other side. q^2 can never equal 2m^2 for any integer value of a or m. Therefore it’s irrational.
This ends the proof much earlier right?