Can you explain this proof for the irrationality of the square root of 2

For the proof of the irrationality of the square root of 2, wikipedia states the following:

If √2 is rational it has the form m/n for integers m, n not both even. Then m^2 = 2n^2, hence m is even, say m = 2p. Thus 4p^2 = 2n^2 so 2p^2 = n^2, hence n is also even, a contradiction.

Please explain the various steps of this proof - starting at: m, n are not both even. Where does that assumption come from?

Assumption 1: The square root of 2 can be expressed as the ratio of 2 integers: m and n
Assumption 2: m and n share no common factors (that is, the ratio of m/n is in its lowest form, i.e. 3/5 or 5/7 as opposed to 9/15 or 20/28)
Axiom 1: All integers are either even or odd. If they are even, then they are divisible by 2

m/n = sqrt(2)
m^2 / n^2 = 2
m^2 = 2n^2

Now, m^2 is an even number, because 2n^2 is an even number. This means that m is an even number as well, because only even numbers (in the case of integers, this is true), when squared, will produce another even number. Therefore, we can state that m = 2a, where "a" is an integer

m^2 = 2n^2
(2a)^2 = 2n^2
4a^2 = 2n^2
2a^2 = n^2

n^2 is an even number, because 2a^2 is an even number. This means that n is an even number as well. However, this contradicts assumption 2 via axiom 1, therefore, assumption 2 must be false. The conclusion is that the square root of 2 cannot be expressed as the ratio of 2 integers, and thus cannot be a rational number.

See, we are first making an assumption that a statement is true and then showing that if the statement was true, then we get an absurd contradiction.
Here, in this case we know that any fraction can be reduced to its simplest form and all that. For example 100/6 = 50/3 .
Therefore, if both the numerator and denominator were even, it is possible to factor 2 out and 'cancel' it i.e. reduce it to a simpler form.