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

If m and n are both even, then m/2 and n/2 are both integers, and
√2 = m/n = (m/2)/(n/2)
is another way to write √2 as a ratio of two integers. If m/2 and n/2 are both even, we can divide by 2 again, and keep going until one of m and n is odd. (More concisely, every fraction can be written in reduced form, which means that m and n have no common factors. Specifically, they can't both be divisible by 2.)

Now we're assuming √2 = m/n. Squaring, we get 2 = m^2/n^2, or equivalently m^2 = 2n^2. This proves that m^2 is divisible by 2. But the square of an odd number is an odd number; if m were odd, then m^2 would also be odd. Thus the fact that m^2 is even proves that m is even. By definition, m=2p for some integer p. Plugging this into m^2=2n^2, we get 2n^2 = (2p)^2 = 4p^2. Therefore n^2 = 2p^2. As before, this proves that n is even.

every rational number is of form p/q where p and q are integers
but 4/6 and 2/3; 10/5 and 6/3; these pairs are same.
so if p and q are relatively prime, i.e. gcd(p,q) = 1 then it is an unique representation.
again -2/3 and 2/-3 are also same. so we have to fix the sign of p or q, then it will be completely unique.
thus rational numbers are of form p/q where p is an integer and q is a natural number and p and q are relatively prime.

so if √2 is rational, it must be of form m/n where gcd(m,n) = 1
and thus it goes.
in the end, we find both m and n are even. thus they have a common divisor which is 2. this is an contradiction.

hope this helps!