Assume that {2 \over 3} \in \mathbb Z. That means \exists x \in \mathbb Z \;.\; x = {2 \over 3}; i.e., 3 x = 2. By the fundamental theorem of arithmetic, each number has a unique prime factorization, which means that both 3x and 2 must have the same prime factors. But 3 is a factor of 3x and not a factor of 2, which is a contradiction.

Because assume that {2 \over 3} \in \mathbb Z led to a contradiction, it must be the case that {2 \over 3} \notin \mathbb Z.

Assume that \sqrt{2} \in \mathbb Q. That means \exists x,y \in \mathbb Z \;.\; {x \over y} = \sqrt{2} where x and y are relatively prime. Rearranging, we have x^2 = 2 y^2. By the fundamental theorem of arithmetic, each number has a unique prime factorization, which means that both x^2 and 2 y^2 must have the same prime factors.

Because x and y are relatively prime, at most one of x and y can have 2 in its prime factorization; we thus proceed by cases:

Case 1: 2 is a factor of x

Then x^2 has 2 as a factor with multiplicity \ge 2. Because 2 is not a factor of y, 2y^2 has 2 as a factor with multiplicity 1. But 1 < 2, which is a contradiction.

Case 2: 2 is not a factor of x

Then x^2 also does not have 2 as a factor, but 2y^2 does, which is a contradiction.

Because both cases led to a contradiction, assuming \sqrt{2} \in \mathbb Q leads to a contradiction in general, which means it must be the case that \sqrt{2} \notin \mathbb Q.

Assume that \log_2(3) \in \mathbb Q. That means \exists x,y \in \mathbb Z \;.\; {x \over y} = \log_2(3) where x and y are relatively prime. Rearranging, we have x = \log_2(3) y = \log_2(3^y), or 2^x = 3^y. By the fundamental theorem of arithmetic, each number has a unique prime factorization, which means that both 2^x and 3^y must have the same prime factors. But all of 2^x’s prime factors are 2s and none of 3^y’s are, which is a contradiction.

Because assuming \log_2(3) \in \mathbb Q leads to a contradiction it must be the case that \log_2(3) \notin \mathbb Q.