PROBLEM 1 We wish to prove $ \notin \mathbb {Q}$ We will do this through a proof by contradiction. Let us assume we can represent $ = {b} a,b \in $. We can reduce the expression ${b}$ completely so that it is identical in value but so that a and b are coprime, sharing no common factors except 1. By the fundamental theorem of algebra, every integer can be expressed as a product of primes, so we will represent these explicitely as a = a₁n₁ × a₂n₂ × . . . × aknk and b = b₁n₁ × b₂n₂ × . . . × bknk. Now if $ = {b}$, then \[15 = ({b})^2 = {b^2} = ^{2n_{1}} \times a_{2}^{2n_{2}} \times \times a_{k}^{2n_{k}}}{b_{1}^{2n_{1}} \times b_{2}^{2n_{2}} \times \times b_{k}^{2n_{k}}} \] \[\Rightarrow 15(b_{1}^{n_{1}} \times b_{2}^{n_{2}} \times \times b_{k}^{n_{k}}) = a_{1}^{n_{1}} \times a_{2}^{n_{2}} \times \times a_{k}^{n_{k}}\] which implies one of the ai is 15. Let us choose a₁ = 15 and reorder as necessary. Then \[15_{1}^{2n_{1}-1} \times a_{2}^{2n_{2}} \times \times a_{k}^{2n_{k}} = b_{1}^{2n_{1}} \times b_{2}^{2n_{2}} \times \times b_{k}^{2n_{k}} \] implying that B is divisible by 15. But then a and b share a multiple of 15, and we already reduced them so they are coprime, which is a contradiction. Therefore $ \notin \mathbb {Q}$ as desired.