Prove a number is irrational (explanation) Here is the solution to a problem in which I am supposed to prove x is irrational. However, I don't understand why it is true that p divides 2 and q divides 1. Where is this coming from? 

 A: If you've taken abstract algebra, you've seen this result before. If you have a polynomial $a_nx^n + \cdots + a_1x+ a_0$ then the set of rational roots are of the form $r/s$ where $r \mid a_0$ and $s \mid a_n$.
A: Suppose $1+\sqrt{1+\sqrt{2}}$ is rational, then $\sqrt{1+\sqrt{2}}$ is rational, from here we conclude $(\sqrt{1+\sqrt{2}})^2=1+\sqrt{2}$ is rational. From here we conclude $\sqrt2$ is rational. Let $\sqrt{2}=\frac{p}{q}$ with $p$ and $q$ coprime, we conclude $2p^2=q^2$, we conclude $q$ is even, so $q=2k$, then $2p^2=4k^2\implies p^2=2k^2$, so $p$ is also even, a contradiction!
A: If $x=p/q$ with non-zero co-prime integers $p, q,$ and $x^4-4x^3+4x^2-2=0$ then $$2q^4=p(p^3q^4-4p^2q^3+4pq^2)$$ so $p| 2q^4,$ and since $\gcd (p,q)=1\implies \gcd (p,q^4)=1,$ we have $p|2.$.... And $$p^4=q(4p^3-4p^2q+2q^3)$$ so $q|p^4,$ and since $\gcd (p,q)=1\implies \gcd (p^4,q)=1,$ we have $q|1.$
Another method is to note that if $A\in Q$ and $B\not \in Q$ then $A+B\not \in Q$ and $\sqrt B\not \in Q .$ Because if $C=A+B\in Q$ then $B=C-A\in Q,$ and if $\sqrt B\in Q$ then $B=(\sqrt B)^2\in Q.$ Knowing that $\sqrt 2\not \in Q,$ we have $$\sqrt 2\not \in Q\implies 1+\sqrt 2\not \in Q\implies \sqrt {1+\sqrt 2}\not \in Q\implies 1+\sqrt {1+\sqrt 2} \not \in Q.$$
