Q:

Alternative proof of $\sqrt{2}$ is irrational assistance.

I was hoping for more insight of this specific proof of the irrationality of $\sqrt{2}$. According to this proof by Alex Healy: 
"Consider the set $W=a+b \sqrt{2}$ , a,b integers. Clearly $W$ is closed under multiplication and addition. Define $\alpha=(\sqrt{2}−1)$, an element of $W$. Obviously, $0 \lt \alpha \lt 1$, so that
$$\alpha^k \to 0 \;\;\text{as}\;\; k \to \infty \tag{1}$$
Assume $\sqrt{2}=p/q$. Since $W$ is closed, $\alpha^k=e+f \sqrt{2} =(eq+fp)/q \ge 1/q$ contradicting $(1)$." 
Here's the link to the proof, it is the 9th one:
http://www.cut-the-knot.org/proofs/sq_root.shtml 
Thanks in advance!!

A:

Consider the set $W=a+b \sqrt{2}$ , a,b integers. Clearly $W$ is closed under multiplication

$(a+b\sqrt{2})(a'+b'\sqrt{2}) = (aa'+2bb') + (ab+a'b)\sqrt{2}$ where $aa'+2bb'$ and $ab+a'b$ are integers.

and addition.

$(a+b\sqrt{2})+(a'+b'\sqrt{2}) = (a+a') + (b+b')\sqrt{2}$ where $a+a'$ and $b+b'$ are integers.

Define $\alpha=(\sqrt{2}−1)$, an element of $W$. Obviously, $0 \lt \alpha \lt 1\,$, so that
$$\alpha^k \to 0 \;\;\text{as}\;\; k \to \infty \tag{1}$$

This requires that for any $\forall \epsilon \gt 0$ there exists $K \in \mathbb{N}$ so that $\alpha^k \in [0, \epsilon)\,$ for all $\forall k \ge K\,$.

Assume $\sqrt{2}=\cfrac{p}{q}$. Since W is closed, $\;α^k=e+f \sqrt{2}= \cfrac{eq+fp}{q}  \ge \cfrac{1}{q}\;$ contradicting $(1)$.

$eq+fp$ is obviously an integer. It cannot be $0$ because that would mean $\alpha^k=0\,$, which is not possible since $\alpha \gt 0\,$. Therefore $\,eq+fp \gt 0\,$ and, since it's an integer, this is equivalent to $\,eq+fp \ge 1\,$. It follows that $\cfrac{eq+fp}{q}  \ge \cfrac{1}{q}\,$, which gives $\alpha^k \ge \cfrac{1}{q}\,$.
But this implies that $\alpha^k \not \in [0,\epsilon)$ for any $\forall \epsilon \lt \cfrac{1}{q}$ and $\forall k$ which contradicts the fact that $\alpha^k \to 0\,$.

