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 $\alpha^k \to 0\,$.
A: Below is an excerpt of my historia-matematica post, June 16, 2001 (or   sci.math May 19,2003) which presents this style of irrationality proof in an elementary form that highlights the contrast between the discreteness of a ring of (algebraic) integers
versus the denseness of a ring of fractions. See also the closely related elegant proofs using  Dedekind's notion of conductor ideal.
Theorem $ $ If an integer $\,b\,$ has
a rational square root: $\sqrt b = c/d\,$ then $\,c/d\,$ is an integer.
Proof $ $ Suppose, as hypothesized $\,w = \sqrt b  = c/d\,$ is rational.
$\,d\,$ is a common denominator for all elements in the ring  $\,R = \Bbb Z[w]\,$
since if  $\,r = m + n\:\!w \in R,\ m,n \in\Bbb Z\,$ then  $\,dr = dm + n(dw)\in\Bbb Z\,$ by $\, dw = c\in\Bbb Z.\,$ So $R$ is a subset of $\,\Bbb Z/d,\,$ i.e. every  $\,r \in R\,$ has form $\,n/d\,$ for an integer $\,n.\,$ Suppose $\,r = n/d\in R\,$ is nonintegral. We will show this leads to a contradiction and, hence, that every $\,r \in R = \Bbb Z[w]\,$ must be an integer, including $\,w = \sqrt b.$
W.l.o.g we may assume $\,0 < r < 1\,$ by taking the fractional part
of $|r|,\,$ since this too lies in $R$ and is nonintegral iff $\,r\,$ is.
But $\,r\,$ has form  $\,n/d\,$  and  $\,0 < r < 1,\,$ so $\, r\,$ is a member of the finite set $\,\{ 1/d,\, 2/d,\ldots,(d\!-\!1)/d\}.\,$  If $\,r\,$ is the smallest element of $R$ in this set then  $\,r^2\,$  is an even smaller such element, since  $\,1 > r > r^2 > 0,\,$ contradicting the minimality of $\,r.\ \ $ QED
Remark $ $ This proof works for any algebraic integer, i.e. any root of monic polynomial with integer coefficients
$\,p(x) = x^n + a x^{n-1} + b x^{n-2} + \cdots + k,\,\  a,b,..,k\in \Bbb Z,\,$ by
using  $\,d^{n-1}$  versus  $\,d\,$  as a common denominator for $\,\Bbb Z[w].\,$
It proves any rational algebraic integer is an integer (compare to the common proof by the Rational Root Test).
The proof above is special case  $\,p(x) = x^2 - b.$
The proof depends crucially on the fact
that $w$ is an algebraic integer, which implies that the ring
$\,\Bbb Z[w] = Z\langle 1,w,w^2,\ldots,w^{n-1}\rangle$  is a finite dimensional $\Bbb Z$-module, i.e.
$\,w^n, w^{n+1},\ldots$ are $\,\Bbb Z$-linear combinations of  $\,1, w, w^2,\ldots, w^{n-1}.\,$
This needn't be true if $w$ is a root of a poly $\,p(x)\,$ whose leading
coef $c\neq 1$ since then  $\,w^n = -c^{-1} (a w^{n-1} + \ldots + k)\,$
so that $c$ may intrude as a nontrivial denominator when using this
equation to rewrite  $\,w^n, w^{n+1},\ldots$ after multiplying two elts.
Essentially the proof shows that if $\,R \supset \Bbb Z\,$ is an ordered ring
extension which adjoins a new finite element then $\,R\,$ is dense,
having infinite descending chains $\, 1 > r > r^2 > r^3 > \ldots > 0\,$
in a nbhd of $0$ (hence everywhere by an additive shift), where
$\,r > 0\,$ is the fractional part of any newly adjoined finite element.
But  $\,\Bbb Z/d\,$  is discrete, not dense, hence the contradiction.
Note that I say "finite" above because it is possible to adjoin
infinite elts alone, e.g. consider $\,\Bbb Z[x]\,$ with $x$ infinite, i.e.
polys in a nbhd of $+\infty$, ordered by declaring
$\,p(x) > q(x)\,$ iff this holds true eventually for all large $x > 0,$
i.e. in a nbhd of $+\infty.\,$  This is same as declaring $p(x)$ positive
or negative as is its leading coef (the leading term eventually
dwarfs others as $\,x\to +\infty)\,$ and  $\,p > q \iff  p - q > 0,\,$ as usual.
This ordered ring extension of $\,\Bbb Z\,$ adjoins no new finite elements.
However, if we additionally adjoin the infinitesimal  $\,1/x\,$  then
we have $\, 1 > 1/x > 1/x^2 > \ldots > 0,\,$ and adding $c$ shifts this
infinitely descending chain to the nbhd of any elt $\,c \in \Bbb Z[x,1/x].$
A: You could note, that if the thing is a polynomial in x, then the powers of x form the span finite set closed under multiplication.
One can then represent the span as a vector, and the vector gives rise to a matrix, whose columns have the form $x^0 v$ to $x^m v$, and if $v\ne 0$ and the polynomial is non-prime, then any non-zero vector is non-zero.
Further note that the span of powers solving a unitary polynomial, for being closed to multiplication, can not include any rational in Q that is not in Z.  So simply demonstrating that an equation has a non-integer solution suffices to show that it has a non-rational one.
The set of solutions of this kind of set is algebraic integers eg J, the intersection between J and Q is Z.
