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Going through "Elementary Number Theory" by Kenneth Rosen, there is a proof showing $\sqrt2$ is irrational using the well-ordering principle. I believe I've understood it correctly, but I'm uncertain. Would one as such kindly read through my understanding of it and correct correct me if / where, I went wrong.

Proof: Suppose $\sqrt2$ is rational. Then there exist positive integers a,b such that

$$\sqrt2 = a/b$$

and

$$a = b\sqrt2$$

Let S be the set of all such positive integers k, $k = m\sqrt2, m \in N$

That is, S holds all values where that is true, for example $k_1=b\sqrt2, k_2=2b\sqrt2$ and so on.

$$S = \{k\}, k = m\sqrt2, m \in N$$

So I'm creating a set containing all possible integer multiples of a from $\frac ab$, where $\frac ab=\sqrt2$

Then $S$ would be a non-empty set of positive integers.

By the well-ordering principle, $S$ has a smallest element. Let $s$ be the smallest such integer of $S$.

By definition of S,

$$ s = t\sqrt2$$

for some $t, t \in N$

I'm now going to start looking for an element smaller than s, that is part of the set.

Since $$\sqrt2 > 1$$

$$s\sqrt2 > s$$

and $$s\sqrt2 - s > 0$$

That is, $s\sqrt2 - s$ is positive.

Now,

$$s\sqrt2 = (t\sqrt2) * \sqrt2$$ $$= t2$$

which is an integer, and as such $$s\sqrt2 - s = t2 -s$$

and $s\sqrt2 - s$ is a positive integer.

So I've now shown that $s\sqrt2 - s$ is a positive integer. I want to show that it is part of S

But, $$s\sqrt2 - s = s\sqrt2 - t\sqrt2$$

$$= (s - t)\sqrt2$$

and since both s and t are integers, and by definition $s = t\sqrt2$, s is larger than t by a factor of $\sqrt2$, and as such $(s - t)$ is a positive integer.

So I've now shown that $s\sqrt2 - s$ can be written as $m\sqrt2$, so it is also part of S

That being the case, $(s -t)\sqrt2$ is also an element of S.

I now need to show that it is smaller than s, contradicting our initial claim

Noting that $$\sqrt2 < 2$$

We have

$$s\sqrt2 < 2s$$ $$s\sqrt2 - s < 2s - s$$ $$s\sqrt2 - s < s$$ $$s\sqrt2 - t\sqrt2 < s$$ $$(s -t)\sqrt2 < s$$

thus contradicting our initial claim that s is the smallest element of S.

By contradiction, our assumption that there exists positive integers a, b such that $a/b = \sqrt2$ does not hold and $\sqrt2$ is as such, not rational.

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    I'm used to seeing a different proof so I could be missing something but I don't follow the part where $k = y\sqrt{2}$. If $k = 1$, then you claim there is a value $y'$ such that $1 = y'\sqrt{2}$. What value $y'$ satisfies this? (Note: You've started with the assumption that *there exists* $a$,$b$ that satisfy $\sqrt{2} = a/b$. This does not imply that you can write every integer as $a = b\sqrt{2}$ for some other integer $b$) – 2017-01-17
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    I have a small doubt. We have, there exists positive integers $a$ and $b$ such that $a = b\sqrt{2}$. Now you're considering the set $S=\{k, k\sqrt{2}\}$ for all positive integers $k$. So, $k\sqrt{2}$ may not be an integer for all $k$. Then why is S a non-empty set of "positive integers"? And so why can we apply well-ordering principle on S? – 2017-01-17
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    Thank you both, I understand what you mean and I've updated my initial proof to fix these issues in my thinking. – 2017-01-17

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Your definition of $S$ is off. If both $k$ and $k\sqrt{2}$ are positive integers, then only $k\sqrt{2}$ goes into $S$, not $k.$ Write instead $S=\{k\sqrt{2} | k \mbox{ and } k\sqrt{2} \in \mathbb{N}\}.$

Just after the place you say "To continue...", you say "...and a member of $S.$" I don't think you've shown that $s\sqrt{2}-s$ is a member of $S$ yet. You need to show that it's 1. an integer (which you did) and 2. It's equal to an integer times $\sqrt{2}$. This second thing is achieved by writing $s\sqrt{2}-s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}.$

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    Thanks a lot, I understood the second paragraph, so I updated to include that. But I don't understand the first paragraph, the book does the same thing but I don't understand why. That is, if S is defined to include only 1 value, how can we use the well ordering principle to state that there is a smaller value? I do understand why my definition of S was off, I've since updated to a different definition of S I think satisfactory? But irrespective, would you kindly explain how it would work if S only has 1 value, cause yes, the book does the same and loses me there. – 2017-01-17