Proving $\sqrt{2}$ is irrational

Question

I had my first lecture the University of Toronto, MAT157 with professor Meinrenken. I am struggling with proof #3 which he briefly walked us through. I do not understand the method of thinking and there are a lot of missing lines. Can someone guide me through how this proof works? I have understood the other two proofs.

If it is any help, I am also reading through Calculus 4e, by Michael Spivak. Any input on this proof is helpful. I understand mostly everything until the last three to four lines. Mainly, how do those statements relate to each other? There are some missing steps?

Proof. Let $$S = \{x\in\mathbb R \mid x=a + \sqrt{2} b\}$$ where both $a,b\in \mathbb Z$.

S is closed under multiplication. If $x_1 , x_2 \in S$ then $x_1 x_2 \in S$

$(a_1 + \sqrt{2} b_1)(a_2 + \sqrt{2} b_2) = (a_1 a_2 + 2b_1 b_2) + \sqrt{2} (b_1 a_2 + b_2 a_1)$

Hence, if $x\in S$, then $x^n= x ... x \in S$ for all $n\in\mathbb N$

If $\sqrt{2} = {p \over q}$ then if $x\in S$, then $qx\in S$.

$x = a + {p \over q} b$ and $qx = qa + qb$

Note $x = \sqrt{2} -1 \in S, 0 < x < 1$.

So $1 \over x$ > 1

Choose $n\in \mathbb N$ with $\left({1 \over x}\right)^n> q, $ so $1 > qx^n > 0 $.

End of proof.

I am struggling specifically with the last three lines of the proof. The rest of the proof is simple for me to understand.

Answer 1

Part of the problem is that you have written the proof in a stream-of-consciousness style; proofs of subclaims are mixed in with the main proof, and variables are used without any explanation of what they stand for. Let me rewrite the proof so that it is easier to understand.

Proof: Let $S = \{x\in\mathbb{R} \mid \text{for some }a,b\in\mathbb{Z}, x = a+\sqrt{2}b\}$.

Claim: $S$ is closed under multiplication.

Proof of claim: Suppose $x_1,x_2\in S$. Then there are integers $a_1$, $b_1$, $a_2$, $b_2$ such that $x_1 = a_1+\sqrt{2}b_1$ and $x_2 = a_2+\sqrt{2}b_2$. Therefore $$ x_1x_2 = (a_1+\sqrt{2}b_1)(a_2+\sqrt{2}b_2)= (a_1a_2+2b_1b_2) + \sqrt{2}(b_1a_2+b_2a_1)\in S. $$ This proves the claim.

Hence if $x \in S$ then $x^n \in S$ for every $n \in \mathbb{N}$.

Suppose $\sqrt{2}$ is rational. Then there are positive integers $p$ and $q$ such that $\sqrt{2} = p/q$.

Claim: For all $x \in S$, $qx \in \mathbb{Z}$.

Proof of claim: Suppose $x \in S$. Then there are integers $a$ and $b$ such that $x = a+\sqrt{2}b = a+(p/q)b$. Therefore $qx = qa+pb \in \mathbb{Z}$. This proves the claim.

Now let $x = \sqrt{2}-1 \in S$. Note that $0 < x < 1$, so $1/x > 1$. Therefore $\lim_{n \to \infty} (1/x)^n = \infty$. Choose $n \in \mathbb{N}$ with $(1/x)^n > q$, so $0<qx^n<1$. But by the claims above, $x^n \in S$ and therefore $qx^n \in \mathbb{Z}$. This is a contradiction, because there are no integers between 0 and 1.

Answer 2

Welcome to Math StackExchange!

So, the third from last line. This is saying that $x = \sqrt{2}-1$, or $x + 1 = \sqrt{2}$. Now, if $x > 1$, then $x+1 > 2$. But, $x +1 = \sqrt{2}$, so that would imply $\sqrt{2} > 2$, or $4 > 2$, which is a contradiction. Therefore, $x<1$. Likewise, if $x<0$, $x+1 <1$, so $\sqrt{2} < 1$, so $2 <1$, which is a contradiction. Therefore, $x<0$, so $0<x<1$ if $x = \sqrt{2} -1$.

Now, the second to last line. If $0<x<1$, then $x$ is positive and $x <1$, so $\frac{1}{x} > 1$.

Finally, the last line. If we choose an $n \in \mathbb{N}$ such that $\left(\frac{1}{x}\right)^n < q$, then $ 1< qx^n$. Since $0<x<1$, $0<x^n<1$, so $0<qx^n<q$. But $qx^n<1$, so $0<qx^n<1$. $x$ was defined as $x = qa + pb$, where $ a,b,p,q \in \mathbb{Z}$, so by the ring axioms, $x \in \mathbb{Z}$. Since $x^n = x...x$, $x^n \in \mathbb{Z}$ by the ring axioms. Since $q \in \mathbb{Z}$, $qx^n \in \mathbb{Z}$. But $0<qx^n<1$, so $qx^n$ cannot be in $\mathbb{Z}$. This is a contradiction, so $\sqrt{2} \neq \frac{p}{q}$ for any $p, q \in \mathbb{Z}$, and thus $\sqrt{2} \not\in \mathbb{Q}$.