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.

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    $\begingroup$ No, $qx=qa+pb$. $\endgroup$ – Wuestenfux Sep 7 '19 at 17:41
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    $\begingroup$ The point is that $0<qx^n<1$, so that $qx^n$ cannot be an integer. $\endgroup$ – Lord Shark the Unknown Sep 7 '19 at 17:53
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    $\begingroup$ @LordSharktheUnknown I keep expecting this to be a proof by contradiction and not finding the contradiction. I wonder if an error in transcription caused something to be lost. (We know the transcription is imperfect because it has $qb$ where it should have $pb$.) $\endgroup$ – David K Sep 7 '19 at 18:04
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    $\begingroup$ The contradiction is that $qx^n$ must be an integer, but as $0<qx^n<1$ it cannot be. $\endgroup$ – Lord Shark the Unknown Sep 7 '19 at 18:06
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    $\begingroup$ The proof is by contradiction. The assumption is that $\sqrt{2}=p/q$ is rational. From this we conclude that $qy$ is an integer for any $y\in S.$ then we show a specific $y\in S$ where $qy$ is not an integer, namely, $y=(\sqrt2-1)^n$ for some large value of $n.$ $\endgroup$ – Thomas Andrews Sep 7 '19 at 18:20

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.

  • $\begingroup$ I understand the clarity of this proof now. I may make you frustrated when I say that I simply don't understand where $\sqrt{2}$ - 1 came from and what the significance of the inequality 0 < x < 1 is. Furthermore, why there is a $1 \over x$ > 1 line. $\endgroup$ – Daniel Mishan Sep 7 '19 at 21:40
  • $\begingroup$ The use of $\sqrt{2}-1$ was a clever choice by the author of the proof. I don't know what made him think of it. But clever choices don't have to be justified in proofs. The inequality $0 < x < 1$ is used to justify $1/x > 1$, which is used to justify $\lim_{n \to \infty} (1/x)^n = \infty$. And that is used to justify the choice of $n$ so that $(1/x)^n > q$. $\endgroup$ – Dan Velleman Sep 7 '19 at 23:49

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 $\sqrt{2} \not\in \mathbb{Q}$.

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    $\begingroup$ $qx$, not $x$, was defined as $qa+pb$. $\endgroup$ – rogerl Sep 7 '19 at 19:01

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