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I'm reading Hardy's "Course of Pure Mathematics" and got stuck in one of his early proofs. I reckon this should be really easy, but it really baffled me:

He supposes $\left(\frac{p}{q}\right)^2 = 2$. So $p^2 = 2q^2$. Then he says it's easy to see that from this it follows that $(2q - p)^2 = 2(p - q)^2$.

I suppose this last part is easy, but I just can't see how he got there. Anyone care to explain his reasoning?

Alain Remillard
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Bruno
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    Expand each side of the last equality – Pedro Mar 01 '20 at 11:15
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    I am not sure whether they actually derived $(2q-p)^2=2(p-q)^2$ from $p^2=2q^2$. The equivalence of those equations is however easy to verfify and I guess the first equation was needed for the proof. – Peter Mar 01 '20 at 11:16
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    I am not sure about the question that you are asking. Is it how did Hardy deduce that $(2q-p)^2=(p-q)^2$ from the equality $p^2=2q^2$? – José Carlos Santos Mar 01 '20 at 11:30
  • Yes, thanks for the answers and for rewriting the question. I should make it more clear that what actually baffled me is how he derived the last equation from the previous one. – Bruno Mar 01 '20 at 11:30
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    That proof goes back to how Greek mathematicians would check that two numbers are commensurable. You subtract the smaller segment from the larger as much as you can. Then do the same with the difference and the smaller segment and repeat until eventually the subtracted segments match. This argument has many similar presentations. See this one. All consist in trying to check commensurability and showing that the segments never match. –  Mar 01 '20 at 11:48
  • I added a very simple proof (that generalizes widely). – Bill Dubuque Mar 03 '20 at 00:30
  • @flan Yes, and this is closely connected with mediant and best rational approximations, Farey series, Stern-Brocot series, etc - see my answer and its links. – Bill Dubuque Mar 03 '20 at 00:34
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    Well the reason is easy (if you know that's where you want to go) $2q^2 = p^2\implies 4q^2 = 2p^2\implies 4q^2-4pq=2p^2-4pq\implies 4q^2-4pq+p^2=2p^2 -4pq +2q^2\implies (2q-p)^2 = 2(p^2-2pq+q^2)=2(p-q)^2$. And it's easier to work backwards $(2q-p)^2=2(p-q)^2\iff4q^2-4pq+p^2=2p^2-4pq+2q^2\iff4q^2+p^2=2p^2+2q^2\iff2q^2=p^2$). The real question is why did he do it and how did he think to do it. – fleablood Mar 03 '20 at 01:19
  • Well, it's much more interesting to see a geometric proof, but from $p^2=2q^2$ you get $(2q-p)^2=\color{green}{4q^2}-4pq+\color{blue}p^2=\color{green}{2p^2}-4pq+\color{blue}{2q^2}=2(p-q)^2$. – egreg Mar 06 '20 at 21:44

3 Answers3

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The equality $(p-q)^2=(2q-p)^2$ is a natural thing to stumble upon if you think long enough about this.

It is possible to get there geometrically. Since $p^2=2q^2$, there is a right triangle such that the length of both legs is equal to $q$, whereas the length of the hypotenuse is equal to $p$:

enter image description here

Now, see it as if it was made of paper and fold its lower left corner as in the picture below. Then you will get a new isosceles right triangle (in red). The length of its hypotenuse is $q-(p-q)=2q-p$ and the length of each of its legs is $p-q$. But then it is similar to the original triangle and therefore $\frac pq=\frac{2q-p}{p-q}$. So,$$\frac{(2q-p)^2}{(p-q)^2}=\frac{p^2}{q^2}=2.$$

enter image description here

José Carlos Santos
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By below: $\ \sqrt{2} = \color{#c00}{\dfrac{p}q} = \color{#0a0}{\dfrac{2q}p}\,$ implies that $\, \sqrt 2 = \dfrac{\color{#0a0}{2q}-\color{#c00}p}{\color{#0a0}p-\color{#c00}q}\,\Rightarrow\,2(p-q)^2 = (2q-p)^2$

Proof $ $ note $\ x = \color{#c00}{\dfrac{p}q} = \color{#0a0}{\dfrac{a}b}\,\Rightarrow\!\!\!\!\!\!\!\!\!\underbrace{\begin{align}\color{#0a0}{bx = a}\\\color{#c00}{qx = p}\end{align}}_{\textstyle (\color{#0a0}b\!-\!\color{#c00}q)\,x = \color{#0a0}a\!-\!\color{#c00}p}\!\!\!\!\!\!\!\!\!\Rightarrow x = \dfrac{\color{#0a0}a-\color{#c00}p}{\color{#0a0}b-\color{#c00}q}$

Geometrically if vectors $(b,a)$ & $(q,p)$ have equal slope then their difference $(b\!-\!q,a\!-\!p)$ has the same slope, which may be viewed as a special case of a mediant approximation, or the secant method.

Bill Dubuque
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I was actually directed to this picture, which probably illustrates where the insight comes from: https://commons.m.wikimedia.org/wiki/File:NYSqrt2.svg#mw-jump-to-license

Bruno
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