Suppose that $a$ and $b$ are positive integers such that $a^2 = 3b^2$. Show that $0 < b < a < 2b$ and $(3b-a)^2 = 3(a-b)^2$

Question

...Use these and the Well-Ordering Principle to prove that no such $a$ and $b$ exist. From this it follows that $\sqrt{3} \notin \mathbb{Q}$

I have no idea where well-ordering principle comes in for this question?

Can you prove $0<b<a<2b$? What have you tried? Is it possible that $0<a<b$? Why not? — Somos, Oct 30 '21 at 03:00
I've tried square rooting both sides and got a is sqr(3) times b, so b must be greater than 0 and smaller than a... a is equal to 1.732b so smaller than 2b. But I don't know where the Well-Ordering Principle — bloomsdayforever, Oct 30 '21 at 03:04

score 1 · Answer 1 · answered Oct 30 '21 at 04:05

fairly common to demand the $a^2 = 3 b^2 $ be the solution in positive integers that has the smallest possible value of $a+b$

But then they point out $(3b-a)^2 = 3 (a-b)^2 .$ You are required to show that $a-b, 3b-a$ are both positive integers, and then show that $$ 3b-a + a-b < a+b, $$ which is a contradiction of the assumption that $a+b$ was minimal

Ng Chung Tak · Answer 2 · 2022-02-04T06:29:32.553

Typical trick due to Hardy:

\begin{align} \sqrt{3} &= \frac{a}{b} \tag{assuming in lowest terms} \\ \frac{3}{\sqrt{3}}&= \frac{3b}{a} \\ \sqrt{3} &= \frac{3b-a}{a-b} \tag{Componendo et Dividendo} \end{align}

Since $1<\sqrt{3}<2$, $$b<a<2b \implies 0<\color{blue}{a-b}<\color{red}{b}$$

Now $\dfrac{3b-a}{\color{blue}{a-b}}$ is lower than $\dfrac{a}{\color{red}{b}}$, contradiction.

See similar trick on another post of mine here.

Also papers here, here and also here.

Suppose that $a$ and $b$ are positive integers such that $a^2 = 3b^2$. Show that $0 < b < a < 2b$ and $(3b-a)^2 = 3(a-b)^2$

2 Answers2