# Suppose that a and b are positive integers such that a 2 </msup> = 3 b

Suppose that a and b are positive integers such that ${a}^{2}=3{b}^{2}$. Show that $0 and $\left(3b-a{\right)}^{2}=3\left(a-b{\right)}^{2}$
Use these and the Well-Ordering Principle to prove that no such a and b exist. From this it follows that $\sqrt{3}\notin \mathbb{Q}$.
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Step 1
Typical trick due to Hardy:
$\begin{array}{rl}\text{(assuming in lowest terms)}& \sqrt{3}& =\frac{a}{b}\frac{3}{\sqrt{3}}& =\frac{3b}{a}\\ \text{(Componendo et Dividendo)}& \sqrt{3}& =\frac{3b-a}{a-b}\end{array}$
Since $1<\sqrt{3}<2$, $b
Now $\frac{3b-a}{a-b}$ is lower than $\frac{a}{b}$, contradiction.

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Step 1
Fairly common to demand the ${a}^{2}=3{b}^{2}$ be the solution in positive integers that has the smallest possible value of $a+b$
Step 2
But then they point out $\left(3b-a{\right)}^{2}=3\left(a-b{\right)}^{2}.$. You are required to show that $a-b,3b-a$ are both positive integers, and then show that $3b-a+a-b, which is a contradiction of the assumption that $a+b$ was minimal