Use strong induction to show the following :
$$\sqrt2\:\text{is an irrational number}$$ $\\$
$\color{red}{\text{Note}}$ : P$(n)\equiv$ $\sqrt{2}$ $\neq \large\frac{n}{b}\small\text{,}\:$$\forall b\in\mathbb{N}$.
I would like to know how to start this proof
Thanks!
The following is a mild variant of the usual proof, given in the strong induction language that is specified in the question. We show by induction that for any $n\ge 1$, there is no integer $b$ such that $\sqrt{2}= \frac{n}{b}$.
The result is true for $n=1$, since $\sqrt{2}$ is not the reciprocal of an integer. Suppose the result holds for all $k\lt n$. We show the result is true for $n$.
Suppose to the contrary that there is an integer $b$ such that $\sqrt{2}=\frac{n}{b}$. Then $n^2=2b^2$, so $n$ is even, say $n=2n'$. Then quickly we find that $b^2=2(n')^2$, so $b$ is even, say $b=2b'$. It follows that $\sqrt{2}=\frac{n' }{b'}$.
This contradicts the induction assumption that $\sqrt{2}$ cannot be expressed as a ratio of integers, with numerator positive and less than $n$.
