# Explanation of this (strong) induction statement

This might seem pretty simple and stupid to some but I am really not able to get it....

I was reading the proof that $\sqrt{2}$ is irrational (using induction) here.

I could understand most of it except when it says:

Then $b^2 = 2a'^2$ or $2 = b^2/(a')^2$; a contradiction of the induction hypothesis since b < n+1.

I don't get two things:

1. When did the text say that $b < n + 1$?
2. How does the above equation seem to be a contradiction to the above inequality?

I have been re-reading this for quite some time but don't seem to grasp it fully. Please explain this concept to me. I am fairly new (just one day) to strong induction (can handle simple induction well)

• Feels very sloppy proof, the induction proof. They say $2 =\frac{(n+1)^2}{b^2}$. So it would be logical why $(n+1)^2 > b^2$ is true, but not immediately apparent why $n+1 > b$ Nov 10, 2015 at 15:34
The induction is based on values of n; after showing that $2 \neq a^2/b^2$ for $a=1$ and for all $b$, i.e. the base step, the inductive step then assumes that the statement holds for all $a$ where $1 \leq a \leq n$ and tries to prove it for $a=n+1$. What the statement says is that, for some integer $a=n$, for any integer b, $2$ cannot be expressed as a ratio between $a^2$ and $b^2$. We then assume this for any $a$ less than or equal to $n$. Thus, 2 cannot be expressed in the form $b^2/a'^2$ either since $b<n+1$.
If you're asking why $b<n+1$, just assume the opposite: say $b \geq n+1$. Then $b^2 \geq (n+1)^2$, and so, $(n+1)^2/b^2 \leq 1$.