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:
- When did the text say that $b < n + 1$?
- 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)
(Please don't use number theory)
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, $(n+1)^2/b^2 \leq 1$.