Induction of $a

Question

Show that for any $a,b\in \mathbb N$ with $b>1$, there exists $n\in \mathbb N$ such that $a<b^n$.

I went on to prove this by induction on $b$.

Base Case: $b=2$: $a<2^n \implies$ Pick $n=a$ since $a<2^a$ is true for all $a\in \mathbb N$. (Do I need to prove this by induction as well?)

Induction Step: Since I assumed that $a<k^n$ is true and I know that $k^n<(k+1)^{n}$, then I can deduce that $a<(k+1)^n$. Therefore $P(k+1)$ is true as well.

So the statement is true. But I feel that something might be wrong in this proof, any help of advice is appreciated!

Answer 1

You should prove $a\lt2^a$ (induction is fine). Then since $b\ge2$, $a\lt 2^a\le b^a$.

Another approach is to use Bernoulli's Inequality (which has a simple inductive proof) to get $$ 1+n(b-1)\le(1+(b-1))^n=b^n $$ Therefore, we just need to choose $$ n\gt\frac{a-1}{b-1} $$

Answer 2

You didn't prove $P(k+1)$. That would be: 'there exists $n\in \mathbb{N}$ such that $a<(k+1)^n$'; not: '... such that $a<k^{n+1}$.'

By the way, the $n$ need not be the same as in $P(k)$.

Answer 3

First of all, the statement is false in the way it is stated: With $a=b=$, there does not exist such a number $n$. The statement needs to be adjusted by adding the condition that $b>1$.

You cannot "Pick $a=n$" because $a$ is given and you are supposed to show the existence of $n$. Did you mean "Pick $n=a$"? Also, if you are going to prove $a<2^a$ (which you'd have to), note that the proof for that can be readily adjusted to show $a<b^a$ directly, i.e., without induction on $b$.

Instead, I'd suggest to try induction on $a$. For this, note that $a<b^n$ quickly implies $a+1<b^{n+1}$.

Answer 4

This is only true for $b>1$. There is no need for induction if you are going to assume $a<2^a$ since then $a<b^a$ for all $b>1$.