$1 < a$ and $b\ne0$ imply $1<a^b$

Question

$1 < a$ and $b\ne0$ imply $1<a^b$ when $a,b$ are arbitrary nonnegative integers.

I've tried to prove it by induction.

I've assumed that $b < a$ (Is valid my assumption?)

I'm using this definition for $a<b$ \begin{align*} a<b \Leftrightarrow (\exists k\in \mathbb N)a + (k + 1) = b\\ \end{align*}

Base P(1)

$0<1\le 1 < a \implies1<a^1$

$1<a^1 \equiv (1+(k+1))^1=a^1$

Hyphotesis P(n)

$0<1\le n < a \implies1<a^n$

$1<a^n \equiv (1+(k+1))^n=a^n$

Induction step

$0<1\le n + 1 < a \implies1<a^{n+1}$

we have $(1+(k+1))^n*(1+(k+1))=a^n*a$ (hyphotesis)

then $(1+(k+1))^n = a^n$, because $(1+(k+1)) = a$ (base)

(here I used the property $ac= bc => a = b$ with $c \ne 0$)

Answer 1

It wasn't valid to assume $b < a$ and you never used it

I can't follow your induction step.

If we assume $1 < a^n$ then $a^{n+1}=a^n*a$ and we want to prove $a^n*a > 1$ or that there is a $k\in \mathbb N$ so that $1 + (k+1)= a^n*a$

We know that $1 < a$ so there is a $j$ so that $1 +(j+1)=a$. And we know that $1 < a^n$ and that there is an $m$ so that $1+(m+1) =a^n$ so $a^{n+1} = (1 + (j+1))(1 + (m+1)) = 1 + (j+m+2) + (j+1)(m+1)$. So let $k = (j+m+2) + (j+1)(m+1)\in \mathbb N$ and we are done.

Answer 2

I'm not sure that induction is required for this proof. Proof by two cases:

Case 1: If $b=1$ then it is obvious that $a^b=a>1$.

Case 2: If $b>1$ then $a^b=a\cdot a^{b-1}\geq a>1$

The first inequality, that $a^{b-1}\geq 1$, is justified by the fact that any power of a nonzero natural number is itself a nonzero natural number (since $\mathbb{N}$ equipped with multiplication has no non-zero zero divisors).