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I want to check if I understand proof by induction, so I want to proof the following:

$a^n<b^n$ for $a,b \in \mathbb{R}$, $0<a<b$, $n \in \mathbb{N}$ and $n>0$

Here's my attempt:

Base Case

If $a=1$ and $b=2$, then $1^n < 2^n$ for any $n$

Induction Step

Now I need to show that if $a^n < b^n$ is true, then $a^{n+1} < b^{n+1}$ is true too.

So, I have:



$a^n < b^n$

So, assume $a^n < b^n$, then $ a^n . a = a^{n+1} < b^n .a \tag{1}$ $a^n . b < b^n . b = b^{n+1} \tag{2}$

I don't know how to bring (1) and (2) together.

Assuming my workings follows the steps for a proof by induction, how do I complete the proof? If this workings is not consistent with a proof by induction, what is the proper way to proof the assertion by a proof of induction?

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For starters, the assertion is only true if $a \lt b$. Also, your base case would be for $n=1$. In other words, Base Case: $a^1 \lt b^1$ which is true. Since the statement is to hold for all values of n the induction is on that variable and not $a$ and $b$. Make sense? – Patrick Oct 1 '13 at 14:45
Your idea behind the induction step and how it should show that the $n+1$-st case should hold is correct. For your (1), consider what I said about the requirement that $a \lt b$. – Patrick Oct 1 '13 at 14:49
Are you guaranteed that $0\lt a\lt b$? Otherwise the statement is not true for all $n$ and all $a,b$. Consider $a=-1, b=1, n=2$. – abiessu Oct 1 '13 at 14:57
Yes, $0<a<b$. Sorry for the many edits. I will remember to be more careful when formulating my own questions next time. – mauna Oct 1 '13 at 15:43
up vote 2 down vote accepted

As it stands, your proposition is still incorrect. Consider for example $a=-2,b=-1$. Then $a<b,$ but $b^2<a^2$. Now, if we make the further assumption that $a,b$ are both positive, then it works out fine.

Your base case is actually immediate, since $a^1=a<b=b^1$. (We must allow $a,b$ to be arbitrary positive numbers with $a<b.$ If we switch to specifics, then we haven't actually proved that proposition.

For your induction step, you're quite close. Assuming that $$a^n<b^n$$ for some $n$, we can multiply by the positive number $a$ to obtain $$a^{n+1}<b^n\cdot a.\tag{$\heartsuit$}$$ On the other hand, we know (or at least, should be able to show) that $b^n$ is positive (since $b$ is), so since $a<b,$ then multiplying both sides by $b^n$ gets us $$b^n\cdot a<b^{n+1}.\tag{$\spadesuit$}$$ Putting $(\heartsuit)$ and $(\spadesuit)$ together finishes the proof.

Edit: Actually, we can even weaken it to the case that $a=0.$ The proof is even simpler, and amounts to showing that $b^n$ is positive for all $n$ when $b$ is positive (which we ended up using in the above proof, if you'll recall).

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Now, suppose the result is true for $n=k$ then


we have to show that the result is true for $k+1$




because $a<b$ so,"a" is replaced by b on rhs


Therefore the result is true for $n=k+1$

$\textbf{For example}$:take $a=2$ and $b=3$ then see what happen

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Since $a < b$, for any number $m>0$, $ma < mb$. In your case, $b^n\cdot a < b^n\cdot b$.

I didn't read your question carefully enough. I assumed you were proving this:

If $a,b\in\mathbb{R}$ and $a<b$, then for every $n\in\mathbb{N}$, $a^n<b^n$.

But the first step in your proof is wrong. You're doing induction on $n$, not on $a$ and $b$, so you should start with $n=1$, not by setting values for $a$ and $b$.

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That is not the case, Neal. Consider $m=-1$ for example. – Cameron Buie Oct 1 '13 at 15:33
@CameronBuie Bah. Thank you. – Neal Oct 1 '13 at 21:09

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