# Math Help - Double Induction for (1+t)^t > 1+nt

1. ## Double Induction for (1+t)^t > 1+nt

Hi, I have the following problem which I think I have solved correctly (it would be great if you could check it), but I don't know how to answer to "simple" questions about my proof... any help would be very much appreciated, thanks!

The problem says:
"Let t>-1 be given with t≠0. Show by induction that for n≥2 we have
$(1+t)^t>1+nt$."

And here are the two questions that I don't know:

"Where is the condition t>-1 used in your proof? Is the result true if instead we had -2<t<-1?"

My solution to the first bit is:

Let $F(n)=(1+t)^n>1+nt$.

Proof by induction on n.

Base n=0. Then $(1+t)^n=(1+t)^0=1=1+0t=1+nt$

Induction. Suppose that there is a number k (k≥2) such thaqt $(1+t)^k>1+kt$ for any t>-1 (t≠0). We need to show that (1+t)^{k+1}>1+(k+1)t for any t>-1, t≠0.

$(1+t)^{k+1}=(1+t)^k*(1+t)$. By the induction hypothesis, $(1+t)^k>1+kt$, so we have:
$(1+t)^k*(1+t) \geq (1+kt)*(1+t)=1+kt+t+kt^2=1+(1+k)t+kt^2$.

Since $t^2>0$ and $k \geq 2, 1+(1+k)t+kt^2 \geq 1+(1+k)t$.

So $(1+t)^{k+1} \geq 1+(k+1)t$

And so the result holds for n=0 and hence for all n≥2 and t>-1,t≠0.

2. Everything looks good, except for one thing: you skipped the n = 1 case! It's a small technicality, but it does actually render your proof invalid. You definitely have the right idea with your proof. I would clean up your notation a bit. Note: it's \geq or \ge, not /geq. That'll make your post a lot more intelligible.

3. Two comments. First, as Ackbeet says, you must start the induction at n=2, not n=0, if you want to get the strict inequality (for $t\ne0$).

The second comment is that in the inductive step you multiplied both sides of an inequality by $(1+t)$. Now you ought to know that an inequality is preserved if you multiply both sides by a positive number, but not if you multiply both sides by a negative number. That should help you to answer the "simple" questions about the proof.