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I am trying to prove that:

$(1+x)^n \ge 1 + nx$ for every $n \in \mathbb N^+ $ and $\ x \in (-1, \infty)$

I have never seen induction on more that one variable.

Since $(-1, \infty)$ has no least element can I even induct on this? Would strong induction be preferable?

Here is my proof for all $n \in \mathbb N^+$ by inducting on $n$.

Proof:

Suppose $P(n) : (1+x)^n \geq 1 + nx$.

$P(1) = 1+x \ge 1+x$

$P(n) \Rightarrow p(n+1)$

$(1+x)^n \ge (1+nx)$

$(1+x) (1+x)^n \ge (1+nx)(1+x)$

$(1+x)^{n+1} \ge 1+ x + nx + nx^2 $

Since $nx^2 \ge 0 $ for all $n,x$ then:

$1+ x + nx + nx^2 \ge 1 +xn + x$ .

Therefore:

$(1+x)^{n+1}\ge 1 +xn + x$

$(1+x)^{n+1}\ge 1 + x (n+1) $

Therefore:

$(1+x)^n \ge (1+nx) \Rightarrow (1+x)^{n+1} \ge (1+(n+1)x)$

Is this correct? Can anyone provide any guidance on how to approach induction on the interval? Is it even possible?

Cheers Guys!

mathamasacre
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1 Answers1

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There is no "double induction". You just do the same induction proof for all $x$. All you are using in your proof is that $x\geq-1$, so any $x\in[-,1\infty)$ would work.

Martin Argerami
  • 205,756
  • But for the base case? If $x \in (−1,∞) $ I can't use P(-1) since it's not in the interval. Do I have to use strong induction on this? – mathamasacre Oct 24 '16 at 22:29
  • Uh? Your proof is correct. Fix any $x\in[-1,\infty)$. The start with $P(0)$ and do the induction thing. End of story. No one is talking about negative $n$ here, so $P(-1)$ is meaningless in this context. – Martin Argerami Oct 24 '16 at 22:32