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Prove by induction that $\forall x \in \mathbb{R}, x \geq -1, \forall n \in \mathbb{N},n \geq 0$ that $$(1+x)^n \geq 1+nx$$ First of all I have a problem with x being a real number, how can I use induction over a real number? Can I simply assume that it holds for all $y \in [-1,x]$ and then show that it also holds for $x + \epsilon, \epsilon \in \mathbb{R},\epsilon \gt 0$?

If that were possible I would try to prove over n. Prove the base case for n, use induction over x and then prove the induction step over n.

But I have some problems with the induction step over n.

Assuming that it works for all $l \in \mathbb{N}, l \leq n-1$. I need to show $$(1+x)^n=(1+x)^{n-1}(1+x) \geq 1+(n-1)x+x$$ which basically results in showing that $$z(1+x) \geq z + x$$ for some quantity z, correct?

But then I'd get $$zx \geq x$$, which would obviously depend on the value of z.

I'm kind of confused at this point, any help would be appreciated.

eager2learn
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    Induction has to be done on $n$, not $x$ – idpd15 Apr 19 '14 at 14:13
  • $(1+x)^{n-1}(1+x) \geq (1+(n-1)x)(1+x) = 1+(n-1)x+x+(n-1)x^2 = 1+nx+(n-1)x^2 \geq 1+nx$ – ah11950 Apr 19 '14 at 14:13
  • @YiyuanLee I think this is not a duplicate, because the post here identifies a particular point at which the argument has got stuck. The question is not how to prove the identity, but how to get past the sticking point. – Mark Bennet Apr 19 '14 at 14:19

2 Answers2

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Once you get to $$(1+x)^n=(1+x)^{n-1}(1+x)$$ you have rearranged the expression in a form in which you can use the inductive hypothesis, because you can assume that $$(1+x)^{n-1}\ge 1+(n-1)x$$

You should substitute that information into the expression you have for $(1+x)^n$ and see where it gets you when you follow it through logically.

So in comments you suggest the base case $n=0$ which is a good start, because you are asked to prove the statement for $n\ge 0$ - so you have to start at the bottom, or deal with special cases, so let's start at the bottom.

You have $$(1+x)^0=1\ge 1+0\cdot x$$You suggest in your comment that the base case does not work for $x\lt 0$, which is not quite correct - the situation is better than that, because you can raise any positive real number to the power zero. So the condition is $(1+x)\ge 0$ which is equivalent to $x\ge -1$, and you are on your way.

Mark Bennet
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  • Thanks for your answer. One more question though. I assume that I don't have to use induction on x and can simply assume x to be an arbitrary real number greater than or equal to -1, right? – eager2learn Apr 19 '14 at 14:24
  • @eager2learn In the question $x$ is defined as a real number $\ge-1$ - so if your proof doesn't work for all those $x$ you need to look for a proof that does! Notice where you are using the condition $x\ge -1$ which is equivalent to $(x+1)\ge 0$. – Mark Bennet Apr 19 '14 at 14:29
  • Ok so could i structure my proof as follows: 1. Base case n = 0, 2. prove that $\forall x \geq -1$ (1+x)^0 = 1 + 0x, if $ x \lt 0$,the base case doesn't hold, 3. induction step for n? – eager2learn Apr 19 '14 at 15:26
  • @eager2learn - I expanded my answer rather than reply in a comment. – Mark Bennet Apr 19 '14 at 16:04
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firstly, we can know it is true for n=1 and n=2,suppose when n=k, it is also true. it means that $(1+x)^k$>=1+kx.Thus,$$(1+x)^{k+1}=(1+x)^k\times(1+x)\ge(1+kx)(1+x)$$ Finally,(1+kx)(1+x)=1+kx+x+$kx^2$>=1+(k+1)x.

python3
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