let $x > -1$ prove for all $n \geq 1$, $(1+x)^n \geq 1+nx$

Question

I tried to prove this by induction:

Base case for n = 1 satisfies $\because$ $1+x$ = $1+x$

I.H: $(1+x)^k \geq 1 + kx$

Inductive Step for $k+1$:

$(1+x)^{k+1} \geq 1 + (k+1)x$

$(1+x)^k (1+x) \geq 1 + kx + x$

$(1+kx) (1+x) \geq 1 + kx + x$ by I.H.

$kx^2 + kx + x + 1 \geq 1 + kx + x$

$kx^2 \geq 0$ which is true for $x>-1$ therefore proved?

Is this the correct way to prove by induction?

That is a collection of statements. You should clarify which statement implies which. — Martin R, Aug 13 '19 at 14:38
For s start, instead of assuming the inequality for $n=k$ and deducing it for $n=k+1$, you seem to be assuming the $n=k+1$ case straight away. — Angina Seng, Aug 13 '19 at 14:39
No. You are assuming what you want to prove and showing you don't get a contradiction. That proves nothing. — fleablood, Aug 13 '19 at 14:54
Unless you state otherwise, the assumption will be that every statement is a result from the previous one and not that every statement is a result of the next one. Thus this seems to be a proof that if you know $(1+x)^{k+1} \ge 1 + {k+1}x$ an $(1+x)^k \ge 1+kx; k \ge 1$ then you have proven that $kx^2\ge 0$. Which is not what was asked to be proven. — fleablood, Aug 13 '19 at 15:04
This is Bernoulli's Inequality. There is a proof by induction at the end of this answer. — robjohn, Aug 13 '19 at 15:08
You can not use induction for all real $n\geq1$. Maybe it means a transfinite induction? If so, your attempt it's something strange. — Michael Rozenberg, Aug 13 '19 at 15:14
Eep! Michael Rozenberg has a point. All these answer do only answer for $n\in \mathbb N$. If you need to prove for all real $n$ then induction will not work. — fleablood, Aug 13 '19 at 15:46
@fleablood: The integer case can be proven inductively. Then the rational case can also be proven inductively. The real case can then be proven by continuity. — robjohn, Aug 13 '19 at 15:57

score 2 · Answer 1 · answered Aug 13 '19 at 14:48

2

The idea here is correct, but the way it is written, it's not clear that this is a proper proof of the inductive step. When proving by induction, you start with the $n=k$ case, and then prove the $n=k+1$ case. If we are to assume the implications are top implies bottom, then what you have done is started with the $n=k+1$ case and then worked down. However, this is precisely what you are trying to prove to begin with.

An example of when this kind of logic doesn't work:

Suppose we want to prove $1=0$. We have

$$\underset{\color{red}X}{(1=0)}\Rightarrow\underset{\color{green}\checkmark}{(0\cdot1=0\cdot0)}\Rightarrow\underset{\color{green}\checkmark}{(0=0)}$$

While the conclusion is in fact true, it does not imply $1=0$.

Instead, start with

$$(1+x)^k\ge1+kx$$

and then work towards proving $(1+x)^{k+1}\ge1+(k+1)x$ from there.

answered Aug 13 '19 at 14:48

Simply Beautiful Art

74,685

so you would start with the I.H.? – ph-quiett Aug 13 '19 at 14:52
Generally speaking yes, or some other statement which you know to already be true (such as $kx^2\ge0$) – Simply Beautiful Art Aug 13 '19 at 14:53
if I want to start with $kx^2 \geq 0$, and work backwards on my proof, how do I show that $kx^2 \geq 0$ is true. Like how would the reader know $kx^2 \geq 0$ is true without me proving it if that makes sense – ph-quiett Aug 13 '19 at 14:55
$k>0$ and $x^2\ge0$ for all real $x$. Clearly you can try going further with that, but I think the reader will understand even if you just stated $kx^2\ge0$. – Simply Beautiful Art Aug 13 '19 at 14:57
if I was to start with the I.H., what would the next couple steps be? – ph-quiett Aug 13 '19 at 14:58
You could, for example, try multiplying both sides by $(1+x)$. – Simply Beautiful Art Aug 13 '19 at 15:00
Well I suppose Axion004's answer shows how to go starting from that. – Simply Beautiful Art Aug 13 '19 at 15:01

score 2 · Answer 2 · answered Aug 13 '19 at 15:43

You can save your proof by added "is implied by" ($\Leftarrow$) marks between lines to indicate the desired direction:

Inductive Step for k+1

:

$(1+x)^{k+1}≥1+(k+1)x\Leftarrow$

$(1+x)^k(1+x)≥1+kx+x\Leftarrow$ by I.H.

$(1+kx)(1+x)≥1+kx+x\Leftarrow$

$kx^2+kx+x+1≥1+kx+x\Leftarrow$

$kx^2≥0$

which is true as $k \ge 1 > 0$.

This would be a correct induction step.

But my advice on a stylish and better induction step would be:

......

We are assuming that $(1+x)^k \ge 1 + kx$ and wish to show this implies $(1+x)^{k+1} \ge 1 + (k+1)x$.

$(1+x)^k \ge 1+kx$ (we are presuming this) and as $x > -1$ we know $1+x > 0$ and thus:

$(1+x)^k(1+x) \ge (1+kx)(1+x)$ so

$(1+ x)^{1+x} \ge 1 + kx + x + kx^2= 1+(k+1)x + kx^2$

And as $k$ and $x^2$ are non-negative, $kx^2 \ge 0$, so

$(1+x)^{1+x} \ge 1+(k+1)x + kx^2 \ge 1+(k+1)x$.

......

but ultimately the proof will need to be written by you.

......

I just realized no-one has yet addressed that claiming $(1+x)^k(1+x) \ge (1+kx)(1+x)$ requires knowing that $1+x > 0$. This is why the condition $x > -1$ is required.

Now prove it holds for $x\ge-2$ :P – Simply Beautiful Art Aug 13 '19 at 15:49 — Simply Beautiful Art, Aug 13 '19 at 15:49

score 1 · Answer 3 · edited Aug 13 '19 at 15:01

1

Theorem: If $x > -1$ then for all $n \geq 1$, $(1+x)^n \geq 1+nx$

Proof: By induction, the base case is $n=1$ where

$$1+x \geq 1+x$$

is true. For the induction hypothesis, we will assume that

$$(1+x)^k \geq 1 + kx$$

for some $k\in\mathbb N$. Then, for the inductive step, we need to show that

$$(1+x)^{k+1} \geq 1 + (k+1)x$$

Starting from the LHS

\begin{align}(1+x)^{k+1}&=(1+x)^k(1+x)\\&\geq (1 + kx)(1+x)\\&=1 + kx + x + kx^2 \\&\geq1 + kx + x\\&=1 + (k+1)x\end{align}

Hence, we see that the theorem holds.

edited Aug 13 '19 at 15:01

Simply Beautiful Art

74,685

answered Aug 13 '19 at 14:59

Axion004

10,056

That does not answer the question “Is this the correct way to prove by induction?” (explicitly tagged with [proof-verification]). – Martin R Aug 13 '19 at 15:03
@MartinR is this not a proof by induction? – ph-quiett Aug 13 '19 at 15:04
Yes, and the above proof is correct. However, I wasn't supposed to write the proof. I was supposed to verify that the proof which you wrote is correct. Let me know if I should delete this answer. – Axion004 Aug 13 '19 at 15:06
Could you elaborate on your third step, how does the inequality changes from $= 1+kx+x+kx^2$ to $\geq 1+kx+x$ – ph-quiett Aug 13 '19 at 15:06
It is because $kx^2\geq 0$. – Axion004 Aug 13 '19 at 15:07

score 1 · Accepted Answer · answered Aug 13 '19 at 15:03

1

then would this be a better proof?:

$\forall x > -1$ we know that $kx^2 \geq 0$

by adding $kx+x+1$ to both sides:

$kx^2+kx+x+1 \geq kx+x+1$

$(1+kx)(1+x) \geq kx + 1 + x$

$(1+x)^k (1+x) \geq kx + 1 + x$ by I.H.

$(1+x)^{k+1} \geq 1 + (k+1)x$

Hence proved?

answered Aug 13 '19 at 15:03

ph-quiett

347

As used in my answer, you want to include \implies between statements (or just write it out in words), to show that one statement leads to the next, and that these aren't just statements thrown together with no relationship between them. – Simply Beautiful Art Aug 13 '19 at 15:05
@SimplyBeautifulArt now I think you are being fussy. If not stated otherwise statement are assumed to follow as conclusions from previous statements. Anyone reading this would assume correctly that the intent is that every statement follows from the one before. – fleablood Aug 13 '19 at 15:08
..... although technically you are correct. But the "list of statement" proofs is a common "style" and usually not viewed as in error. Although technically it is better to have few cues or words as "so" and "therefore" to make it clear. – fleablood Aug 13 '19 at 15:11
@fleablood I do agree that it is implied to read the implications in that order, but adding \implies does not hurt, and it certainly removes any chance of ambiguity. – Simply Beautiful Art Aug 13 '19 at 15:12
1

Actually, reading your answer and the conversation you had with the OP about better proof writing, I change my mind. You are trying to explain to the OP (who has essentially figured out why it is true) how to express it as a good proof and ... yes, it becomes a better proof if statements of implication are made clearer. – fleablood Aug 13 '19 at 15:16
then does @Axion004 prove with implications? as in, is that the "format" per say – ph-quiett Aug 13 '19 at 15:24
They are implicitly taking advantage of $a\ge b\ge c\Rightarrow a\ge c$. – Simply Beautiful Art Aug 13 '19 at 15:33

let $x > -1$ prove for all $n \geq 1$, $(1+x)^n \geq 1+nx$

4 Answers4