Demonstrate by induction: $(1+a)^n ≥1+an$ is true, given a real number $a$, for any $n ∈ \mathbb N$. With $a > 0$

I need to demostre this using the induction principle. My doubt is in the second part of the demonstration.

This is what I have:

In order to demonstrate the predicate these two points must be true:

  1. $P(1)$ must be True.
  2. $∀ n ∈ \mathbb N , P(n) => P(n+1)$ is true.

We demonstrate the first point:

$(1+a)^1 ≥ 1+a*1$, $1+a ≥ 1+a$ So it is true

Now, the second part is where I have the problem. I do not know what to do. I understand the theory but I don't know how to apply it.

I tried this:

$(1+a)^{n+1} ≥ 1+a(n+1)$

But I don't see that as useful.

Any tips?

  • $\begingroup$ If $n$ is odd, it isn't true for all $a$, just for $a \geq -1$. $\endgroup$ – tpv Jan 28 '11 at 16:51
  • $\begingroup$ oh, sorry, I forget to add that A is positive. $\endgroup$ – Nerian Jan 28 '11 at 16:55
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    $\begingroup$ Just an FYI, the inequality in question is known as Bernoulli's Inequality. $\endgroup$ – user6459 Feb 1 '11 at 2:15

Ross has pretty much given you all the hints necessary to answer this particular question, so let me instead give you some general advice on proofs by induction.

In doing the induction, especially at first and until you get very comfortable with induction, be very explicit. Label the base Base. Label the inductive step as "Inductive step." State explicitly what the Induction Hypothesis is, state explicitly what you want to prove. Use this as a way to organize your thoughts and have the information at the ready.

Second: The key to most proofs by induction is to take the case you need to prove, and to somehow reduce it to "the previous case plus something extra"; then one applies the inductive hypothesis to simplify/answer the part that is the previous case, and then deal with the "something extra."

Here's an example different from the one at hand, so you can see what I mean.

Consider the following:

Prove that for all natural numbers $n$, $$1\cdot 2 + 2\cdot 3 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}.$$

Proof. We proceed by indution on $n$.

Base. We prove the statement for $n=1$: indeed, $1\cdot 2 = \frac{1(2)(3)}{3}$.

Inductive step.

Induction Hypothesis. We assume the result holds for $k$. That is, we assume that $$ 1\cdot 2 + 2\cdot 3 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3}$$ is true.

To prove: We need to show that the result holds for $k+1$, that is, that $$ 1 \cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2) = \frac{(k+1)\bigl((k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}.$$

(Now comes the point where we take $1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2)$ and try to think of it as "the $k$-case plus something extra").

\begin{align*} &{1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2)}\\ \quad &= \Bigl(1\cdot 2+ 2\cdot 3+\cdots + k(k+1)\Bigr) + (k+1)(k+2) &&\mbox{(associativity of $+$)}\\ \quad&= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) &&\mbox{(This is the induction hypothesis!)}\\ &= \frac{k(k+1)(k+2)}{3} + \frac{3(k+1)(k+2)}{3} &&\mbox{(Just algebra)}\\ &= \frac{(k+1)(k+2)\bigl(k+3\bigr)}{3} &&\mbox{(factor out $(k+1)(k+2)$)}\\ &= \frac{(k+1)\bigl( (k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}. \end{align*} Thus, $1\cdot 2 + 2\cdot 3 + \cdots + (k+1)(k+2) = \frac{(k+1)\bigl((k+1)+1\bigr)\bigl((k+1)+2\bigr)}{3}$. This proves the inductive step.

Since the statement holds for $n=1$, and if it holds for $k$ then it holds for $k+1$, then by mathematical induction we conclude that $$1\cdot 2 + 2\cdot 3 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}$$ for all natural numbers $n$. QED

So: be very explicit and clear, to help organize your thoughts. As you get more comfortable with induction, you can leave off makign explicit statement of what you need to prove, etc., but until you are very comfortable, better to be explicit than to be confused.

  • $\begingroup$ I.e. in radix $3:\ \ (3^n-1)/2\ =\ 11\cdots 1\ >\ 1+1+\cdots +1\ $ ($n$ times). $\ $ See my answer. $\endgroup$ – Bill Dubuque Jan 28 '11 at 17:56
  • $\begingroup$ @Bill: Sigh; that was silly. It was the same problem, which is precisely what I was trying to avoid; I've changed it to something completely different. My goal was to exhibit the "be very explicit, be very clear, be very organized", not to discuss the particular problem. $\endgroup$ – Arturo Magidin Jan 28 '11 at 18:18
  • $\begingroup$ I used your example to redo the problem. I am still trying it. $\endgroup$ – Nerian Jan 28 '11 at 18:36
  • $\begingroup$ @Arturo: I reach to the point of 'k-case plus something extra'. I write: $(1+a)^k+1 = (1+a)^k*(1+a)$ . I see that the extra is (1+a). So I just add that to the other side? That's $(1+ak)(1+a)$. So I have: $(1+a)^k * (1+a) ≥ (1+ak)(1+a)$ . I move the (1+a) to the other side and that's: $(1+a)^k ≥ (1+ak)$ Is that right, that proofs the inductive step? $\endgroup$ – Nerian Jan 28 '11 at 18:51
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    $\begingroup$ Nerian: you don't want to move the $(1+a)$ to the other side here; that's undoing exactly what you did in the first place! Instead you want to work some algebra on that side and see if you can get to your 'target' statement (in this case, that $(1+a)^{k+1} \geq 1+a(k+1)$.) $\endgroup$ – Steven Stadnicki Jan 28 '11 at 19:24

Hint: Now you assume that $(1+a)^n \ge 1 + an$ and need to prove that $(1+a)^{(n+1)} \ge 1 + a(n+1)$. So $(1+a)^{(n+1)}=(1+a)^n*(1+a) \ge $ what?

Added:$(1+a)^{(n+1)}=(1+a)^n*(1+a) \ge (1+an)(1+a)$ by the inductive hypothesis $(1+an)(1+a)=1+a(n+1)+a^2n \ge 1+a(n+1)$ So, as Arturo showed in his example, we have proven that if the relation is true for $n$, it is also true for $n+1$. Given that you verified the base case of $n=1$, it is true for all $n$

  • $\begingroup$ [what]: 1+a(n+1) // 1+an+a // (1+a)+an ? $\endgroup$ – Nerian Jan 28 '11 at 16:54
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    $\begingroup$ @Nerian: Those three are all equal and are what you want the the right side of the greater than sign. So substitute in what you know-what replacement is there for (1+a)^n that will help? $\endgroup$ – Ross Millikan Jan 28 '11 at 16:58
  • $\begingroup$ @Ross: So the right side is OK. We need to transform the left side. A replacement for $(1+a)^n$...mmm. Well, I know that $(1+a)^n≥ 1+an$, which is part of the statement that we want to prove, is true. So if I take that out and just leave the rest, we have: $(1+a) ≥ a$ which is true. Is that what you meant? $\endgroup$ – Nerian Jan 28 '11 at 17:19
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    $\begingroup$ @Nerian: You are assuming that $(1+a)^n \geq 1+na$ as your induction hypothesis. You want to prove that from this you can conclude that $(1+a)^{n+1}\geq 1+(n+1)a$. So take $(1+a)^{n+1}$ and write it as $(1+a)^{n+1}=(1+a)(1+a)^n$. Now, by the induction hypothesis, you know that this entire product is greater than or equal to <fill in the blank>; now continue. $\endgroup$ – Arturo Magidin Jan 28 '11 at 17:33
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    $\begingroup$ @Nerian: I added the finish. $\endgroup$ – Ross Millikan Jan 31 '11 at 19:42

HINT $\ $ Put $\rm\ \ x = 1+a\ \ $ in $\rm\displaystyle\ \ \frac{x^n -1}{x-1}\ =\ x^{n-1}+\:\cdots\: +x+1\ \ge\ n\ $ for $\rm\ x\ge 1$

Many induction problems are best solved in this manner, i.e. by transforming them into a form that makes the induction completely trivial.

  • $\begingroup$ I don't understand. Where is that fraction coming from? $\endgroup$ – Nerian Jan 28 '11 at 17:15
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    $\begingroup$ @Nerian: $\rm\ (1+a)^n\ >\ 1+an\ \iff\ ((1+a)^n-1)/a\ >\ n\:,\ $ i.e. $\rm\ (x^n-1)/(x-1)\ >\ n\ \ \ \ \ \ \ \ $ $\endgroup$ – Bill Dubuque Jan 28 '11 at 17:20

Before you do anything else, ask yourself: do you even believe that this is true? Have you tried any examples with specific numbers? Fixing the value of $a$ (especially taking $a$ to be very small) and varying the value of $n$, perhaps? Does the result seem plausible after doing this? Do your experiments suggest anything about what's going on?

  • $\begingroup$ (1 + a)^2 = 1 + 2a + a^2 would be a good start $\endgroup$ – Simpson17866 Sep 8 '15 at 13:25

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