# Proof by Induction

1. Dec 7, 2008

### kathrynag

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution
I want to show for P(k+1)
(1+x)^(k+1)$$\geq$$1+kx+x
(1+x)^k*(1+x)$$\geq$$1+kx+x

2. Dec 7, 2008

Use this:

3. Dec 7, 2008

### kathrynag

I don't really understand how I use that...

4. Dec 7, 2008

### mutton

You want to use an inequality involving $$(1 + x)^k$$ to derive an inequality involving $$(1+x)^k (1+x)$$, which is a multiple of it.

An inequality still holds if you multiply both sides by a positive number. If you multiply both sides of an inequality by a negative number, then you have to flip the sign from $$\le$$ to $$\ge$$ or vice versa.

5. Dec 7, 2008

### kathrynag

Ok, so:
(1+x)^k(1+x)>(1+kx)(1+x)
>(1+2kx+x)

6. Dec 7, 2008

### mutton

Right idea, but there is an error in your expansion.

Does the question give any restrictions on x? Is what you wrote still true of 1 + x < 0?

7. Dec 7, 2008

### kathrynag

No restrictions.
1+kx+x+kx^2

8. Dec 7, 2008

### mutton

When k = 3 and x = -4,

$$(1 + x)^k = (-3)^3 = -27 < -11 = 1 - 12 = 1 + kx$$

which makes the statement false.

9. Dec 7, 2008

### kathrynag

So, we have to assume 1+x>0

10. Dec 7, 2008

### kathrynag

so, x>-1

11. Dec 8, 2008

### Staff: Mentor

I believe x has to be strictly within 1 unit of 1; i.e., |1 + x| < 1, which means that 0 < x < 2.

12. Dec 8, 2008

### kathrynag

Once I get here I'm unsure where to go

13. Dec 8, 2008

### mutton

Look at your first post; you need to show that that is $$\ge 1 + kx + x$$.