Understanding the Inductive Proof Method for Mathematical Statements

[zeion]

Homework Statement

We'll show that, if x => -1, then (1 + x)^n => 1 + nx for all positive integers n.

Solution ...
The text goes on to explain that if n = 1 is true, we assume n = k is true, and n = k + 1 is true.

I understand everything until this part:

Since

(1 + x)^k+1 = (1 + x)^k(1 + x) => (1 + kx)(1 + x)

How does 1 + nx where n = (k+1) become (1 + kx)(1 + x)?

Homework Equations

The Attempt at a Solution

 

[LCKurtz]

"The text goes on to explain that if n = 1 is true, we assume n = k is true, and n = k + 1 is true."

That shows a confused understanding of what your text is doing. You don't "assume n = k is true". You assume the proposition you are trying to prove is true for n = k. Then what you are trying to prove is that the proposition is true for k + 1.

So your assumption is: (1 + x)^k ≥ 1 + kx

What you are trying to prove is: (1 + x)^k+1 ≥ 1 + (k+1)x

So you might start by multiplying both sides of your assumption by (1 + x). That will get you the left side of what you are trying to prove and it is your job to show what you have satisfies the right side.

 

[zeion]

So I need to show that

(1 + kx)(1 + x) = 1 + (k+1)x

 

[VietDao29]

Homework Statement

We'll show that, if x => -1, then (1 + x)^n => 1 + nx for all positive integers n.

Solution ...
The text goes on to explain that if n = 1 is true, we assume n = k is true, and n = k + 1 is true.

I understand everything until this part:

Since

(1 + x)^k+1 = (1 + x)^k(1 + x) => (1 + kx)(1 + x)

How does 1 + nx where n = (k+1) become (1 + kx)(1 + x)?

Well,

(1 + x)^{k + 1} = (1 + x)^k(1 + x), right?

And from your Induction Hypothesis, we have: (1 + x)^k >= 1 + kx, right? So, in conclusion:

(1 + x)^{k + 1} = (1 + x)^k(1 + x) >= (1 + kx)(1 + x).

 

[LCKurtz]

So I need to show that

(1 + kx)(1 + x) = 1 + (k+1)x

You can't show that because those aren't equal. But wouldn't ≥ do? Can you show that? Then write the full inequality up nice.

 

[zeion]

Assume:
(1 + x)^k >= (1 + kx)
Then:
(1 + x)^k+1 which equals to (1 + x)^k(1 + x) >= (1 + kx)(1 + x)
since I multiplied (1 + x) to (1 + x)^k to make (1 + x)^k+1, I have to do the same to the other side.

So I have to show that
(1 + kx)(1 + x) >= 1 + (k+1)x

1 + x + kx + kx^2 >= 1 + (k+1)x
1 + (1 + k)x + kx^2 >= 1 + (k+1)x

So the left side is +kx^2 more than the right side

 

[LCKurtz]

Yes. You now have the argument down and understand the steps. And you have what I would call a good "scratch paper" writeup. Having that all figured out, a nice writeup of the induction step argument you have would be to write:

Assume (1 + x)^k ≥ 1 + kx

Then:

(1 + x)^(k+1) = (1 + x)^k(1+x) ≥ (1 + kx)(1 + x)

= 1 + x + kx + x² = 1 +(k + 1)x + kx²

≥ 1 + (k + 1)x

And you might want to note why it is important x > -1.

 

[zeion]

Ok thanks for the help.