# How to prove by mathematical induction?

1. Feb 3, 2013

### James2

How do I prove a formula/rule or something by mathematical induction? Please give me a few examples or resources and explain it as best you can because I think I'm messing up some how.

2. Feb 3, 2013

### Number Nine

What are you not getting, exactly? If you just don't know what induction is, surely a google search would be faster than starting a new thread.

3. Feb 3, 2013

### James2

Everything I read confuses me, it tells me to do something different everytime...

4. Feb 3, 2013

### Number Nine

We can't help you if you don't explain what's confusing you. Try posting your attempt at solving an induction problem and explain where you get stuck.

5. Feb 3, 2013

### James2

I have an equation, (5n + 2) = 2[(5/2)n + 1] I know this is true from the basis step. Then I asume n = k now I must prove n = k + 1. So, (5k + 2) = 2[(5/2)k + 1]

Alright then, I try to substitute k + 1 in and add it or something so I get... 2[(5/2)k + 1] + [5(k + 1) + 2] = 2 [(5/2)(k + 1) + 1]

Simplifying, I get 10k + 2 + 5k + 5 + 2 = 10(k + 1) + 2

And finallly, 15k + 9 =/= 10k + 12

SO.... whaaaat? What happened here?

6. Feb 4, 2013

### Ferramentarius

(5n + 2) = 2[(5/2)n + 1]

n = 0:
5*0 + 2 = 2[(5/2)*0 + 1]

The case is true for 0.

Suppose the case is true for n = k.
Now we can use (5k + 2) = 2[(5/2)k + 1].

n = k + 1:
5(k + 1) + 2 = (5k + 2) + 5 = 2[(5/2)k + 1] + 5 = 2[(5/2)k + 1 + 5/2] = 2[(5/2)(k+1) + 1]

The case n = k + 1 follows from the case n = k.
With case n = 0 true the equation therefore works for all non-negative integers.