# Discrete Math Proof n^2 > n +1

1. Aug 5, 2012

### Xuanwu

1. The problem statement, all variables and given/known data
For any given n, where n is an element of the natural number set, prove n^2 > n +1, for all n > 1.

2. Relevant equations
This week in lecture we defined the greater than relationship as:

Let S = Natural numbers
Let R = {(a,b): $\exists$ c: a = b+c}
then aRb

3. The attempt at a solution

My first thought was show a general expression of either an odd of even number (n = 2x + 1 or n = 2x), but the resultant (4x^2 + 4x +1 > 2x + 2 and 4x^2 > 2x +1 for odd/even respectively) doesn't really give me anything useful as far as I can see.

I can see that it would be true, as n = 2 would be 4 > 3, and n= 3 would be 9 > 4, I'm just not sure on how to get started.

Some literature on approaches to getting started on proofs would be great, as I find I'm fairly hit/miss. I can either see the approach or I stand there going "I know it's true, but I can't show it's true".

Last edited: Aug 5, 2012
2. Aug 5, 2012

### MathematicalPhysicist

Here's one approach.

n^2-n = n(n-1)

n>1 thus (n-1)>=1, now n(n-1)>= n >1

3. Aug 5, 2012

### Xuanwu

Sorry, still not sure how having the form,

n(n-1) > 1 for all n > 1,

helps me in this?

4. Aug 5, 2012

One way to do it would be to rewrite it as n^2 - n -1 > 0, and show that that is true for n > 1

5. Aug 5, 2012

### Xuanwu

Hmm.. explain it to me like I'm 5. I last did math proof type subjects in '99, prior to this my degree has had calculus and stats which I had no issue with.

6. Aug 5, 2012

### ehild

See the previous post #2. The original inequality can be written as n2-n>1, as you can subtract the same amount from both sides of an inequality: for example, if a>b, a-1>b-1.

n2-n=n(n-1). So you have to prove that n(n-1)>1 if n>1 integer.
If n>1 n-1≥1. You multiply n with a number greater then 1. Will the result greater or less than n?

ehild

7. Aug 5, 2012

### Xuanwu

Alright, that makes sense. Issue is I'm not sure if this is the sort of proof I'm meant to be doing for it.

I think I'll go bang my head on a book.

Thanks.

8. Aug 5, 2012

### ehild

It is a proof and simple enough. If you want to apply the definition find that "c" >0 that makes (n+1)+c=n2. It is just the difference n2-(n+1), isn't it? You need to prove that n2-n-1>0.

ehild

EDITED!

Last edited: Aug 5, 2012
9. Aug 5, 2012

### MathematicalPhysicist

ehild, you meant n^2-n-1>0 not n^2-n+1.

You can do this also by induction, though it's really easy without.

10. Aug 5, 2012

### ehild

Thank you, I corrected it.

ehild