Discrete Math Proof n^2 > n +1

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): [itex]\exists[/itex] 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".

MathematicalPhysicist

Here's one approach.

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

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

Xuanwu

Sorry, still not sure how having the form,

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

helps me in this?

Bread18

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

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.

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

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.

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!

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.

ehild

Thank you, I corrected it.

ehild