Discrete Math Proof n^2 > n +1

[Xuanwu]

Homework Statement

For any given n, where n is an element of the natural number set, prove n^2 > n +1, for all n > 1.

Homework 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

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]

Sorry, still not sure how having the form,

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

helps me in this?

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]

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.

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]

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

Thank you, I corrected it.

ehild