Is √(n-1) + √(n+1) Always Irrational for n≥1?

[choirgurlio]

Homework Statement

Prove that √(n-1)+√(n+1) is irrational for every integer n≥1.

Homework Equations

Proofs i.e. by contradiction

The Attempt at a Solution

2n + 2√(n^2-1) = x^2

so

√(n^2-1) = (x^2-2n)/2

Now if x is rational then so is (x^2-2n)/2 so this says that √(n^2-1) is rational.

But the square root of an integer is rational if and only if that integer is a perfect square so we have

n^2 - 1 = m^2 for some integer m.

Then (n-m)(n+m) = 1, so since these are integers we conlclude that n+m = 1 and n-m = 1. But this is only possible if n = 1 and m = 0.

Since n = 1 does not satisfy "√(n-1) + √(n+1) is rational", we have proven the claim for all n >= 1.

***Is my way to solve this correct? Also, are there any other relevant ways to solve this, better ways?

Thank you for your help!

 

[╔(σ_σ)╝]

If n-1 is a perfect square n+1 is not. That is what I would do.

 

[choirgurlio]

If n-1 is a perfect square n+1 is not. That is what I would do.

Thank you for your reply! Ok, going too far back into Algebra world; could you explain a bit more what you mean? Also, do you think my way is adequate? Would you add or take away anything?

Thanks again!

 

[╔(σ_σ)╝]

To be honest I do not think you proved anything. I don't even follow your logic.

Where did you get the first equation from? The 2n + blah blah = x^2?

Anyway following through to the last line of your proof I think you are trying to proof by contradiction. This is fine it is probably a good idea.

So basically you assume the opposite and show that it false of n=1 but the error in this is that you have only proved the initial statement of n=1 not for all n. Your conclusion is not logically sound.