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**quiney** Prove that given a nonnegative integer n, there is a unique nonnegative integer m such that m^2≤n<(m+1)^2

I started like this:

There are two cases to consider: either n is a perfect square or it is not.

Case 1:

Assume n is a perfect square. For the existence part of the proof, let m=√n, where √n is a positive root of n. Then (√n)^2=n, so (√n)^2≤n. Assume for reductio that n≥(√n+1)^2. Factor the right hand side of the inequality and get n≥n+2√n+1. Therefore -1≥2√n, and -1/2≥√n. But √n is a positive root, so we have a contradiction.

Therefore

(i) (√n)^2≤n<(√n+1)^2.

So there exists an m such that m^2≤n<(m+1)^2.

Now I prove that m is unique. I do this by showing that m cannot be greater or less than √n. To show it is not greater, assume without loss of generality that there is a nonnegative integer m' such that m'=m+1. Therefore, m'=√n+1. Then, for reductio, assume (√n+1)^2≤n<(√n+1+1)^2. The left side of the inequality, (√n+1)^2≤n, contradicts the right hand side of (i): n<(√n+1)^2.

Now assume that there is a nonnegative integer m' such that m'=m-1. So, m'=√n-1. Now assume (√n-1)^2≤n<(√n-1+1)^2. The right side becomes n<(√n)^2 or n<n, which is a contradiction. Thus, m is unique.

How do I show this for the case where n is not a perfect square? I know that if n is not a perfect square, then m^2≠n, so I have to prove m^2<n<(m+1)^2. I don't know where to go from here though.