Proving an Increasing Sequence (a question about the answer)

[student34]

Homework Statement

Question from my professor: "Consider the sequence {a(n)} (from n=1 to ∞) defined inductively by a(1) = 0, and a(n+1) = √(a(n) + 2) for n ≥ 1. Prove that {a(n)} (from n=1 to ∞) is increasing".

Here's the first part of the answer from my professor: "Consider a(n+1)^2 − (a(n))^2 = a(n) + 2 − (a(n))^2 = −(a(n)^2 − a(n) −2) = −(a(n) − 2)(a(n) + 1). Note that a(n) ≥ 0 for all n. For a(n) ∈ [−1,2] we see that −(a(n) −2)(a(n) + 1) ≥ 0. The initial element a(1) = 0 belongs to the interval [0,2]".

How does he know that a(n) ≥ 0 for all n? Is this a given of some sort?

 

[CompuChip]

All a(n) are defined as the square root of their predecessor, and the square root is always non-negative.

If you want to be completely rigorous, you can prove it inductively: assume that a(k) ≥ 0 for all k ≤ n. Then in particular a(n) ≥ 0, therefore a(n) + 2 ≥ 2 (> 0), so a(n + 1) is the square root of a non-negative number which is itself non-negative. And the base step -- a(1) ≥ 0 -- is trivial.

 

[student34]

All a(n) are defined as the square root of their predecessor, and the square root is always non-negative.

If you want to be completely rigorous, you can prove it inductively: assume that a(k) ≥ 0 for all k ≤ n. Then in particular a(n) ≥ 0, therefore a(n) + 2 ≥ 2 (> 0), so a(n + 1) is the square root of a non-negative number which is itself non-negative. And the base step -- a(1) ≥ 0 -- is trivial.

Why did you start off with this, a(k) ≥ 0 for all k ≤ n? I generally understand induction, but this use of k is unfamiliar to me.

 

[CompuChip]

Actually "assume that a(n) ≥ 0 for some n" is enough.

 

[student34]

Actually "assume that a(n) ≥ 0 for some n" is enough.

Thank-you very much for helping me