Could someone explain this step in this proof to me?

[flyingpig]

Homework Statement

I already have the solutions, just don't understand one step

Let $$(n_k)_{k=1}^{\infty}$$ be a sequence of integers

$$n_1 < n_2 < n_3 < ...$$

Then $$n_k \geq k$$ $$\forall k = 1,2, ...$$

In particular $$n_k \to \infty$$ when $$k \to \infty$$

Prove this is true


2. The solution

Employ Induction

(1) Base Case: k = 1

$$n_1 \geq 1$$

(2) Induction: Assume $$n_k \geq k$$ for the case k. We show k + 1 case

$$n_{k+1} > n_k > k \implies n_{k+1} > k \implies n_{k+1} \geq k + 1$$

Hence the result holds

Question

How on Earth did we get to $$n_{k+1} \geq k + 1$$ from $$n_{k+1} > k$$

 

[Ryker]

Because n_k+1 is strictly greater than k, and both are integers. So n_k+1 is either k + 1 or an integer even greater.

 

[flyingpig]

Because n_k+1 is strictly greater than k, and both are integers. So n_k+1 is either k + 1 or an integer even greater.

I don't follow that

Before we had >, then ≥

Counterexample

5 > 4

5 + 1 > 4 + 1

6 > 5

6 ≠ 5, ever

 

[Ryker]

Hmm, first of all, if in your "counter-example", you took n_k+1 = 5 and k = 4, then 5 ≥ 4 + 1 = 5. Secondly, just because 6 ≠ 5, that doesn't mean 6 ≥ 5 is not true. ≥ is just less strict than > and it allows equality, but doesn't fail even if the equality is never reached. It's still true that 6 is greater than or equal to 5.

 

[flyingpig]

Hmm, first of all, if in your "counter-example", you took n_k+1 = 5 and k = 4, then 5 ≥ 4 + 1 = 5. Secondly, just because 6 ≠ 5, that doesn't mean 6 ≥ 5 is not true. ≥ is just less strict than > and it allows equality, but doesn't fail even if the equality is never reached. It's still true that 6 is greater than or equal to 5.

Wow I hadn't even thought of it that way. I mean not the ≥ part, but 5 = 5.

Thanks problem solved