Page 168 - 'Blast_Into_Math

Blast into Math! Analatic nummer theora: ants, ghosts and giants

Proof: First, let’s assume the sequence is increasing: the ants are marching to the right. If the sequence
is not bounded above, this means that for any rational number Q, and for any N ∈ N , there exists
n> N such that

x n >Q.

Since N ⊂ Q , this means that for any M ∈ N , and N ∈ N there exists n ∈ N such that

x n >M and n> N.

If we imagine the sequence of ants on the number line, for each natural number, the ants eventually
march past it. So, the sequence cannot converge! Let’s prove this by contradiction. We assume the
sequence does converge, and show that this will lead to something impossible. Let’s call the limit of the
sequence L. By the definition of convergence, for any > 0, there exists a giant number N ∈ N such
that for all n> N ,

|x n − L| <.

Well, since 1 > 0, we can let 1 play the role of the ghost number > 0he definition of limit. Then,
in t
there exists a giant number N ∈ N such that for all n> N

|x n − L| < 1,

which means that the distance between x n and L on the number line is less than 1. Therefore,

x n <L +1 ∀n> N.

But, since L is a real number, it is either a rational number or the least upper bound of a set of rational

numbers. Now we can use the Friendly Q Lemma. Since

L +1 <L +2,

there exists a rational number M such that

L +1 <M <L +2.

Then, for all n> N ,

x n <M.

But this is a contradiction!

Exercise: Why is this a contradiction?

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