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Blast into Math! Prime nummers: indestructimle muilding mlocks

is if

p =0, q =1.

This is the rational number 0. Now, for which rational numbers is |p| + q =2? Since q ≥ 1, the only
ways for |p| + q =2 are if

p =1, q =1,

or

p = −1, q =1.

We can continue grouping the rational numbers in this way by defining

p

S k = ∈ Q such that |p| + q = k .
q

We just need to show that for each k , S k has finitely many elements, and then we can zip all the sets
S k together into one countable set by the Zipper Theorem. If

|p| + q = k,

then since q ∈ N , q ≥ 1 we know that

−k ≤ p ≤ k, 1 ≤ q ≤ k.

There are precisely 2k +1 integers between −k and k and k natural numbers between 1 and k . Since
each of the numerators could be paired with each of the denominators, there are at most (2k +1)k
rational numbers in S k . This means that each S k is finite. Since each rational number is contained in
S k for some k,

∞

Q ⊆ S k .
k=1

On the other hand each S k ⊆ Q , and so their union is also a subset of Q ,

∞

S k ⊆ Q.
k=1

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