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

Lemma 7.2.4 (Friendly Q Lemma). For any two real numbers x and y such that

x< y,

there exists a rational number z ∈ Q such that

x< z< y.

Proof: If x and y are both rational numbers, then since x< y ,

x + y
x< <y,
2
and

x + y
∈ Q.
2

In this case the role of z in the lemma is played by x+y . If y/∈ Q , then y is the least upper bound of
2
a non-empty set S ⊂ Q . Since x< y , x is not an upper bound for S , because y is the least upper
bound of S . So, there is some s ∈ S ⊂ Q with

x<s< y.

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