Math 201 - The square root of 2 exists
Theorem There exists α ∈ (0, ∞) such that α2 = 2.
Proof. Define
A = {x ∈ R : x2 < 2}
Clearly, 1 ∈ A so A 6= ∅ and eg 7 is an upper bound for A. By the completeness axiom, A has a least upper bound which we will call α. Our goal
is to prove α2 = 2. To do this, we study the inequalities α2 < 2 and α2 > 2
and prove that both of these are impossible. In case α2 < 2, define
β=
2
α
Directly from dividing the inequality α2 < 2 by α, we get
α < β.
Let y be the arithmetic mean of α and β, y = (α + β)/2. Then
α<y<β
From the AGM inequality (note that we don’t take square roots, because
we are at this point not yet sure that they exist!)
y 2 > αβ = 2
We play the same trick on y tfhat we did with α:
2
2
4
4
=
< =2
2
y
y
2
and so 2/y ∈ A. But
2
2
> =α
y
β
which contradicts our definition of α (α is supposed to be an upper bound
of A).
Now assume α2 > 2. We define β and y in the same way as before. This
time, β < α and
β < y < α.
As before, the AGM inequality gives y 2 > 2. So for any x ∈ A,
x2 < 2 < y 2
which implies x < y since all variables here are positive (taking square roots
is justified here since both sides in x2 < y 2 are obviously squares). This
means y is an upper bound for A, contradicting that α should be the least
upper bound. Together, the two contradictions we have obtained prove that
α2 = 2 is the only remaining possibility.