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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.