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Unformatted text preview: t S is bounded above; for example, it is not diﬃcult to show that if x ∈ S , then we must have x ≤ 2; for if x > 2, then we must have x2 > 4, so that x ∈ S . Hence S is a non-empty set of real numbers and S is bounded above. It follows from the Completeness axiom that there is a real number M satisfying conditions (S1) and (S2). We now claim that M 2 = 2. Suppose on the contrary that M 2 = 2. Then it follows from axiom (O1) that M 2 < 2 or M 2 > 2. Let us investigate these two cases separately. If M 2 < 2, then we have (M + )2 = M 2 + 2M +
2 <2 whenever < min 1, 2 − M2 . 2M + 1 This means that M + ∈ S , contradicting conndition (S1). If M 2 > 2, then we have (M − )2 = M 2 − 2M +
2 >2 whenever < M2 − 2 . 2M This implies that any x > M − will not belong to S , contradicting condition (S2). √ Note that M 2 = 2 and M is a real number. It follows that what we know as 2 is a real number. 1.5. The Complex Numbers It is easy to see that the equation x2 +...
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