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Math 316, Intro to Analysis: Proving that 2 is a real number.
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During this “Quiz” you will
√ prove that there is a positive real number s such that s = 2.
That is you will prove that 2 ∈ R. I expect you to be able to complete at least pages 1 and
2 in the hour. You may freely use any basic arithmetic without justification.
Let X ⊆ R be given by X = {x ∈ R : x2 ≤ 2}
(1) Prove that 2 is an upper bound on X. Conclude that X is bounded above.
(2) Prove that 0 ∈ X. Conclude that X is non-empty.
(3) Prove that X has a supremum.
Let s be the supremum of X.
(4) Pick One: Use (2) to prove that 1 ≤ s OR use (1) to prove that s ≤ 2. You will use
both of these conclusions later on.
Intuitively, s should be the square root of 2. In the next two pages, you will prove
that s2 = 2
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(5) Suppose for the sake of contradiction that s2 < 2. Let E = 2 − s2 . Show that E > 0.
(6) Let be the minimum of 1 and E/5. You may assume that s ≤ 2, that 0 < ≤ 1 and
that ≤ E/5. Prove that
(s + )2 ≤ s2 + E
Hint: Expand (s + )2 to get s2 + (2s + ) and show that (2s + ) ≤ E
(7) Use the result of the previous exercise to show that s + is in X. (Hint: What is
s2 + E?)
(8) Why is this a contradiction? What can you conclude?
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(9) Suppose for the sake of contradiction that s2 > 2
Let E = s2 − 2. Show that E > 0.
(10) Let be the minimum of 1 and E/4.
You may assume that s ≤ 2, that 0 < ≤ 1 and that ≤ E/4. Prove that
(s − )2 ≥ s2 − E
Hint: Expand (s − )2 to get s2 − (2s − ) and show that (2s − ) ≤ E
(11) Show that s − ≥ 0.
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(12) Use things you know to conclude that s − is an upper bound for X. You may freely
use basic properties of inequalities in ordered fields.
(13) Why is this a contradiction? What can you conclude?
(14) Write a few sentences concluding that s ≥ 0 and that s2 = 2.
(15) Do you think that this argument could run for an arbitrary positive number a ∈ R in
place of 2?