Page 166 - HOW TO PROVE IT: A Structured Approach, Second Edition

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152 Proofs
Solution
Theorem. For every real number x, if x = 2 then there is a unique real number
y such that 2y/(y + 1) = x.
Proof. Let x be an arbitrary real number, and suppose x = 2. Let y =
x/(2 − x), which is defined since x = 2. Then
2x 2x
2y 2−x 2−x 2x
= x = = = x.
y + 1 + 1 2 2
2−x 2−x
To see that this solution is unique, suppose 2z/(z + 1) = x. Then 2z =
x(z + 1), so z(2 − x) = x. Since x = 2 we can divide both sides by 2 − x
to get z = x/(2 − x) = y.

The theorem in Example 3.6.1 can also be used to formulate strategies for
using givens of the form ∃!xP(x). Once again, statement 3 of the theorem is
the one used most often.

To use a given of the form ∃!xP(x):
Treat this as two given statements, ∃xP(x) and ∀y∀z((P(y) ∧ P(z)) →
y = z). To use the first statement you should probably choose a name, say x 0 ,
to stand for some object such that P(x 0 ) is true. The second tells you that if
you ever come across two objects y and z such that P(y) and P(z) are both
true, you can conclude that y = z.
Example 3.6.4. Suppose A, B, and C are sets, A and B are not disjoint, A and
C are not disjoint, and A has exactly one element. Prove that B and C are not
disjoint.
Scratch work
Givens Goal
A ∩ B = ∅ B ∩ C = ∅
A ∩ C = ∅
∃!x(x ∈ A)
We treat the last given as two separate givens, as suggested by our strategy.
Writing out the meanings of the other givens and the goal, we have the following
situation:
Givens Goal
∃x(x ∈ A ∧ x ∈ B) ∃x(x ∈ B ∧ x ∈ C)
∃x(x ∈ A ∧ x ∈ C)
∃x(x ∈ A)
∀y∀z((y ∈ A ∧ z ∈ A) → y = z)
To prove the goal, we must find something that is an element of both B and
C. To do this, we turn to the givens. The first given tells us that we can choose a