Page 141 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Conjunctions and Biconditionals 127
we have two new given statements: k ∈ Z, and x = 2k.
Givens Goal
2
x ∈ Z ∃k ∈ Z(x = 2k)
k ∈ Z
x = 2k

The goal starts with ∃k, but since k is already being used to stand for a
particular number, we cannot assign a new value to k to prove the goal. We
must therefore switch to a different letter, say j. One way to understand this is
2
to think of rewriting the goal in the equivalent form ∃ j ∈ Z(x = 2 j). To prove
this goal we must come up with a value to plug in for j. It must be an integer,
2
and it must satisfy the equation x = 2 j. Using the given equation x = 2k,we
2
2
2
2
see that x = (2k) = 4k = 2(2k ), so it looks like the right value to choose
2
2
for j is j = 2k . Clearly 2k is an integer, so this choice for j will work to
complete the proof of our first goal.
2
To prove the second goal (x is even) → (x is even), we’ll prove the con-
2
trapositive (x is not even) → (x is not even) instead. Since any integer is
either even or odd but not both, this is equivalent to the statement (x is odd) →
2
(x is odd).
Givens Goal
2
x ∈ Z x is odd
x is odd
The steps are now quite similar to the first part of the proof. As before, we
begin by writing out the definition of odd in both the second given and the
goal. This time, to avoid the conflict of variable names we ran into in the first
part of the proof, we use different names for the bound variables in the two
statements.

Givens Goal
2
x ∈ Z ∃ j ∈ Z(x = 2 j + 1)
∃k ∈ Z(x = 2k + 1)

Next we use the second given and let k stand for a particular integer for
which x = 2k + 1.

Givens Goal
2
x ∈ Z ∃ j ∈ Z(x = 2 j + 1)
k ∈ Z
x = 2k + 1