Page 146 - HOW TO PROVE IT: A Structured Approach, Second Edition

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132 Proofs
Multiplying out both sides gives us:
2
2
2
2
2
2
(a + b) − 4(a − b) = a + 2ab + b − 4(a − 2ab + b )
2
2
=−3a + 10ab − 3b ;
2
2
2
2
(3b − a)(3a − b) = 9ab − 3a − 3b + ab =−3a + 10ab − 3b .
Clearly the two sides are equal. The simplest way to write the proof of this
2
2
is to write a string of equalities starting with (a + b) − 4(a − b) and ending
with (3b − a)(3a − b). We can do this by copying down the first string of
equalities displayed above, and then following it with the second line, written
backward.
Solution
Theorem. For any real numbers a and b,
2 2
(a + b) − 4(a − b) = (3b − a)(3a − b).
Proof. Let a and b be arbitrary real numbers. Then
2
2
2
2
2
2
(a + b) − 4(a − b) = a + 2ab + b − 4(a − 2ab + b )
2
=−3a + 10ab − 3b 2
2
2
= 9ab − 3a − 3b + ab = (3b − a)(3a − b).
We end this section by presenting another proof without preliminary scratch
work, but with a commentary to help you read the proof.
Theorem 3.4.6. For every integer n, 6 | n iff 2 | n and 3 | n.
Proof. Let n be an arbitrary integer.
(→) Suppose 6 | n. Then we can choose an integer k such that 6k = n.
Therefore n = 6k = 2(3k), so 2 | n, and similarly n = 6k = 3(2k), so 3 | n.
(←) Suppose 2 | n and 3 | n. Then we can choose integers j and k such
that n = 2 j and n = 3k. Therefore 6( j − k) = 6 j − 6k = 3(2 j) − 2(3k) =
3n − 2n = n,so6 | n.

Commentary. Thestatementtobeprovenis∀n ∈ Z(6 | n ↔ (2 | n ∧ 3 | n)),and
the most natural strategy for proving a goal of this form is to let n be arbitrary
and then prove both directions of the biconditional separately. It should be clear
that this is the strategy being used in the proof.
For the left-to-right direction of the biconditional, we assume 6 | n and then
prove 2 | n and 3 | n, treating this as two separate goals. The introduction of