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74 The Nuts and Bolts of Proof, Third Edition
If we combine these two equalities, we obtain 5n = 7m.
As 5 and 7 are prime numbers, 5n is divisible by 7 only if n is divisible
by 7. Thus, n = lk for some integer number k.
Therefore, x = 5n = 5{lk) — 35k for some integer number k. This means
that X is a multiple of 35.
Part 2. {x G Z I X is a multiple of 35} c^ n B.
Let X be a multiple of 35. Therefore, x = 35t for some integer number t.
Thus, X is divisible by 5 (so x G A) and it is divisible by 7 (so x G B).
This implies that x e AnB.
Therefore, the two sets are equal. •
We will consider another set. The complement of a set A is the set of all
elements that belong to the universal set U, but do not belong to the set A.
The complement of the set A can be denoted by a variety of symbols. The
most commonly used are A\ C{AX and A. We will use A\ Therefore,
A' = {xe
Ulx^A}.
The part shaded in the following diagram represents the complement of
the set A:
EXAMPLE 5. Let A c U and B c U. Then,
(A n By = A'U B'.
(This is known as one of De Morgan's laws. The proof of the other law—
namely, {A U B)' = A' f\ B'—is left as an exercise. August De Morgan
(1806-1871) was one of the first mathematicians to use letters and symbols
in abstract mathematics.)
Proof:
Part L {AOB)' ^A'VJW.
Let X e{AC\ B)'. By definition of complement of a set this implies that
x^(^ nB). Therefore, either x^A ov x^B. (Indeed, if x was an element of