Page 321 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Equinumerous Sets 307
finite set A to be the unique n such that I n ∼ A. This number is also sometimes
called the cardinality of A, and it is denoted |A|. Note that according to this
definition, ∅ is finite and |∅|= 0.
The definition of equinumerous can also be applied to infinite sets, with
results that are sometimes surprising. For example, you might think that Z +
could not be equinumerous with Z because Z includes all the positive inte-
gers, plus all the negative integers and zero as well. But consider the function
f : Z → Z defined as follows:
+
n
⎧
⎪ if n is even
⎪
2
⎨
f (n) =
⎪ 1 − n
if n is odd.
⎪
⎩
2
This notation means that for every positive integer n,if n is even then
f (n) = n/2 and if n is odd then f (n) = (1 − n)/2. The table of values for f in
Figure 1 reveals a pattern that suggests that f might be one-to-one and onto.

Figure 1

To check this more carefully, first note that for every positive integer n,if
n is even then f (n) = n/2 > 0, and if n is odd then f (n) = (1 − n)/2 ≤ 0.
Now suppose n 1 and n 2 are positive integers and f (n 1 ) = f (n 2 ). If f (n 1 ) =
f (n 2 ) > 0 then n 1 and n 2 must both be even, so the equation f (n 1 ) = f (n 2 )
means n 1 /2 = n 2 /2, and therefore n 1 = n 2 . Similarly, if f (n 1 ) = f (n 2 ) ≤ 0
then n 1 and n 2 are both odd, so we get (1 − n 1 )/2 = (1 − n 2 )/2, and once again
it follows that n 1 = n 2 . Thus, f is one-to-one.
To see that f is onto, let m be an arbitrary integer. If m > 0 then let n = 2m,
an even positive integer, and if m ≤ 0 then let n = 1 − 2m, an odd positive
integer. In both cases it is easy to verify that f (n) = m. Thus, f is onto as well
+
as one-to-one, so according to Definition 7.1.1, Z ∼ Z.