320 Functions, graphs, and numbers
14 Prove the Archimedes property of numbers: if e > 0 and a > 0, then there
is an integer n for which ne > a. Solution: Suppose ne 5 a for each n. Then
the set S of numbers e, 2e, 3e, is nonempty and has an upper bound. Hence
S must have a least upper bound M. There must be an integer m for which
me > M - e. Then (m + 1)e > M, and hence M is not an upper bound of S.
This contradiction proves that there is an n for which ne > a.
15 Prove that each nonempty set of positive integers contains a least element.
Solution: Let S be a set of positive integers. Since S is nonempty and has the
lower bound 1, S must have a greatest lower bound in. Let 0 < e < '. The
interval m < x < in + e must contain an integer n in S, since otherwise m + e
would be a lower bound greater than in. Since the interval has length less than
4 and cannot contain two integers, it follows that n is the one and only integer
in S which is less than m + e. Therefore, n is the least integer in S. Remark:
The fact that each nonempty set of positive integers contains a least element
will now be used to prove the following principle of mathematical induction. If
a particular assertion involving a positive integer k is true when k = 1, and if the
assertion is true when k = n + I provided n? I and it is true when k = n, then
the assertion is true for each positive integer. Let T be the set of positive integers
for which the assertion is true, and let F be the set for which the assertion is
false. If F is nonempty, then F must have a least element m which is a positive
integer greater than 1. Then m - 1 must be in T and our hypothesis gives the
conclusion that m is in T. Thus m is in both F and T and (on the basis of the
tacit assumption that we are dealing with statements that are either true or
false but not both) we have a contradiction that proves that F is empty. There-
fore, T contains each positive integer and the assertion is therefore true for each
positive integer.
16 Prove that if x is a number, then there is an integer n for which n 5 x <
n + 1. Remark: This property of numbers was mentioned in Section 1.1, and
the integer n is [x], the greatest integer in x. Solution: Suppose first that x > 2.
Using the Archimedes property of real numbers (Problem 14) with e = 1 and
a = x shows that the set S of integers greater than x is a nonempty set of positive
integers. Hence, as Problem 15 shows, S must have a least element m. Then
x < in, but the inequality x < m - 1 must be false because otherwise m would
not be the least element of S. Therefore, in - 1 S x < in and we obtain our
result by setting it = m - 1. In case x <= 2, we can choose an integer k such
that x + k > 2. Letting m be an integer for which in < x + k < m + 1, we
find that m - k is an integer n for which n S x < n + 1.
17 Prove that if x is a number and n is a positive integer, then there is an
integer k for which
k<x<k-i 1
n n
Remark: In case m is a nonnegative integer and n = 10-, the result shows how
x is related to "finite decimals." Proof: Problem 16 shows that there is an
integer k for which
k5nx<k+1,and the result is obtained by dividing by n.