10 1. From the Real Numbers to the Complex Numbers
This proposition provides a method of proof, called induction, surely known to
many readers. For each n N, let Pn be a mathematical statement. To verify that
Pn is a true statement for each n, it suffices to show two things: first, P1 is true;
second, for all k, whenever Pk is true, then Pk+1 is true. The reason is that the set
of n for which Pn is true is then an inductive set containing 1; by Proposition 3.1
this set is N.
0 1 2 x x + 1
Figure 1.3. Induction.
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Exercise 1.8. Apply the principle of mathematical induction to establish the
well-ordering principle: every nonempty subset of N contains a least element.
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Exercise 1.9. It is of course obvious that there is no natural number between
0 and 1. Prove it!
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Exercise 1.10. For a constant C put f(x) = x + C. Find a formula for the
composition of f with itself n times. Prove the formula by induction.
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Exercise 1.11. For nonzero constants A and B, put f(x) = A(x + B) - B.
Find a formula for the composition of f with itself n times. Prove the formula by
induction. Find a short proof by expressing the behavior of f in simple steps.
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Exercise 1.12. For constants M, C with M 6 = 1 put f(x) = Mx + C. Find a
formula for the composition of f with itself n times. Suggestion: Write f in the
notation of the previous exercise.
We close this section by proving a precise statement to the effect that many
small things make a big thing. This seemingly evident but yet surprisingly subtle
property of R, as stated in Proposition 3.2, requires the completeness axiom for
its proof. The proposition does not hold in all ordered fields. In other words,
there exist ordered fields F with the following striking property: F contains the
natural numbers, but it also contains super numbers, namely elements larger than
any natural number. For the real numbers, however, things are as we believe. The
natural numbers are an unbounded subset of the real numbers.
Proposition 3.2 (Archimedean property). Given positive real numbers x and ,
there is a positive integer n such that n x. Equivalently, given y 0, there is
an n N such that
1
n
y.
Proof. If the first conclusion were false, then every natural number would be
bounded above by
x.
If the second conclusion were false, then every natural number
would be bounded above by
1
y
. Thus, in either case, N would be bounded above.
We prove otherwise. If N were bounded above, then by the completeness axiom N
would have a least upper bound K. But then K - 1 would not be an upper bound,
and hence we could find an integer n with K - 1 n K. But then K n + 1;
since n + 1 N, we contradict K being an upper bound. Thus N is unbounded
above and the Archimedean property follows.
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