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.