MAT 300 Mathematical Structures
John Quigg
Spring 2012
revised April 16, 2012
Notes on: Section 6.4
1. Strong induction
To prove the statement ∀n ∈ NP (n) by strong induction, do the following:
h
i
• Prove ∀n ∈ N ∀k < nP (k) → P (n) .
Remark 1. The above method deserves some explanation. First look a little more closely at
exactly what the method requires us to prove. We must show that for any natural number
n, if we assume that P (k) is true for every natural number k smaller than n, then we can
deduce that P (n) is true. Note that in the statement of the strong induction method we
wrote “∀k < n” when we actually meant that k must also be a natural number. This is one
of those times when we are supposed to infer from the context that the variable k is meant
to have to same universe of discourse as n, namely N . To state it more precisely would make
it look somewhat messier: ∀n ∈ N[(∀k ∈ N(k < n → P (k))) → P (n)].
We are calling it a form of induction, so we should examine the similarities and differences
between strong induction and (ordinary) induction.
(1) Induction has a Base Case, but strong induction does not. Why not? Suppose we
successfully apply the method of strong induction. How can we be sure that P (0) is
true? Because when n = 0 we have proven that if for all k < 0, P (k) is true then
P (0) is true. Remembering that the universe of discourse of k is N, note that the
set {k ∈ N | k < 0} is empty, so the universally quantified statement “for all k < 0,
P (k)” is vacuously true, and hence the truth of the conditional statement “if for all
k < 0, P (k) is true then P (0) is true” guarantees that P (0) is true (because of the
tautology [Q ∧ (Q → R)] → R).
(2) In strong induction, to prove, say P (3), we are allowed to assume that P (0), P (1),
and P (2) are all true. On the other hand, in (ordinary) induction, to prove P (3) we
are only allowed to assume P (2). Thus the induction hypothesis in strong induction
is stronger (and this is why it’s called “strong” induction!). In some circumstances
we have a choice whether to use induction or strong induction, however we’ll see
examples where strong induction works but induction would not be good enough to
prove the desired theorem. Interestingly, however, it turns out that the two forms of
induction are equivalent, although we will not need this.
Example 2. This is Example 6.4.1 in the book, and is a fundamental result in number theory.
Theorem 3 (Division Algorithm). Let n ∈ N and m ∈ Z+ . Then there exist q, r ∈ N such
that n = mq + r and r < m.
Remark 4. Before we begin the proof, it’s perhaps good to comment that the Division
Algorithm is actually true for all n ∈ Z, provided we modify the statement to “q, r ∈ Z” and
“0 ≤ r < m”.
2
Proof. Let m ∈ Z+ . We will prove by strong induction that for all n ∈ N there exist q, r ∈ N
such that n = mq + r and r < m. Let n ∈ N. Assume that for all k < n there exist
q, r ∈ N such that k = qm + r and r < m. We must show that there exist q, r ∈ N such that
n = qm + r and r < m.
Case 1. n < m. Then we can take q = 0 and r = n, because 0, n ∈ N, n = 0m + n, and
n < m.
Case 2. n ≥ m. Let k = n − m. Then k ≥ 0, so k ∈ N. Choose q 0 , r ∈ N such that
k = q 0 m + r and r < m. Then we can take q = q 0 + 1, getting q, r ∈ N, r < m, and
n = k + m = q 0 m + r + m = (q 0 + 1)m + r = qm + r.
Remark 5. Study the choices of notation in the above proof carefully, so that you understand
why we used both q 0 and q, but we only used r and didn’t need r0 (as the textbook used).
This is not a big deal, but I wanted to point out an alternative choice for the notation.
Example 6. This is Example 6.4.2 in the book. In this example, just as we did with induction
in Section 6.1, we will use the fact that strong induction can be applied to sets closely related
to, but slightly different from, N. In this case the set will be {n ∈ Z | n > 1}.
We will need to prove something about prime numbers, so it is probably a good idea to
recall the official definition used in this textbook:
Definition 7. Let n be an integer larger than 1. Then n is prime if it is not a product of
two smaller positive integers.
In more detail: let n ∈ Z, and assume that n > 1. Then n is prime if and only if for all
a, b ∈ Z+ , if n = ab then a ≥ n or b ≥ n. It will be convenient to have an explicit statement
of the negation of this: n is not prime if and only if there exist a, b ∈ Z+ such that n = ab,
a < n, and b < n.
Before stating the theorem, it will be convenient to be perfectly clear about a certain way
of phrasing things: when we say “n is a product of primes”, we mean that there exist prime
numbers p1 , p2 , . . . , pk such that n = p1 p2 · · · pk . Note that this includes the possibility that
n itself is prime, in which case k = 1 and n = p1 . In the following proof, we will use the
obvious fact that for two positive integers a and b, if each of a and b is a product of primes,
then so is ab.
Theorem 8. Every integer larger than 1 is a product of primes.
Proof. Let n be an integer larger than 1. We will use strong induction. Assume that for
every integer k larger than 1, if k < n then k is a product of primes.
Case 1. n is prime. Then there is nothing to prove.
Case 2. n is not prime. Choose a, b ∈ Z+ such that n = ab, a < n, and b < n. Note that
since n = ab > b and b > 0, we have a > 1. Similarly, since n = ab > a and a > 0 we have
b > 1. Thus we can apply the induction hypothesis: each of a and b is a product of primes,
and hence so is n, since n = ab.
Example 9. In this example we’ll prove Theorem 6.4.5 in the book.
We’ll need to prove something about irrational numbers, so it’s probably prudent to record
the definition of rational:
Definition 10. Let x ∈ R. Then x is rational if there exist n, k ∈ Z such that n =
k 6= 0). x is irrational if it is not rational.
n
k
(and
3
Remark 11.
• Thus, we use “irrational” to mean the negation of rational.
• In the above definition we put the condition k 6= 0 in parentheses, because we introduced the fraction nk , which would not make sense for k = 0, so the requirement
k 6= 0 can be safely inferred.
• When we express a rational number as a quotient nk of two integers, it’s clearly ok,
and frequently convenient, to assume that the denominator k is positive, i.e., k ∈ Z+ .
• For the next proof it will be convenient to observe that if x ∈ R then x is rational if
and only if there exists k ∈ Z+ such that kx ∈ Z.
One more thing before the next proof: we skipped the concept “smallest element” in
Chapter 4, and we need it now, so we should record the definition, but we only need it for
subsets of R:
Definition 12. Let S ⊆ R, and let x ∈ S. Then x is a smallest element of S if for all y ∈ S
we have y ≥ x.
We will need the following, which is equivalent to (both forms of) induction, and is Theorem 6.4.4 in the book:
Theorem 13 (Well-Ordering Principle). Every nonempty subset of N has a smallest element.
The textbook offers a proof of the Well-Ordering Principle, but we won’t, because if we
were going to do that, then we should really show that it is actually equivalent to both
forms of induction. Since the Well-Ordering Principle seems intuitively obvious, we’ll use it
without proof.
√
Theorem 14. 2 is irrational.
√
Proof. We argue by contradiction. Assume that 2 is rational. Let
√
S = k ∈ Z+ 2k ∈ Z .
Then S 6= ∅ by assumption.
By the Well-Ordering Principle, we can choose a smallest
√
element k of S. Let n = 2k. Then n ∈ Z, and n2 = 2k 2 , so n2 is even. By an earlier result,
n is even. Choose s ∈ Z such that n = 2s. Then
2k 2 = n2 = 4s2 ,
so k 2 = 2s2 . Thus k 2 is even, so k is even. Choose t ∈ Z such that k = 2t. Note that t > 0
since k > 0. Thus t ∈ Z+ . Also,
√
√
2k
n
2s
2t =
= =
= s,
2
2
2
√
so 2t ∈ Z. Thus t ∈ S. But this is a contradiction, since k = 2t > t.