Strong Induction
Reading
The reading is section 5.2. The section on “Using strong induction in computational geometry” is optional. After
you read example 2, you should also go read the proof of the uniqueness assertion of the fundamental theorem of
arithmetic in 4.3, if you haven’t already.
Summary
Let L be some fixed integer. It’s common to take L = 0 or L = 1. Strong induction is a variant of induction. It is
also method of verifying that some predicate P whose domain all integers n ≥ L is actually true for all n ≥ L. To
prove that P (n) is true for all n ≥ L by strong induction, you need to verify two things.
(i) (Base case) Show that P (L) is true.
(ii) (Inductive step) Show that if P (1), . . . , P (k) are all true for some integer k ≥ L, then P (k + 1) is also true.
Comments
(1) Like induction, this is meaningless in the abstract, so focus on the examples.
Examples
One kind of strong induction that shows up often is one where we use both P (2) and P (k) to conclude P (k + 1).
In this case, our base case has to be both P (1) and P (2). Here is an example.
Proposition 1. If p is a prime integer and p | a1 · · · an for some integers a1 , . . . , an , then p | ai for some i = 1, . . . , n.
Proof. Let P (n) be the statement that whenever there are n integers a1 , . . . , an and p | a1 · · · an , then p | ai for
some i = 1, . . . , n. This is clearly true for n = 1, and I proved this for the case n = 2 in my notes on primes from
7/8.
Now suppose we know P (1), . . . , P (k) for some k ≥ 2. Let a1 , . . . , ak+1 be k + 1 integers and suppose p |
a1 · · · ak+1 . Then p divides the product of a1 · · · ak and ak+1 , so using P (2), we know that either p | ak+1 or
p | a1 · · · ak . If p | ak+1 , we are done. If p - ak+1 , then p | a1 · · · ak . Using P (k), we see that p | ai for some
i = 1, . . . , k, so again we are done. This proves P (k + 1), completing the induction.
We can now complete the proof of the fundamental theorem of arithmetic.
Theorem 2 (Fundamental theorem of arithmetic). Every integer n > 1 can be written uniquely as a product of
primes.
Proof of existence. Let P (n) be statement that n can be written as the product of primes. Clearly n = 2 can be
written as a product of primes, so this shows P (2).
Suppose we know P (2), . . . , P (k). Then k + 1 is either composite, or else prime. If it is prime, then clearly it
can be written as a product of primes, so we are done. If it is composite, then there exist integers a and b such that
k + 1 = ab, where 2 ≤ a, b ≤ k. Since P (a) and P (b) are both true, we know that a = p1 · · · pr and b = q1 · · · qs
for primes p1 , . . . , pr , q1 , . . . , qs , which means that k + 1 = p1 · · · pr q1 · · · qs is also a product of primes. This shows
P (k + 1), completing the induction and showing that every integer n > 1 can be written as a product of primes.
Proof of uniqueness. Let P (n) be the statement that n can be written uniquely as the product of primes. The
base case, P (2), is clear. Suppose we know P (2), . . . , P (k). Let k + 1 = p1 · · · pr and k + 1 = q1 · · · qs , where
p1 ≤ p2 · · · ≤ pr and q1 ≤ · · · ≤ qs . We are trying to show that these prime factorizations are the same. Then
p1 | (k + 1), so by proposition 1, we must have p1 | qj for some j = 1, . . . , s. But qj is prime, so in fact we must
have p1 = qj . Similarly, we know that q1 | (k + 1), so q1 | pi for some i = 1, . . . , r, so q1 = pi . But then we have
p1 ≤ pi = q1 ≤ qj = p1
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so in fact we must have p1 = pi = qj = q1 . Now consider the integer m which is (k + 1)/p1 , or equivalently,
(k + 1)/q1 . If m = 1, then we had n = p1 = q1 , so the two prime factorizations are equal. So m > 1. Then we know
that P (m) is true, so since m = p2 · · · pr = q2 · · · qs , we must have r = s and pi = qi for all i = 2, . . . , r = s. Since
we also know that p1 = q1 , this completes the induction.
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