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Section 2.4: The Principle of Mathematical Induction
1. Prove by induction that for all natural numbers n, the quantity 32n − 1 is divisible by
8.
Proof. When n = 1, we have 32n − 1 = 32 − 1 = 8, which is certainly divisible by 8. So
suppose 32n − 1 is divisible by 8 for some natural number n, and consider the quantity
32(n+1) − 1. We have
32(n+1) − 1 = 32n+2 − 1
= 32n 32 − 1
= 9 · 32n − 1
= (8 + 1)32n − 1
= 8 · 32n + (32n − 1).
Note the first term 8 · 32n is obviously divisible by 8, while the bracketed expression
(32n − 1) is divisible by 8 by our induction hypothesis. Thus their sum, which equals
32(n+1) − 1, is divisible by 8, which completes the induction step and the proof.
2. Consider the statement “For all integers n, n2 ≥ n”. Explain why this statement can
not be proven by induction.
Solution. To use induction, there must be a well-defined “first number” for which the
theorem applies. Since there is no “first” integer, this theorem can not be proven by
induction.
3. Consider the statement “For all real numbers x ≥ 1, x2 ≥ x”. Explain why this
statement can not be proven by induction.
Solution. Unlike the previous problem, here there is a well-defined first number for
which the theorem applies (x = 1). However, another requirement for using induction
is that there is a well-defined notion of the “next number”. The integers have this
property (given some integer n, which integer comes next? It’s n + 1. No ambiguity
about that!). But the real numbers (and the rational numbers, and the irrationals,
and the primes, etc.) lack this property. Given any real number x, there is no answer
to the question “What real number comes next?” Thus induction can not be used to
prove this statement.
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4. Imagine a simplified version of American Football where a team can score in one of two
ways: a 7-point touchdown and a 3-point field goal. Note that under these rules it is
impossible for a team to score, for example, exactly 4 points. Investigate which scores
are achievable. If you look long enough, you’ll find a number n such that it appears
that every score n or greater is achievable. What’s this value of n? Prove that every
score n or greater is achievable.
Theorem. Any score 12 or greater is achievable.
Proof. A team can achieve exactly 12 points with 4 field goals. So suppose it is possible
to achieve exactly n points, where n is some integer greater than or equal to 12. We’ll
demonstrate why a score of n + 1 is also achievable.
One way a score of n can be turned into a score of n + 1 is by removing two field goals
(worth 6 points) and adding one touchdown (7 points). But what if the only way to
achieve a score of n involves fewer than two field goals? In that case, since n is at least
12, the team scoring n points must have scored at least two touchdowns. So remove
two touchdowns (worth 14 points) and add five field goals (15 points). In either case,
if a score of n is achievable, so is a score of n + 1. This completes the induction step
and the proof.
5. Repeat the previous exercise, but now suppose that a field goal is worth 4 points (a
touchdown is still worth 7 points).
Proof. The proof is very similar to the previous one (but 12 won’t be the magic number
here). Give it a shot!
6. The product of two odd numbers is odd (we already proved this in Theorem 2.6).
Likewise, the product of any three odd numbers should be odd (or the product of any
four odd numbers, or the product of any ten million odd numbers). There’s a theorem
lurking here; find it, state it clearly, and prove it by induction.
Theorem. The product of any n ≥ 2 odd numbers is odd.
Proof. We already proved the base case in Theorem 2.6 (i.e., we’ve already proven that
the product of any two odd numbers is odd). So suppose the product of any n numbers
is odd.
Now consider the product of n + 1 odd numbers; call them a1 , a2 , . . . , an , an+1 . Their
product is
a1 a2 · · · an an+1 = (a1 a2 · · · an )an+1 .
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The factor (a1 a2 · · · an ) is a product of n odd numbers, which is odd by our induction
hypothesis. So let (a1 a2 · · · an ) = b. Then the product of all n + 1 odd numbers reduces
to b · an+1 , which is a product of two odd numbers, hence is odd. This completes the
induction step and the proof.
7. Prove that for all natural numbers n, the quantity n3 + 5n is divisible by 6.
Proof. We’ve already proved n3 + 5n is divisible by 3 for all n ∈ N (this was Theorem
2.28). To show n3 + 5n is divisible by 6, it suffices to prove this quantity is always
even, because any integer divisible by both 2 and 3 must be divisible by 6 (if you’re
skeptical, come see me and I’ll show you why that’s true).
So our goal is to prove n3 + 5n is even for all natural numbers n. We could do this by
induction, but just to mix it up, we’ll prove it by considering two cases.
If n is even, then n = 2k for some k ∈ Z. Then
n3 + 5n = (2k)3 + 5(2k) = 8k 3 + 10k = 2(4k 3 + 5k).
Since k is an integer, so too is (4k 3 + 5k). This proves n3 + 5n is even.
If n is odd, then n = 2j + 1 for some integer j. Then
n3 + 5n = (2j + 1)3 + 5(2j + 1) = 8j 3 + 4j 2 + 2j + 1 + 10j + 5
= 8j 3 + 4j 2 + 12j + 6
= 2(4j 3 + 2j 2 + 6j + 3).
Since j is an integer, so too is that bracketed factor. Thus n3 + 5n is even.
Since n3 + 5n is both even and divisible by 3 for all natural numbers n, it follows that
n3 + 5n is divisible by 6 for all natural numbers n.
8. This problem involves calculus (sorry, but you can only go so far in life without running
into calculus). Compute the first, second, and third derivatives of the function f (x) =
xex . If you clean up after each step, you should observe a pattern forming. State the
general rule for the nth derivative of xex , and prove it by induction.
Theorem. If f (x) = xex , then the nth derivative of f is f (n) (x) = (x + n)ex .
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Proof. When n = 1, we have
f (n) (x) = f 0 (x) = xex + ex (1) = (x + 1)ex
by the Power Rule. This shows our claimed formula for f (n) is correct when n = 1,
which completes the base case. So suppose f (n) (x) = (x + n)ex for some n ≥ 1. The
(n + 1)st derivative is found by differentiating the nth derivative, so we have
f (n+1) (x) =
d
d (n)
f (x) =
(x + n)ex = (x + n)ex + ex (1) = (x + n + 1)ex .
dx
dx
Thus the claimed formula is correct for n + 1, which completes the induction step and
the proof.
9. Repeat the previous exercise for the function f (x) = x2 ex .
Proof. I’d love it if somebody gets this one. If you think you’ve got it (or would like
to get it but need some assistance), please please please come talk to me. It would
make me very happy.
10. We proved in Theorem 2.27 that 8n − 3n is divisible by 5 for all natural numbers n.
But if you consider the values of 8n − 3n for n = 1, 2, 3, . . ., you notice something
more specific happening: the last digit of 8n − 3n is always a 5, never a 0. Is this a
coincidence, or do you think 8n − 3n always ends in a 5? Investigate, form a conjecture,
and then prove it.
Theorem. For all n ∈ N, 8n − 3n is divisible by 5 but not by 10.
Proof outline. To prove 8n − 3n is not divisible by 10, we must prove 8n − 3n is odd
(because if it were even it would be divisible by both 2 and 5, hence it would be divisible
by 10).
So why is 8n − 3n always odd? Short answer: 8n is always even, and 3n is always odd,
and we’ve already proven that “even plus odd equals odd” (Theorem 2.6). The longer
answer involves explaining why 8n is always even and why 3n is always odd. Exercise 6
shows why 3n is always odd. As for why 8n is always even, just note 8n = 8 · 8n−1 , and
we proved in Theorem 2.6 that if one factor in a product is even, the whole product is
always even.
11. Generalize Theorem 2.27 by proving the following: if a − b = c, then an − bn is divisible
by c for all natural numbers n. (assume a, b, and c are natural numbers, otherwise
none of this makes sense . . . it might not make sense anyway)
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Proof. Suppose a, b, and c are natural numbers with a − b = c. We must prove an − bn
is divisible by c for all n ∈ N. We will prove this by induction.
When n = 1, we have an − bn = a1 − b1 = c, which is obviously divisible by c. So
suppose an − bn is divisible by c for some n ≥ 1, and consider the number an+1 − bn+1 .
We have
an+1 − bn+1 = a · an − b · bn
= (b + c)an − b · bn
= b · an + c · an − b · b n
= c · an + b(an − bn ).
The first term c · an is obviously divisible by c, while the second factor is divisible by c
by our induction hypothesis. Thus their sum, which equals an+1 − bn+1 , is also divisible
by c, which completes the induction step and the proof.
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