MI 4
Mathematical Induction
Name ________________
The Seamy Underbelly of Induction
Induction lays many traps for the unwary. See if you can spot the error in each of the following
induction “proofs”—all of which assert things that are clearly wrong.
1.
All positive integers are equal. For given any two positive integers x and y, we will prove
that they are equal. Let the maximum of x and y be called m.
Base case: if m = 1, then clearly x = y = 1, so x and y really are equal.
Induction: by induction, let’s say we know the statement to be true if the maximum, m, is equal
to k – 1, and we wish to show that it is true when m = k. Well, if the maximum of x and y is k,
then the maximum of x – 1 and y – 1 is k – 1, so we know that x – 1 = y – 1. Add one to both
sides to see that x = y. What went wrong?
The trap is when one was subtracted from x and y. When using the inductive step, we needed x – 1 and y – 1 to be positive integers, because the statement only applies to positive integers. But if x = 1, then x – 1 = 0 is not positive. So the inductive hypothesis is false—we don’t know that x – 1 and y – 1 are equal after all. Watch out! Sometimes when you work backwards you end up going down below the base cases you have proved! 2.
In any set of horses, all the horses in that set are the same color. For if you line up all the
horses in that set in a row, then:
Base case: if there is only one horse in that row, it is the same color as itself, so all the horses in
the set (all one of them!) are the same color.
Induction: let’s say we know, per the inductive hypotheses, that any set of k – 1 horses are all
the same color. If our set has k horses in it, lined up in a row, the left-most k – 1 horses must all
be the same color, and the rightmost k – 1 horses must be the same color. So the horse on the left
is the same color as the horses in the middle, and so is the horse on the right, so all the horses
must be the same color. What went wrong?
The error here is similar but slightly more subtle. What if there aren’t any horses in the middle? The base is correct, but the induction step is not correct in going from 1 to 2. When k = 2, the leftmost k – 1 horses is only the one horse on the left, and the rightmore horses is only the one horse on the right, and there aren’t any in the middle to be the same color to bridge the left to the right. Make sure your inductive step explains how to get from k – 1 to k regardless of the value of k! Induction 5.1
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MI 4
Mathematical Induction
Name ________________
3.
Let’s say you have a bunch of coins that appear identical, except that it is known that one
is counterfeit and weighs slightly less than the others. You have a balance scale. Claim: you can
find the counterfeit coin in at most two weighings, no matter how many coins you have!
Base cases: if there is only one coin, it must be the counterfeit; if you have two, clearly you can
just compare them and the counterfeit coin will be the lighter one. So you can solve the problem
in at most one—and certainly at most two—weighings for one or two coins.
Induction: now say that you can solve the problem for k – 1 coins using no more than two
weighings. If you have k coins, set one aside. Now use your two weighings to determine which
of the coins in the remaining k – 1 is counterfeit. If none are, then the one you set aside must be
the counterfeit, and you don’t need an extra weighing to see that! So at most two weighing
suffice for any number of coins! It turns out you can find the counterfeit coin if you have any
number of coins up to nine (can you figure out how?) but with ten or more coins at least three
weighings are needed. Where did the “proof” go wrong?
This time, the induction tried to use more than the induction hypothesis allowed. Since the claim is you can find a counterfeit coin with two weighings, that is all you are allowed to do. It does not allow you to determine that there is no counterfeit coin, only to find one when you already know that there is one. The induction, when it pulled out one coin, ignored this. The coin that was pulled out might have been the counterfeit, but then the two weighings can’t necessarily determine that the remaining coins are all true. Be careful to only use what the inductive hypothesis actually states! 4.
If n is a non-negative integer and x ≠ 0, then xn = 1.
Base case: we already know that for any non-zero x, that x0 = 1.
Induction: let’s say that it is true for all the power up to and including k – 1. Then
x k−1 ⋅ x k−1 1⋅1
xk =
=
= 1 so it is true for the kth power also. So if it is true up to k – 1 you can
x k−2
1
continue and it is true up to k. So by induction it is true up to any number at all! What went
wrong?
This is very similar to the first problem. The base is OK, but when we set k = 1, we see that we need xk – 2 = x–1 which is not a non-­‐negative integer exponent. So we don’t know that it equals one, which is needed for the computation. Induction 5.2
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MI 4
Mathematical Induction
Name ________________
Even though the particular argument in the last problem has a hole in it, the idea of using all of
the numbers up to k – 1 is a nifty trick.
5.
Prove that every positive integer, 2 or larger, can be factored into a product of primes.
Note: 1 is not prime. It is not composite, either. It is a special case. (In fact, if you think about
what multiplying zero numbers together means—think about x0—we could even write 1 as a
product of zero primes!)
a.
Base case: I’ll make it easy for you: 2 is already the product of (just one—itself) primes.
b.
Induction step: We’re going to have to be sneaky. Instead of S(k – 1) being “k – 1 can
be factored into primes,” use the statement “All the integers up to, and including, k – 1 can be
factored into primes.” This is still just a single statement, which is either true or false, about the
number k – 1, so it can serve just as well as our inductive hypothesis. Now what do you know
about the number k? What if it is prime? What if it isn’t?
If the number k is prime, it is already factored into the product of (just one—
iteslf) primes. Otherwise, k is composite, and k = ab where a and b are both numbers between 1 and k, exclusive. Since both a and b are at most k – 1 but at least two, the fit into the category of “all the integers up to, and including, k – 1” so they can be factored into primes. So take the primes that make up a and the primes that make up b, and we have factored k. The reason we had to use this trick was that knowing how to factor k – 1 doesn’t tell us how to factor k. The number k breaks down into smaller numbers, but we don’t know what those numbers are going to be. So we take the precaution of making our statement not about the one number k – 1 but about all the numbers up to k – 1. This sneaky technique of including all the numbers up to k – 1, instead of just using k – 1 itself,
is often referred to as the second principle of mathematical induction or strong induction even
though it is really just a variation of regular induction.
Also, note that this did not prove the factorization is unique only that it can be done in at least
one way! The uniqueness requires different tools.
Induction 5.3
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MI 4
Mathematical Induction
Name ________________
6.
Try another one. Pick any number, n (we will use the example of 143). Create a list
starting with your number. Find the next number on the list as follows. If the current number is
prime, subtract one. If it is not prime, divide by its largest prime factor. Starting at 143, we
would then get 11 (13 is the largest prime factor of 143), then 10, then 2, then 1. Since 1 is
neither prime nor does it have any factors, we must stop. Prove that no matter what number n
you start with, the process stops at 1.
Base case: the number 1 aleady is at 1, so comes to an instant stop. Induction: let’s say we knew that all the numbers less than k lead to lists that stop at one. Starting from k, we either subtract one, or divide it by a prime number (which is larger than one), so either way, the next number on the list starting with k will be less than k. But we already know that if a number is less than k it leads to a list that stops at one. Since k leads to this number, k also leads to a list that stops at one. 7.
In the game of nim there are a number of piles of beans. Two players take turns taking
any positive number of beans from one of the piles (including taking the whole pile, if desired).
The person who gets the last bean wins. For instance, if there is only one pile, the first player
easily wins just by taking the whole pile! Prove that if there are two equal piles to start, the
second player should always win no matter how many beans are in the two piles.
If there is only one bean in each pile, the first player takes one pile, and the second players takes the other and wins. Now let’s say that the second player will win if there are two piles with 1, 2, 3, …, k – 1 beans. If there are two piles of k beans, then: a) the first player might take all the beans in one pile, then the second player takes all the beans in the other and wins, or b) the first player might take some beans from one pile. If the second player then takes the same number of beans from the other, she leaves two equal piles with at most k – 1 beans, and she is again the second player in the smaller game, which we know she will win! Induction 5.4
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MI 4
Mathematical Induction
Name ________________
7.
Are you up for a real challenge? Prove Zeckendorf’s Theorem: Every positive integer
can be written in a unique way as the sum of non-consecutive Fibonacci numbers. (For example,
86 = 55 + 21 + 8 + 2. This is unique, because 55 + 21 + 5 + 3 + 2 has consecutive Fibonacci
numbers 2 and 3 (and also 3 and 5).)
Base case: 1 is already a Fibonacci number, and there is certainly only one way to write 1 as a sum of positive numbers. Induction: Using the strong form of induction, we can assume that all the numbers from 1 up to k – 1 can be written, uniquely, as sums of non-­‐consecutive Fibonacci numbers. If f is the largest Fibonacci that is less than or equal to k, we can write k = f + r. If k is already Fibonacci then f = k and r = 0 and we’re done. Otherwise, r > 0, but r < k, and r can be written, uniquely, as the sum of non-­‐
consecutive Fibonacci numbers. Now the worry is that the largest Fibonacci in r and the number f might be consecutive, messing up the fact that we need non-­‐consecutive numbers. For instance could f be 13 and r = 10 = 8 + 2 as a sum of Fibonaccis? No—this can’t happen. For if r had this next Fibonacci down (call it g) then k = f + r ≥ f + g. But f and g are two consecutive Fibonacci numbers, so f + g is a Fibonacci number larger than f. But this can’t be because f is the largest Fibonacci that is not bigger than k. Finally, how do we know that this is the only possible way? Well, a sum of Fibonacci’s that adds to k must contain f—it is not possible to write k as a sum of non-­‐consecutive Fibonacci’s without the largest Fibonacci not exceeding k. This is because the largest you could get with non-­‐consecutive Fibonaccis is either F1 + F3 + F5 +  + F2n – 1 = F2n (which you proved on the Fibonacci page) or it is F2 + F4 +  + F2n = F2n + 1 – 1 (which is proved in the same way). So unless you include the largest Fibonacci not exceeding k, the sum of the rest of the non-­‐consecutive Fibonacci’s can at most get up to this one you aren’t using—which is less than k. So you must use this largest one, f, as stated. Then, by the inductive hypothesis there is only one way to write r, so k must be f + the only way to write r, and thus k’s expression is unique. Induction 5.5
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MI 4
8.
Mathematical Induction
Name ________________
As a final challenge, see if you can figure out what happens in the following situation:
The evil Dr. Fogg tells his MI4 class that they will have a pop quiz one day of the semester, but
they will not know before the day of the quiz what day it is going to be. It will truly be a surprise
quiz. Having just completed the induction unit, his students reason as follows:
Base case: the quiz can’t be on the last day of class, because if it hasn’t been given by the
second-to-last day of class they will then know beforehand that it must be on the last day.
Induction: let’s say, inductively, that Dr. Fogg’s pop quiz can’t be on any of the last k – 1 days
of the semester. Then, if tomorrow is the kth-from-last day of class and there hasn’t been a pop
quiz yet, it must be tomorrow, because it can’t be after that either. But then the class would
know which day it was going to be before the day of the quiz, so it can’t be on the kth from last
day either! So by induction, Dr. Fogg can never give a pop quiz.
Oddly, Dr. Fogg gave a pop quiz on the fifth Thursday of class. No one knew beforehand when
it was going to be, and it was truly a surprise. What went wrong?
The explanation is actually not agreed on by mathematicians and philosophers! There is actually nothing mathematically wrong with the induction. The fallacy is in the the whole set-­‐up. A simple explanation is that, based on the students having convinced themselves that there can’t be a pop quiz, no matter what day you have one it would be a surprise. A more complicated way of thinking is that Dr. Fogg made a statement that cannot be true. He is evil, after all, and what’s a little lie among the fiendish? If he is lying, then what he says about the timing of the quiz conveys no useful information, so there may or may not be a pop quiz, which may or may not be figured out in advance. There are plenty of English sentences which simply cannot be true. “There are six words in this sentence.” “This sentence is a lie.” Some sentences are lies because they reference things in the real world which are not factual (“Stop signs are green.”). They could be true in some alternate universe, though. Other sentences reference themselves and assert something about the sentence that leads to a paradox. These sentences can’t be true in any universe. Dr. Fogg’s statement is such a sentence—in a sneaky way it references itself, because it essentially says “from the information you have about this pop quiz—namely this sentence—you can’t tell when the pop quiz will happen.” So the sentence says something that cannot be true. Be glad you’re not in his class! Induction 5.6
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