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MI 4
Mathematical Induction
Name ________________
Induction—Notes and Practice
A couple of notes about principle of mathematical induction and its use:
• One thing that is often very difficult for students to come to terms with when starting to
work with proofs by induction is that the inductive step often starts by assuming the thing
you are trying to prove is actually true for the number k – 1. Wait—assume the thing you
are trying to prove? That doesn’t sound quite right…. But notice the difference between
k and n. The inductive hypothesis—the assumption that the thing we want is actually true
for the specific (though unknown!) number k – 1—is just that: an assumption that may or
may not actually be true about a particular number. We don’t actually care whether it is
true or not. What induction needs is the strategy for getting from k – 1 up to k. It is n
that will be the variable that changes to take on any value. This is just like the hypothesis
in any other theorem. If a triangle is isosceles, then its base angles are equal. We don’t
know that any particular triangle is isosceles. We are doing the same thing during an
induction proof. We don’t know that S(k – 1) is true. What we are trying to show that in
case it is true, then S(k) is also true. Maybe Dr. Fogg was lying about having the best
solution—he is evil after all! But we don’t care as far as induction is concerned. We just
need to know how to get from the solution he claims to have (that is, we assume he has)
for one case up to the next case. That, together with the basis case to get you started, then
allows us to count through all the positive integers and know that S(n) is true no matter
what n is. So keep in mind: you aren’t assuming the thing you want is already true; you
are assuming some particular case in order to prove the next particular case.
• Notice that in the case of the lines cutting through the plane, we actually could have
started at n = 0 instead of n = 1, because the formula worked in case n = 0 as well.
Inductions can actually start anywhere, and then the principle of induction proves that
your statement is true for all n from your base case on up. Starting at n = 1 gives you a
statement that is true for all positive integers. Starting at n = 0 makes the statement true
for all whole numbers. Inductions also sometimes start at n = 2 (because 1 is a special
case that actually doesn’t work), or even higher. This is because the technique for getting
from k – 1 to k you use doesn’t work for small numbers.
• Since induction is not the easiest proof technique to master, at first you will get some nice
hints at how to complete each part of the proof. There will gradually disappear.
Induction 3.1
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MI 4
Mathematical Induction
Name ________________
1.
Use induction to prove that if there are n people in a room, and each shakes hands with
n(n − 1)
each other person, then there are a total of
handshakes.
2
a.
Base case: what choice for n is the base case here? What is the smallest number of
people you can have in a room? How many handshakes occur with that number of people?
It is reasonable to start with a base case of one person in the room, in which case there are no handshakes. It is also reasonable to start with two people, the smallest number of people needed to have a handshake. It is even reasonable to start with zero people—where of course there are no handshakes! b.
Induction step: what does the induction step ask us to prove here? It asks us to
substitute the particular number k – 1 for n, and assuming we have the right formula in that case,
to demonstrate we also have the right formula in case k people are in the room. So assume that
(k-­‐1)(k-­‐2)/2 there are __________ handshakes in a room with k – 1 people. Now when one more person
walks into the room, so that there are k people, the original k – 1 people used those handshakes to
all shake hands with each other already, so we only need the new handshakes that involve this
k – 1 new person. With how many people must the new person shake hands? __________.
Combine
these new handshakes with the handshakes that have already taken place, and simplify:
(k – 1)(k – 2)/2 + (k – 1) = (k – 1)[(k – 2) / 2 + 1] = (k – 1)k / 2 (k − 1)(k − 2)
2
So, assuming a room with k – 1 people takes __________ handshakes, we learn that a room with
k(k − 1)
k people requires __________ handshakes.
2
Since the formula worked in your base case, and it worked for k whenever it worked for k – 1,
then by the principle of mathematical induction, it works no matter how many people are in the
room.
Note: some people may have n = 1 as their base—no handshakes. Some might have had n = 0 as
their base—no people, no handshakes! And it is also reasonable to start with n = 2, because that
is the lowest number that actually has any handshakes. Whichever you used in the base is OK!
Induction 3.2
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MI 4
Mathematical Induction
Name ________________
n(n − 3)
diagonals. (A diagonal will mean
2
a line segment between two vertices that are not adjacent to each other, because that would be a
side of the polygon!)
2.
Prove that a convex polygon with n sides her
a.
Base case: What is the smallest number of sides a polygon can have? How many
diagonals does this polygon have? Does it fit the formula?
The smallest number of sides a polygon can have is three. A triangle doesn’t have any diagonals, and if n = 3, then n(n – 3)/2 = 3⋅0/2 = 0, so the formula does work in this case. b.
Induction step: How could we go from a k – 1-sided polygon to a k-sided one? We
could add a side to the polygon by doing this (the dashed lines represent diagonals):
A
A
B
B
X
All of the diagonals of the old polygon are also diagonals of the new one. How many additional
diagonals are there that start from X? (Remember, the new polygon has k sides!) Are there any
other diagonals we need to include?
The new vertex X must be connected to all the previous vertices except A and B, of which there are k – 3. But now A and B are not adjacent any more, so we have one more diagonal connecting them. So we have a total of k – 2 new diagonals. (k – 1)(k – 1 – 3)/2 k – 2 Now if we assume the old polygon had __________ diagonals, then we add __________
new
diagonals to this to get a total of:
(k − 1)(k − 4)
(k − 1)(k − 4) + 2k − 4 k 2 − 5k + 4 + 2k − 4 k 2 − 3k k(k − 3)
+ (k − 2) =
=
=
=
2
2
2
2
2
Since the formula worked in the base case, and it worked for k whenever it worked for k – 1, by
the principle of induction the formula works for any polygon.
Induction 3.3
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MI 4
Mathematical Induction
Name ________________
3.
Sneaky math trick! Explain why, if you knew the formula for the number of handshakes
from the first problem, that you don’t actually have to do the second proof (or vice versa—if you
knew the number of diagonals, you could easily figure out the number of handshakes).
Mathematicians often exploit tricks like this to save work—if you can find a similarity between
two problems you don’t have to solve both!
Put the people at the vertices of the n-­‐gon. Then all the handshakes that are not with someone to their immediate left or right would be along diagonals of the polygon. So we eliminate those n handshakes (the n edges of the polygon) to obtain a total of: handshakes – n = n(n – 1)/2 – n = n⋅[(n – 1)/2 – 1] = n(n – 3)/2 diagonals. 4.
Prove that 1 + 2 + 4 +  + 2n = 2n+1 – 1 for all positive integers n. Note: any time you
see “” in mathematics, it’s a good bet that induction is lurking!
a.
Base case: What is the base case here? What should we plug in for n? Check that the
formula works out for this choice of n.
Since we’re being asked to prove something for all positive integers, we should start with the smallest one, n = 1. In this case, the left-­‐hand side is just 1 + 2, while the right-­‐hand side is 21+1 – 1 = 4 – 1 = 3. b.
Induction step: What do you do to 1 + 2 + 4 +  + 2k – 1 to get to 1 + 2 + 4 +  + 2k?
Do this to the formula and simplify!
Starting with 1 + 2 +  + 2k – 1, to get to 1 + 2 +  + 2k we are simply adding 2k. So we take 2k – 1 + 1 – 1 = 2k – 1 and add 2k, giving 2⋅2k – 1 = 2k + 1 – 1. So if the formula works when we substitute k – 1 for n it also is true when we substitute k for n. Induction 3.4
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MI 4
Mathematical Induction
Name ________________
5.
Prove that the interior angles of an n-sided polygon (whether or not it is convex!) add up
to (n – 2)⋅180°. (You may use, without having to prove it, the fact that a triangle’s angles add up
to 180°.)
a.
Base case: what is the base case here?
The base case is a triangle—we start with n = 3 because there aren’t any polygons with fewer sides. Fortunately, (3 – 2)⋅180° = 180°, which we know is the correct total angle for the triangle. b.
Induction step: How can you go from a k – 1-sided polygon to a k-sided one? For a hint,
look back at problem 2.
Look back at how a k – 1-­‐sided polygon became a k-­‐sided polygon in problem 2. We added new vertex X and sides AX and BX where side AB used to be. Let’s say the old polygon had total interior angles adding to T. All of those angles are in the total for the new polygon except A and B. In the new polygon, A has been replaced by (A + m∠BAX), and B is replaced in the sum by (B + m∠ABX). Finally, we also add the new interior angle m∠AXB. In other words, we have the old sum plus three new parts, and these three new parts just happen to be the interior angles of triangle ΔAXB, which total 180°. So the new total is T + 180°. If the old total was (k – 1 – 2)⋅180°, the new total is this plus 180°, or (k – 2)⋅180°. So if the formula works for k – 1 it also works for k, and induction has done its job. c.
What can you prove about the sum of the exterior angles of a polygon?
The exterior angles of any polygon always add to 360°. In the base case of triangle ΔABC, the exterior angles are (180° – m∠A), (180° – m∠B), and (180° – m∠C), so the total is 540° – (m∠A + m∠B + m∠C) = 540° – 180°, which is 360° as advertised. In the induction, similar to above, the exterior angle at A is reduced by m∠BAX, the exterior angle at B is reduced by m∠ABX, and there is a new exterior angle at X whose measure is 180° – m∠AXB. So the total exterior angle changes by –m∠BAX – m∠ABX + 180° – m∠AXB. But this total change is zero because those three angles are the interior angles of ΔAXB so they cancel out the 180°. Thus, the total always changes by zero—so always stays the same 360°. Induction 3.5
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MI 4
6.
Mathematical Induction
Name ________________
An L-triomino is three squares glued together along their edges in a triangle fashion, like
this:
. By tiling a figure with L-triominoes, it is meant that you place copies of the
triomino, with no overlaps, to exactly cover up the figure. For instance, you cannot tile a 4 × 4
chessboard with triominoes because there are 16 squares on the chessboard which is not a
multiple of three, the number of squares in the triomino. But if we remove one square from the
chessboard, we can tile it. For example,
. If the square that had been removed
were in a different place then by rotating the triomino at the bottom right, and/or rotating the
whole figure, we could still tile the resulting chessboard with one square removed.
Using induction, prove that this is true for any 2n × 2n chessboard—if you remove any
square the resulting deleted chessboard can be tiled by L-triominoes.
a.
Base case: write down what you have to show in the base case. The proof is pretty easy
here!
In the base case we have a 21 × 21 chessboard which is missing a square. But a 2 × 2 chessboard missing one square is an L-­‐triomino! b.
Induction step: How can you get from a 2k-1 × 2k-1 chessboard to the 2k × 2k? Hint: cut
the 2k × 2k chessboard into four pieces—how big are they? The missing square can only be in
one of those pieces. Can you arrange to place one or more triomonoes on the rest of the pieces
so that you can apply the k – 1 solution to the rest?
As indicated in the hint, cut the 2k × 2k chessboard into quadrants. One of those quadrants is a 2k-­‐1 × 2k-­‐1 chessboard missing a single square, so apply the “Dr. Fogg technique”—we can assume we know how to tile this by the induction hypothesis. For the other three pieces, put a single triomino “in the middle” as shown. Then they, too, will be 2k-­‐1 × 2k-­‐1 chessboards missing a single square and can be tiled. Done! Induction 3.6
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