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Review and examples for writing proofs in math.
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Questions and Answers about proof by induction
September 16, 2005
I will only hand out the first four pages of these questions and answers in class.Print out the rest at home; you can find it at:
http://www.math.lsu.edu/ verrill/teaching/discrete2020/
Contents
1 Questions about proof in general 1
2 Questions about proof by induction 2
3 Questions about Assumptions 6
4 Induction and Sums 7
5 Where do the formulas for sums come from? 10
6 Induction problems involving exponents 11
7 Strong induction and double induction 13
8 Other examples of proof by induction 14
9 I don’t understand 17
10 Comments on group work 17
1 Questions about proof in general
1. Q: Can you give us a one or two class crash course on “everything we need to know about proofs”?
A: This whole course should be giving you lots of things you need to know about proofs. We are startingoff with some strategies for proving results; the first is proof by induction. We will cover more strategieslater. “Everything you need to know” is too much for one or two classes.
A few points about theorems and proofs:
The statement of a theorem generally has hypothesis and conclusion. You need to assume the hypoth-esis, and show how these lead to the conclusion.
A proof is a sequence of true statements, one of which follows from the next via a logical argument. Astatement is a sentence which can be either true or false, e.g., “The sky is orange”, “I like turnips”, and“47 > 90” are all statements (though they might not be true, or provable). We also call sentences like“x is a perfect square” and “x > y” statements, though whether or not they are true depends on what xand y actually are. But “is 5 bigger than 3?”, “x + y”, “the yellow of the sunset”, “Baton Rouge” and“Hello” are not statements.
Logical arguments usually follow by using some kind of algebra, or geometry, or some other tools in someparticular area of mathematics.
The hypothesis is a statement H, maybe H(x), and the conclusion is also a statement. A typical theoremmight have the form:
Theorem: If H(x) holds, then C(x) holds
1
This theorem can also be written as “H(x) ⇒ C(x)”. For an example, we might have
H(x): “x is a real number and x > 1”C(x): “x2 > x.
In this case, the theorem is true, which means that whenever H(x) is true C(x) is also true.
Some people find it useful to write truth tables. Here’s one with H(x) and C(x) as above.
H(x) C(x) H(x) ⇒ C(x) examples with these truth valuesT T T x = 7T F F no examples, since for this case H(x) ⇒ C(x) is always trueF F T x = 0F F T x = −5
Here T means “true”, and F means “false”.
A truth table can help in understanding the logic of a proof, but a truth table is not a proof.
Sometimes it might be necessary to use more complicated logical constructions, and we’ll talk a bit aboutthis when we cover another proof strategy, called “proof by contradiction”.
We will talk more about proofs as the course progresses.
Meanwhile, there are some useful web pages about proofs. Have a look at:
http://www.math.ucsd.edu/∼ebender/proofs.html
http://pass.maths.org.uk/issue7/features/proof1/index-gifd.html
http://en.wikipedia.org/wiki/Mathematical induction
http://www.math.csusb.edu/notes/logic/lognot/node1.html
2. Q: Is there a certain set of words to be used when writing proofs? Like a standard example you couldshow us, where we could just go back and fill in the blanks of a particular problem?
A: There are many different kinds of proofs, so it would be possible to give a “fill in the blanks” for somekinds, but there are too many to give such a form for all proofs. However, for proof by induction for sums,we have given a “fill in the blanks” method in class.
3. Q: Do you always use words in proving or is there a shorter way to prove it by only numbers?
A: Usually you have to use words and numbers. Quite often the words can be minimal, since you canassume the reader knows what proof by induction is, so you might just write, “Proof by induction”, “basecase:” “induction step”, and “QED”.
Writing clearly is very important, especially for anyone who wants to become a teacher.
4. Q: After doing base cases and finding patterns, I don’t know how to use induction to prove a case. I don’tknow how to put answers in mathematical terms.
A: I think this is a question of practice and experience. The more examples you read through, and themore exercises you try, the easier this will become. There are quite a lot of examples in this hand out.
2 Questions about proof by induction
5. Q: What exactly is proof by induction? We are learning how to solve, but what is the formal definition?
A: Let P (n) be a statement involving a positive whole number. (See Question 1 for a definition of astatement; see the previous hand out on induction for three examples.)
A proof by induction of the statement “P (n) for all n ≥ n0” is a proof that
(a) demonstrates that P (n0) is true (this is called the “base case”)
(b) Shows how to prove P (n) ⇒ P (n + 1) (this is called the “induction step”)
2
This is a proof of the result for all n. E.g., if you take any given value of n, e.g., n = 17, then, supposingthat n0 = 3.
By substituting n = 3 into (b), we have P (3) ⇒ P (4)
By substituting n = 4 into (b), we have P (4) ⇒ P (5)
By substituting n = 4 into (b), we have P (5) ⇒ P (6)
And so on, so we get
P (3) ⇒ P (4) ⇒ P (5) ⇒ P (6) ⇒ P (7) ⇒ P (8) ⇒ P (9) ⇒ P (10) ⇒
P (11) ⇒ P (12) ⇒ P (13) ⇒ P (14) ⇒ P (15) ⇒ P (16) ⇒ P (17)
So, because we showed that P (3) is true (step (a), the base case), this means that P (17) is true.
6. Q: How can you prove something by induction?
Do you mean in general, or in particular?
See Question 5 for a formal description.
The idea of proof by induction can be written even more formally, as follows, where P (n) is a statementinvolving an integer n.
(
“P (n0)” ∧ “P (n) ⇒ P (n + 1)∀n ≥ n0”)
⇒ “P (n)∀n ≥ n0”
Here, ∧ means “and”, and ∀ means “for all”, and ⇒ means “implies”.
If this doesn’t mean anything to you, just use the definition at the the beginning of this answer.
There are examples of proof by induction on the next few pages.
Note that “proof by induction” is not the same as “inductive reasoning”.
“Inductive reasoning”, also called the “inductive method” or the “Scientific Method” takes many manyexamples, and uses these to make a guess about what a result might be.
Inductive reasoning can be a very useful method of working out what you want to prove. But even if youhad many examples where you had checked that 1 + 2 + · · · + n = n(n + 1)/2, just showing me all theexamples would not convince me that the result was always true.
E.g., defined two functions, f and g on positive integers, by
f(n) = n
g(n) = “the remainder when you divide n by 8”
You can check that g(n) = f(n) for n = 1, 2, 3, 4, 5, 6 and 7, but this does not imply that g(8) = f(8).
7. Q: Can you give us a formal/Layman definition of the word “induction”?
A: Do you want the formal definition, or the layman’s one?
For formal definitions, see Questions 5 and 6 above.
There are many analogies used to describe what a proof by induction means. Here are some (the first isvery standard; I made up the others, hopefully they help too).
Domino analogy
If I have an infinite line of dominoes, if I knock over the first (base case), and if the dominoes are closeenough so that when I knock down domino n it knocks over domino n+1 (induction step), then I’ll knockdown all the dominoes.
3
dominoes stretching
to infinity
same distance between
each pair of dominoes
Shorter than a domino
If I knock down the first one, everything falls
It it necessary for
the first one to fall
(base case)
(induction step)It is necessary the dominoes are close enough together
Steps analogy
If I am trying to climb some stairs, as long as I can get to the first step (base case), and if I know that Ican always get from step n to step n + 1, then I know I can climb as far as I want.
Ladder analogy
If I can reach the first rung of a ladder, and if the rungs are close enough so that once I’m at rung n it’spossible to reach rung n + 1, then I know it’s possible to climb the whole ladder.
8. Q: If you know how to prove something starting with the nth case, do you still need to start with thebase case?
A: Yes, since if the base case is false, even if P (n) ⇒ P (n + 1), you could still have everything false.
E.g., here’s an example.
I’m going to define a new kind of triangle, call it the 1 − 2-Pascal triangle (this is a made up name); thisis like Pascal’s triangle, but it starts off with a row with 1 and 2, instead of starting at 1. The rules forgetting to the next row are the same. So, I define
b1,0 = 1
b1,1 = 2
b1,i = 0 for i 6= 0, 1
bn,r = bn−1,r−1 + bn−1,r for n > 1
Here are the first few rows:1 2
1 3 21 4 5 2
1 5 9 7 21 6 14 16 9 2
It would be a good project to investigate some patterns in this table—what can you see?
Anyway, one statement which is not true isP(n): The sum of the elements in row n of the 1-2-Pascal triangle is 2n
For some notation, let
f(n) = bn,0 + bn,1 + · · · + bn,n =
n∑
i=0
bn,i
So P (n) says f(n) = 2n.
You can check that from the definition, it follows that the sum of row n + 1 is twice as much as the sumof the elements in row n. (The details are one of the project questions.) So we have
P (n) ⇒ P (n + 1).
So the induction step is true. But f(1) = 3, so the base case is false. So we can’t prove this result byinduction.
4
9. Q: Is it always necessary to have g(f(n + 1)) in the right side of the inductive step, even if the problemis not a sum?
A: I think you mean g(n + 1) here.
No, this is not necessary. This method (described in class) is only for proving results where P (n) sayssomething like
f(1) + f(2) + · · · + f(n) = g(n)
10. Q: Why is proof by induction better for solving some problems rather than using a different method?
A: This depends on the problem.
Some problems can be solved in many different ways, and induction, even if it can be used, is not alwaysthe best way.
E.g., for proving that 1 + 2 + · · ·n = n(n+1)2
We can use Gaussian pairing (see page 75 of the text book; also done in class).
My personal taste is that if possible, very often I’d prefer a proof that does not use induction. E.g., Ithink the Gaussian pairing is much nicer than the induction proof of the same result. Usually inductiondoes not tell us anything about how to get the result, where it comes from, or “why” it’s true. But thisalso depends on the situation. For some problems induction might be the best way.
Question 37 gives three different proofs of one result. Some people collect different proofs of the sameresult. E.g., look at:
http://www.cut-the-knot.org/pythagoras/index.shtml
for 9 different proofs of the Pythagorean theorem.
Which one you like best is a question of personal taste, like which kind of music do you prefer.
11. Q: Is it possible to prove a false statement true by induction? If so, what is the most likely flaw?
A: If induction is applied correctly, then you can’t prove something false is true.
However, if you missed out a step, you might have this happen. Probably the most likely flaw would be tonot do the base case, like in example 8. But sometimes you might think you’ve proved P (k) ⇒ P (k + 1),but there is some flaw in the argument.
Here is an example (this is well known; I don’t know who first thought of it):
Theorem
If you have set of horses, then they are all the same color.Proof
The statement we want to prove for all n is:P (n): in a set of n horses, all horses have the same color
Base case: P (1) is true, since there is only one horse.Inductive step: Suppose P (k) is true.Suppose we have a set of k + 1 horses. Remove one horse. By induction, the remaining k horses areall the same color. Now put that horse back, and remove a different horse; the remaining horses areagain all the same color, so the horse you added back has the same color as all the others. Henceall horses in the set of size k + 1 have the same color. We’ve shown that P (k) ⇒ P (k + 1).Thus by induction all horses in any set of horses are the same color.
What do you think is wrong? See
http://www.personal.kent.edu/∼rmuhamma/Algorithms/MyAlgorithms/DeductInduct.htm
for an explanation.
12. Q: Do we always have to show that the previous case is true when writing our proof?
A: No, you always assume the previous case, i.e., if the statement is P (n), then you assume that P (n) istrue, and show this would imply P (n + 1). But the only case you need to prove is the base case.
E.g., Suppose A1 is a certain animal, and if n > 1, then An+1 is the mother of An.
Suppose P (n) is the statement “An is a rabbit”. Then (assuming no genetic engineering or evolution, andthat the world has existed for an infinite length of time), we have
P (n) ⇒ P (n + 1).
5
So, if P (1) is true, then all P (n) are true. I.e., if A1 is a rabbit, all the ancestors of A1 are also rabbits.But if the base case is not true, then even though we still have P (n) ⇒ P (n + 1), all P (n) are false.
The statement “If my dog was a rabbit, her mother would have been a rabbit”, is true even though mydog is not actually a rabbit.
13. Q: The pattern of perfect squares by induction.
A: This is not written as a question. Do you mean you want to know what happens when you add up theperfect squares? Or do you want a sum that adds up to perfect squares?
14. How can you prove a power series by induction?
A: I don’t understand this question.
You can only prove statements by induction. A statement is a sentence that can be true of false.
“A power series” is not a statement, it is a thing; it can’t be true or false. Maybe you want to prove someresult about a power series? See Question 1 for a definition of a statement. Give me a question, and I’lltry and give an answer (or at least try). Give me a statement, and I’ll try and work out if it is true orfalse.
15. Q: Is there a case where the base case f(1) is true, and f(n + 1) exists, but f(n) doesn’t?
A: I’m not 100% sure if you’ve written this down properly – Did you mean“for a statement P (n), can P (1) and P (n) be true, but P (n + 1) false?”
If you meant this, then yes, there are lots of cases where the base case is true, but the other cases arefalse.
E.g., let P (n) be the statementP (n) : 2n = n2
Suppose we thought the base case was n = 2. Then P (2) is true. But for n > 1 the result is false. P (3)is false.
3 Questions about Assumptions
16. Q: Do you ever assume things when doing induction?
A: See question 18 below.
17. Q: What can we assume the person knows?
A: See question 18 below.
18. Q. What is safe to assume (axioms, theorems, etc.) when using induction?
A. Axioms are something that are always true and can be assumed. E.g., in Euclidean geometry, it is anaxiom that “for any two points in the plane, there is a straight line between these two points”. You don’tneed to prove this.
But you should clearly state what axioms are used.
If you use a theorem, you should clearly state what you are assuming. You need to state what thehypothesis of the theorem is, what the result is, and make sure the theorem applies to the problem youare working on. In most cases, for any theorem you use, you should also be able to prove it, so ideallyyou should also give the proof of the theorem, unless the question says you may assume a certain result.
E.g., in most cases, you can assume anam = an+m, without even mentioning that you are using a theorem,since this is a fairly basic result, which “everyone knows”. But if a question asks you to prove this, thenyou should give the details. And in any case, you should know how to prove this result, if asked. (SeeQuestion 30 for a proof).
19. Q: Is the key to proof by induction using the characteristics and definitions of a function?
A: This depends on what you are trying to prove, and whether there is a function or not. But yes, ingeneral, whatever objects your proof is about, using the definition of that object is often the key (whetherthe object is a polygon with n-sides, or a function, or something else).
Definitions are important in all proofs, not just proof by induction. The definitions are even more basicthan your axioms. They are your starting point on which you build.
6
4 Induction and Sums
20. Q: How can we prove that
1 + 3 =
2∑
i=1
(2i − 1)
by induction?
A: This is not really an induction problem, since the left hand of this statement just means
(2 × 1 − 1) + (2 × 2 − 1)
and we can see this is the same as 1 + 3 just by algebra.
The induction version of this problem would be to show that
n∑
i=1
(2i − 1) = n2
This is saying that the sum of the first n odd numbers is n2
I’ll leave this for an exercise, since this is one of the projects. To do this by induction we can just workout what f(i) and g(n) are, and plug into the method given in class.
But here is an alternative “picture proof”:
1 1
5
11
3 3
5
7
3
The areas of each region is written in it
To get to the next square in each case, we
just add the next odd number. E.g., to get
to 52, we’d have to add a region of
area 9 around the edge of the box.
To make this into a proof still needs a bit more work, mainly that
(n + 1)2 − n2 = 2n + 1
This shows that the difference between any two successive squares is an odd number. I’ll leave it as anexercise to fill in the details.
21. Q: Find a formula for 11 + 22 + · · · + nn and prove by induction.
A: This looks hard!
The first few terms are:n nth term of sum sum1 1 12 4 53 27 324 256 2885 3125 34136 46656 500697 823543 8736128 16777216 176508289 387420489 40507131710 10000000000 10405071317
I have no idea what or whether there is a simple formula for this. Maybe there isn’t!
I looked up in sloane’s integer sequences data base, at
7
http://www.research.att.com/∼njas/sequences/index.html
which does not give any formula, (though it does say something about the ratio of the successive sums).
You could investigate this sequence further as a project.
22. Is the nth term given?
I don’t know what this question means. The nth term of what? If you’re talking about proving a formulalike
1 + 22 + · · · + m2 =m(m + 1)(2m + 1)
6
then the nth term on the left side would be n2, though this is not explicitly stated; you’re supposed tolook at the pattern, and work that out.
23. What is the sum of all negative numbers between 0 and 1000? Prove by induction.
A: This is a wonderful trick question – I might use this on students in the future.
There are no negative numbers between 0 and 1000, and so the sum of them is 0!
However, I’m not sure this is what you meant?
Maybe you want to find the sum of the positive numbers between 0 and 1000?
The answer is:
1 + 2 + · · · 1000 =1000 × 1001
2= 500500
To prove this by induction, we’ll prove the more general result
P (n) : 1 + 2 + · · ·n =n × (n + 1)
2,
and then plug in n = 1000.
I’m not going to write out the proof by induction here, since we have done this in class.
You can find lots of examples of formulas for sums at:
http://mathworld.wolfram.com/PowerSum.html
try proving as many from there as you can.
24. Q: Prove by induction
1 + 2 + 3 + · · · +(n2 + 2)2
3=
n(n + 1)(2n + 4)
4
A: First we have to work out what the general term is.
It says that the last term is (n2+2)2
3 . Let’s call this f(n), and make a table of some values, to see what’sgoing on:
n (n2+2)2
3
1 (1+2)2
3 = 32
3 = 3
2 (22+2)2
3 = (4+2)2
3 = 62
3 = 36/3 = 18
3 (32+2)2
3 = (9+2)2
3 = 1213 = 40 1
3
4 (42+2)2
3 = (16+2)2
3 = 182
3 = 108
Well, this is a bit funny, since the person wrote that the first three terms were 1, 2, 3, which does notseem to be the case.
So, let’s just assume the question should have been:
3 + 18 +121
3+ · · · +
(n2 + 2)2
3=
n(n + 1)(2n + 4)
4
8
We better check this is the right result! What do we get when we add up the first n terms?
Let’s write
g(n) =n(n + 1)(2n + 4)
4=
n(n + 1)(n + 2)
2
Let’s try adding up the f(i) and comparing with the g(n):
n (n2+2)2
3 sum of first n terms g(n)
1 3 3 3
2 18 19 12
3 40 13 59 1
3 30
4 108 167 13 60
This doesn’t seem to work at all, except when n = 1! I wonder what the question was supposed to be?
I will prove a different result instead of the one in the question:
Theorem
For all positive integers n, we have
3 + 9 + 18 + 30 + 45 + 63 + 84 + · · · +3
2n(n + 1) =
1
2n(n + 1)(n + 2)
Proof
Note that the sum on the left side can also be written as s(n), where
s(n) =n∑
i=1
3
2i(i + 1).
Base case: We have s(1) = 321 × 2 = 3, and g(1) = 1
21(1 + 1)(1 + 2) = 3, so the statement is true for n = 1.Induction step: Assume the result for n = k. So we assume that
s(k) = 3 + 9 + 18 + 30 + 45 + 63 + 84 + · · · +3
2k(k + 1) =
1
2k(k + 1)(k + 2) = g(k). (1)
Now we have
s(k + 1) = 3 + 9 + 18 + 30 + 45 + 63 + 84 + · · · +3
2k(k + 1) +
3
2(k + 1)(k + 2)
By substitution from (1), we have
s(k + 1) = s(k) +3
2(k + 1)(k + 2) = g(k) +
3
2(k + 1)(k + 2). (2)
Some algebra gives:
g(k) +3
2(k + 1)(k + 2) =
1
2k(k + 1)(k + 2) +
3
2(k + 1)(k + 2)
=(k + 1)(k + 2)
2k +
(k + 1)(k + 2)
23 =
1
2(k + 1)(k + 2)
(
k + 3)
(3)
We also have
g(k + 1) =1
2(k + 1)(k + 1 + 1)(k + 1 + 2) =
1
2(k + 1)(k + 2)(k + 3) (4)
Putting (2) (3), and (4) together gives us:
s(k + 1) = g(k + 1)
9
25. Q: More examples would help me understand induction.
A: Are there enough above? Let me know if you still need more! Would you prefer these on a handout,on the board, both?
Have you looked at the examples in the text book, and these on web sites?
(a) http://bigcheese.math.sc.edu/∼sumner/numbertheory/induction/Induction.html
(b) http://www.cc.gatech.edu/people/home/idris/AlgorithmsProject/ProofMethods/Induction/UnderstandingInduction.html
5 Where do the formulas for sums come from?
26. Q: Where do you get formulas to prove?
A: This often comes from trial and error and experimentation. Induction will not give you the formula,that has to come from somewhere else. See the next example.
27. Q: Can you use induction to find the formula for sums, or do you only prove a known formula usinginduction?
A: Generally, induction does not give you what the formula is. You have to try and guess that. The guesswill only be a guess until you actually prove it, and induction is one way to prove your guess was correct.
28. Q: I don’t understand how to prove by induction when only one formula is given. It makes sense when
1 + x2 + x3 + · · · + n2 = formula
but not when it says solve by induction, and only one formula is given.
A: This is a good point, though I’m not sure what example in particular you are talking about... Maybeyou meant
1 + x2 + x3 + · · · + xn = formula?
Anyway, you’re right, induction will never tell you what the formula is. For this, you have to try severalcases, and collect some data. Then in the data, you should hopefully see a pattern.
Then if your pattern is correct, you should be able to prove the result by induction.
E.g., We have
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
You probably see a pattern here – the sums are always squares.
So you might guess that1 + 3 + · · · + (2n − 1) = n2
Then you can prove this by induction.
Finding the formula in the first place is good practice, and helps you understand better what the formulameans.
29. Q: What is the point of induction if you are already given a proved formula.
A: This is a good question! (compare with Question 28 where someone complained about not being givena proved formula).
There are lots of puzzles and problems for which noone yet knows the solutions. For these, if you couldguess a solution, you would then have to prove it was true, and induction might be one way. Perhapssomeone in this class will solve such a problem in the future.
Here is an example of a simple problem that no-one yet knows a formula for:
10
How many ways are there to fold a map? That is, given an n × m grid, with crease lines dividing thepaper into 1 × 1 squares, how many ways are there to fold the map so that the resulting folded map hasa 1 × 1 area?
Here’s a fun special case of this:
Take some letter paper, and make creases as in the following picture. Can you fold along these in such away that when folded, each numbered sheet lies directly on top of the sheet with the previous number?
1
34
2
65
78
For lots and lots of interesting sequences, take a look at Sloane’s integer sequences data base.
http://www.research.att.com/∼njas/sequences/index.html
You can find what’s known about map folding there.
6 Induction problems involving exponents
30. Q: What is another way to write (1)n+1? 1n(1) = 1n
A: I’m not 100% sure what you mean. The answer, is of course 1. To prove this by induction, you havethe correct induction step, that 1n+1 = 1 × 1n. If 1n = 1, then this implies that 1n+1 = 1. You just haveto check the base case, 11 = 1, which is easy.
Here is a similar problem:
Definition: For a real number a and a positive integer n, we define
a1 = a, (5)
ak+1 = ak × a for all positive integers k (6)
Note that this is called a recursive definition, which means that one case depends on the previous case,i.e., the definition of an+1 depends on the definition of an.
Theorem:
For positive integers n and m, we have
am+n = aman
Proof:
Do this by induction on n, i.e., we fix m, and prove for any fixed m, the result holds for all n.I.e., for a fixed m, the statement P (n) is:
P (n) : am+n = aman
Base case:am+1 = am × a
This is true by the definition, part (6), plugging in k = m.P (n) says:
am+n = am × an
We assume this is true. We want to show that this implies P (n + 1), which says:
am+n+1 = am × an+1
We have
am+n+1 = am+n × a by plugging k = m + n into (6)
= am × an × a by applying P (n)
= am × an+1by applying (6) with k = n
This shows that P (n) ⇒ P (n + 1)So by induction we conclude the result holds for all m,n.
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31. Q: How can you make 1m+1 disappear? Take natural log.
A: I’m not 100% sure what you mean. Perhaps this is a fragment of a bigger statement or proof?
32. Q: Prove using induction that 5n is divisible by 5 for all n ∈ Z
A: This is not true for all n ∈ Z, e.g., if n = −1, we get 5−1 = 15 , and usually we only talk about divisibility
for integers.
We’ll assume that n ≥ 1.
You don’t really need induction to prove this result, since
5n = 5 × 5n−1.
For a number x to be divisible by 5 means that there is some integer y such that
x = 5y
So, here we just take y = 5n−1. QED.
Do you really want to force me to prove this by induction?
Here’s another way to prove it:
Check the result for the base case n = 1. This gives 51 = 5, which is divisible by 5.
Now assume that 5n is divisible by 5. So we assume we can write 5n = 5x for some x.
Now5n+1 = 5 × 5n = 5 × (5x)
Since (5x) is divisible by 5, any integer multiple of 5x is also divisible by 5x. I.e., k5x is divisible by 5 forall integers k. In particular, this is true with k = 5. So we have that 5n+1 = 5(5x) is divisible by 5.
I hope this is clear - sometimes proving something too easy by induction might make things unnecessarilycomplicated.
33. Q: How do you show that 7n − 1 is divisible by 6 for all positive integers n?
A: Since this is the same as a homework question, I’ll give a similar, but slightly different example.
Theorem For all positive integers m, we have that if m is odd, then 3m + 4m is divisible by 7.Proof We prove this by induction. If m is odd, then for some n, we have m = 2n + 1, so the Thestatement we want to prove for all n is
P (n) : n ∈ N ⇒ 32n+1 + 42n+1 is divisible by 7
Base case: P (0) says that 3 + 4 is divisble by 7, which is easy to see is true.Induction step.Suppose for some value of n, say n = k, we know that P (k) is true. This means that there is somenumber qk (it depends on k what this is) such that
32k+1 + 42k+1 = 7qk
Now for n = k + 1, the statement P (k + 1) will be true if we can show that 7 divides 32(k+1)+1 +42(k+1)+1, so we will try and write this in terms of 32k+1 + 42k+1. We have
32(k+1)+1 + 42(k+1)+1 = 32k+3 + 42k+3 = 9 × 32k+1 + 16 × 42k+1
= 9 × 32k+1 + (9 + 7) × 42k+1
= 9 × 32k+1 + 9 × 42k+1 + 7 × 42k+1
= 9(×32k+1 + ×42k+1) + 7 × 42k+1
= 9 × 7 × qk + 7 × 42k+1
= 7 ×(
9qk + 42k+1)
So, assuming P (k), we’ve shown that P (k + 1) is true.By induction P (n) is true for all n ∈ N
Note, in this question, N means the set of non-negative whole numbers.
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7 Strong induction and double induction
34. Q: What is “double induction”? Could you give an example of double induction?
A: I use this term to describe proofs by induction where there are two variables, and you need an inductionthat works for both. The statement has the form P (n,m). This is a common kind of problem. This use ofthe term “double induction” is also used in a good book on proofs, “How to prove it” by Daniel Velleman.
Some examples with two variables are Question 35 and Question 30 below.
35. Q: When can you use strong induction?
A: If you want to prove a statement P (n), sometimes it’s not enough to assume just P (n − 1).
Here is a problem where you can apply this method:
3 breaks 4 × 4 = 16 breaks
What’s the least number of
breaks needed to break a
n × m bar of chocolate
into individual 1 × 1 pieces?
E.g., for 4 × 5, we cen do it in
19 steps as below.
Can you do it in any less? How many?
Example:
Theorem
Given an n × m rectangle with n,m positive integers, and lines dividing the rectangle into 1 × 1squares, then however the rectangle is cut into individual squares, it requires exactly nm − 1 cuts.Proof
Base case: for m = n = 1 no cuts are needed, and mn − 1 = 0. You can’t have less than 0 cuts, sothis is the least number necessary. So the statement of the theorem is true when n = m = 1.Induction step: Given an n×m rectangle. Assume the result is true for k1 ×k2 rectangles wheneverk1k2 < nm.How can we break the n × m bar? There must always be a first break, which splits the rectangleinto two parts.We have two possible cases, depending on which direction the first break is:
n
r s
r
s
n
m m
Each case is the same, so I’ll just discuss the first. We now have made one cut, and have rectanglesof sizes n × r and n × s, with r + s = m.Now by the inductive hypothesis, we need exactly nr − 1 and ns − 1 cuts to break these smallerrectangles up.So, the total number of breaks must be:
1 + (nr − 1) + (ns − 1) = nr + ns − 1 = n(r + s) − 1 = nm − 1
So, by assuming the result for smaller rectangles, we’ve also proved it for the n × m rectangle.So by induction the result is true for all integers sided rectangles.
Notice that here we had to use strong induction, since the first break might be anywhere, we don’t know,so we have to assume the result for all smaller rectangles, not just those of size n × (m − 1) for example.
There is a java applet to describe this problem, at the bottom of the page at:
http://www.cut-the-knot.com/ctk/memes.shtml
click on lines to make the rectangle split along that line. This site gives more explanation at:
http://www.cut-the-knot.com/ctk/chocolate.shtml
Another example of strong induction is given in the last proof in Question 37, though in that case proofby usual induction is also sufficient, as the first example shows.
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8 Other examples of proof by induction
36. Q: Prove by induction that the sequence of numbers in the first diagonal of Pascal’s triangle is 1, 2, 3 · · · ,I.e., an,1 = n.
37. Q. How can you prove a triangle has 180◦ using induction? (If you can at all)
A. Make sure to be precise in your questions. I think you mean“how can you prove that the sum of the internal angles in a triangle is 180◦?”
This is not really a question to prove by induction, since there is no varying number n. This statementwould be proved by using methods of geometry. There are several different ways to prove this.
One method is based on the following picture:
α
β
γ
α
β
γ
γ
Add a line
parallel to one
side, through
opposite vertex
α
There are many web sites with more details, e.g.,:
http://mathworld.wolfram.com/Triangle.html
http://www.mathsrevision.net/gcse/pages.php?page=17
Also, you will probably see a complete proof of this result in a geometry course, such as MATH4005.
There might be a way to prove this by induction, but I can’t think of one, because I can’t easily see wherethe “n” would be.
To make this into an induction question, you need to make a statement about an infinite sequence ofthings. Here is one possibility:
Theorem: The sum of the internal angles in an n-gon add up to 180(n − 2)Proof:
The base case is n = 3, which is a result about triangles. This result is true for triangles by theargument given above. Note that to prove the base case you can’t use induction; this is part of theinduction, and might need various methods. Mostly the base case is fairly simple, but this is notalways true.How to go from n = k to n = k + 1:Assume the result is true for all k-gons. This is the inductive hypothesis.Give me any k-gonI can split a triangle off it:
10-gon split into
a 9-gon and a triangle
By induction, the result is true for the k − 1-gon, and we know it’s true for the triangle.By induction the k − 1 gon has internal angles summing to 180(k − 3). The angles of the trianglesum to 180So the internal angles of the k-gon sum to
180(k − 3) + 180 = 180(k − 2)
So the result follows by induction.
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We can use strong induction to prove the same result:
Theorem: The sum of the internal angles in an n-gon add up to 180(n − 2)Proof:
The base case is n = 3, which is a result about triangles. This result is true for triangles by theargument given above. Note that to prove the base case you don’t use induction; this is part of theinduction, and might need various methods. Mostly the base case is fairly simple, but this is notalways true.How can we prove the statement for n = k if we know it for smaller values of n?Assume the result is true for all r-gons for r < k This is the inductive hypothesis.Give me any k-gon. Break it into two smaller polygons, e.g.:
10-gon split into
a 5-gon and a 7-gon
It’s always possible to do this if the polygon has more than 3 sides.The smaller polygons have k1 sides and k2 sides.Since k1 < k and k2 < k, the result is true for these polygons.So they have internal angles adding up to 180(k1 − 2) and 180(k2 − 2).So, the polygon with k sides has sum of internal angles
180(k1 − 2) + 180(k2 − 2) = 180(k1 + k2 − 4)
Now notice thatk1 + k2 = k + 2,
which is because there is an extra edge on the two smaller polygons which was not an edge of theinitial polygon.So, by substitution, the sum of internal angles of the initial polygon is 180(k − 2).So, we’ve proved by induction that the result always holds.
Here’s a completely different proof, without using induction at all:
Theorem: The sum of the internal angles in an n-gon add up to 180(n − 2)Proof: You can divide the n-gon up into n − 2 triangles:
10-gon split into
8 triangles
Assume the result for triangles. Then the total sum of the internal angles is
sum of angles in a triangle× number of triangles = 180(n − 2).
38. Q: Can you prove the minimum number of steps to move n disks for the tower of Hanoi from one peg toanother?
A: Yes, but since these notes are getting rather long, I’m going to refer you to some web pages:
http://www.cl.cam.ac.uk/Teaching/1998/InroAlgs/notes98/node3.html
http://www.cut-the-knot.com/recurrence/hanoi.shtml
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9 I don’t understand
39. It needs more explaining, still unsure how it’s done.
40. I’m still very confused on the steps of induction.
I hope some of the above examples help. Please read through everything on these sheets carefully, andlet me know where you get stuck. Try and locate the step you don’t understand.
10 Comments on group work
The comments about group work:For
1. Groups
2. I like groups
3. I like working in groups because we put our heads together
against
1. More examples on board
2. Need more work on board
3. Too many groups
4. I think we’re doing too many groups
conclusion: Perhaps this means I should do more work on the board! I will give at least one class nextweek with no group work.
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