Download On mathematical induction

On mathematical induction
Stefano Ferri
29.02.2016
17 Phagun 547
Abstract
A short note on mathematical induction for the course MATE–1214 of Los Andes
University. Use it as you want but if you take it as it is, please, mention where it comes
from and who wrote it. Any comment, please contact [email protected]
1
Introduction
In the last few classes we have met certain sequences which are defined by a recursive
formula. For example we can define a sequence (xn )∞
n=0 of real numbers by the following
rules:
x0 = 2
xn+1 = 41 xn +
3
4
for n = 1, 2, 3, . . .
(1.1)
For this type of sequence in order to compute xn we need first to compute all of the
terms xk for k < n. For this reason in general it is rather difficult to apply what we know
about limits of functions of real variable to these sequences in order to compute their limit.
A natural thing to do could be to use the Monotone Convergence Theorem.
Theorem 1.1. If (xn )∞
n=0 is a sequence or real number which is either increasing and
bounded above or a decreasing and bouded below, then the sequence has a limit in R.
However, trying to prove directly that a sequence such as (1.1) is monotone and bounded
is also not easy.
Another natural thing to try is to compute some terms of the sequence and try to see
whether it is possible to find a formula for xn as an explicit functions of n. This is not
always possible and sometimes, even when it is possible, it may be utterly difficult to find
such a function. However, there are cases (like the one we are examining) in which this can
be done relatively easily.
We begins by calculating the value of the first few terms of the sequence as follows:
x1 = 41 x0 +
3
4
=
1
4
·2+
3
4
= 45 ;
x2 = 41 x1 +
3
4
=
1
4
·
5
4
3
4
=
x3 = 41 x2 +
3
4
=
1
4
·
17
16
+
3
4
=
65
64 ;
x4 = 41 x3 +
3
4
=
1
4
·
65
64
+
3
4
=
257
256 ;
+
17
16 ;
1
2
Induction
x5 = 41 x4 +
3
4
=
1
4
·
257
256
+
3
4
=
1025
1024 .
At this point probably you have already formulated the conjecture that a possible formula
22n +1
for the n’th term of the sequence (xn )∞
n=0 is perhaps xn = f (n) = 22n .
Of course, just checking the formula for a few values of n is not enough, so the next
thing we have to do is to look for a proof that xn = f (n) for all values of n.
In order to do this we can first check the equality xn = f (n) holds for n = 0. (Of course
it does, since we have formulated our conjecture exactly by looking at the first values of xn ,
0
but in case you are not sure you can repeat the computation: x0 = 2 2+1
= 2 = f (0).) Next,
0
we check that f (n) satisfy the recursion. Namely, that, for all n = 1, 2, 3, . . . , we have that
f (n + 1) = 14 f (n) + 43 . This is done quickly by observing that:
1
3
1 22n + 1 3
22n + 1 3
22n + 1 + 3 · 22n
22n+2 + 1
f (n) + = ·
=
=
+
+
=
= f (n + 1).
4
4
4
22n
4
22n+2
4
22n+2
22n+2
At this point we can conclude that the equality must hold for every value of n = 1, 2, 3, . . . .
We can conclude this by reasoning as follows. We know that x0 = f (0) and also that
x1 = 14 x0 + 43 and we also know that f (1) = 41 f (0) + 34 , hence we can conclude that
x1 = f (1). Then we start the same reasoning again starting from x1 . We have that
x1 = f (1), x2 = 14 x1 + 43 and f (2) = 14 f (1) + 43 , so x2 = f (2). Then again we can apply
the same reasoning again starting from x2 and so on repeating the same process infinitely
many times we can conclude that xn = f (n) for all n = 1, 2, 3, . . . .
2n
Of course, once we know that xn = 2 22n+1 it is very easy to compute the limit of the
sequence. Namely:
22n + 1
= 1.
lim xn = lim xn = lim
n→∞
n→∞
n→∞
22n
2
Introducing mathematical induction
The process we just used to prove that xn = f (n) in the previous section can be formalised
as follows.
Principle of Mathematical Induction.
Suppose that we have a sequence of propositions {Pn }∞
n=0 . (In the previous example the
2n
proposition Pn stated that xn = 2 22n+1 . In our course usually a proposition will be some
sort of simple mathematical statement which is either true or false. If the idea of having
infinitely many propositions confuse you, just think to a proposition which involves functions
of integer numbers, for example in the previous example, you can think that you want to
prove the infinitely many propositions of the list {x0 = f (0), x1 = f (1), x2 = f (2), . . . } or
that you want to prove that xn = f (n) for all n = 1, 2, 3, . . . .)
Suppose that the following two conditions are satisfied:
(1.) P0 is true;
(2.) For all integer k, if we assume that Pk is true this implies that Pk+1 is also true.
Then all the propositions in the list are true. (Or, if you prefer the formulation using only
one proposition which involves functions of n, you conclude that this proposition is true for
every value of n = 1, 2, 3, . . . )
3
Stefano Ferri
Example 2.1. Let us try to prove again that if
x0 = 2
xn+1 = 41 xn + 34 for n = 1, 2, 3, . . .
then we have that
22n + 1
22n
using the formalism of mathematical induction.
xn =
Proof. First we write clearly what the propositions Pn are.
Pn : “an =
22n + 1
”.
22n
0
0
2 +1
We then prove that P0 is true. The proposition P0 states that x0 = 2 2+1
=
0 , since
20
2 = a0 we can readily check that P0 is indeed true.
Finally, we fix k ∈ N. We assume that Pk is true and we try to show that we can prove
that Pk+1 is true. In general, it is not always easy to see how proposition Pk is related to
Pk+1 but when we need to prove something about recursively defined sequences usually the
recursive formula (the second line of Formula (1.1) in our example) gives you the relation
you need. To be more explicit, in this case we simply plug the formula of xk (which we
assume to be true) into the recursive formula which gives us xk+1 and see whether we get
the same formula for xk+1 .
3
1 22n + 1 3
22n + 1 3
22n + 1 + 3 · 22n
22n+2 + 1
1
+ = 2n+2 + =
=
= f (n+1).
xk+1 = xn + = ·
2n
2n+2
4
4
4 2
4
2
4
2
22n+2
Now we can coclude that by mathematical induction we have that xn = f (n) for every
n ∈ N.
Remark 2.2. Notice that we have really just repeated what we had done before, except
that now we have formalised the way to write this down. Since similar proofs appear many
times we have done this once and for all and we do not need to reinvent the wheel each
time.
Remark 2.3. Notice that the same technique can be helpful also if we want to use the
Monotone Convergence Theorem.
Example 2.4. Use the Monotone Convergence Theorem in order to prove that Sequence
(1.1) is convergent, then compute its limit.
Solution. We prove that (xn )∞
n=1 is decreasing and bounded from below. Both things are
proved by mathematical induction. We first prove that it is decreasing. In this case we have
that:
Pn : “xn ≥ xn+1 ”.
We have that P0 : “x0 ≥ x1 ” is true, since x0 = 2 ≥ 54 = x1 .
Next we fix k ∈ N and assume that Pk is true, namely that xk ≥ xk+1 . We want to use
this assumption to prove that xk+1 ≥ xk+2 . As before the recursion formula gives us the
implications:
1
1
3
1
3
1
xk ≥ xk+1 ⇒
xk + ≥ xk+1 +
⇒ (xk+1 ≥ xk+2 ) .
(xk ≥ xk+1 ) ⇒
4
4
4
4
4
4
4
Induction
Next we prove that the sequence is bounded from below. In this case it is not difficul to
see that all its terms are positive, nevertheless we prove this using mathematical induction.
Now Pn : “xn ≥ 0”. Since 2 ≥ 0 it is clear that P0 is true. Next fix k ∈ N and assume
that Pk is true. This means that xk ≥ 0. From this it follow that:
1
3
1
xk ≥ 0 ⇒
xk + ≥= ⇒ (xk+1 ≥ 0).
(xk ≥ 0) ⇒
4
4
4
We have proved by mathematical induction that for all n ∈ N the term xn ≥ 0. Now by
the Monotone Convergence Theorem we know that the sequence is convergent.
In order to compute the limit we can simply take limit to both sides of the second line
of Formula (1.1).
3
1
xn +
.
L = lim xn+1 = lim
n→∞
n→∞ 4
4
Now using basic properties of limits and the fact that lim xn+1 = lim xn we get:
n→∞
L = lim xn+1 = lim
n→∞
n→∞
3
1
xn +
4
4
n→∞
=
3
1
lim xn + =
4 n→∞
4
1
3
L+ .
4
4
=
From this we get L = 14 L + 43 , which implies that L = 1.
Remark 2.5. Notice that we cannot take limits before we know that limits exist and are
real number. For example, if we consider the sequence
x0 = 1
xn+1 = −xn for n = 1, 2, 3, . . .
This is just the sequence (1, −1, 1, −1, . . . ) which does not converge. If we take limits to
both sides of the recursive formula as we did before we get L = −L which would imply that
L = 0. Of course, this is wrong because L simply does not exist.
Remark 2.6. Sometimes it is not easy to find what a possible bound is. You can argue
as follows. First prove that the sequence is monotone. For instance, in our example prove
that it is decreasing. At this point you know that either the sequence has no lower bound
or it has a limit in R. If it has a limit this is also a lower bound. So compute a candidate
for the limit as we just did (in our example that would be 1). Finally, try to to prove that
what you have is really a bound (you have to do this!).
3
Why induction works
So far we have given examples of proof by induction and we tried to justify heuristically
that they were proofs, however we did not really explain why induction works.
Induction is based on a property of natural numbers called the well ordering principle.
Definition 3.1. We say that an ordered set A is well-ordered if every nonvoid subset B ⊆ A
has a smallest element.
Stefano Ferri
5
The fact that natural numbers are well-ordered allows us to prove that induction works.
Proof. Suppose that we have a sequence of propositions {Pi : i = 1, 2, 3, . . . } and that the
following holds.
(1.) P0 is true;
(2.) For all integer n, assuming that Pn is true implies that also Pn+1 is true.
We want to prove that all the Pi are true. Suppose, by contradiction, that this is not the
case, then the subset:
B = {i : Pi is false}
in nonvoid. Hence, by the well ordering principle, B has a smallest element which we call
m. By (1.) m 6= 0. Moreover, since m is the smallest element of B and m 6= 0 we have that
m − 1 ≥ 0 and that Pm−1 is true. But then by (2.) we have that Pm is true, reaching a
contradiction.
4
More examples and concluding remarks
More example not necessarily related to sequences. Even more examples can be found
among the exercises.
Example 4.1. Prove by mathematical induction that, for every natural number n, we have
that
n · (n + 1)
.
1 + 2 + 3 + ··· + n =
2
Solution.
Here
n · (n + 1)
”.
Pn : “1 + 2 + 3 + · · · + n =
2
First we check that the proposition is true for n = 1. Notice that, if the sequence of
propositions is enumerated starting from 1 (or from 2 or whatever) we start from the first
number of the enumeration. It does not have to be 0. We have that 1·(1+1)
= 22 = 1, so the
2
proposition is true for n = 1.
Next we assume that the proposition is true for k and we would like to prove that, under
this assumption, we have that the proposition is also true for k + 1. Since we assume that
the proposition is true for k what we assume is that
1 + 2 + 3 + ··· + k =
k · (k + 1)
.
2
In this case there is not a recursion formula to guide us and we have to examine Pk and
Pk+1 for a moment in order to determine what we have to do to pass from one to the next
one. The right hand side of the formula is maybe confusing but it is not hard to see that
the left hand side of the formula in Pk becomes the left hand side of the formula in Pk+1 if
we sum k + 1. For this reason we sum k + 1 to both sides of the equality and get:
1+2+3+· · ·+k+(k+1) =
k · (k + 1) 2 · (k + 1)
(k + 2) · (k + 1)
k · (k + 1)
+(k+1) =
+
=
.
2
2
2
2
A moment’s reflection should suffice to see that what we got is the formula we are trying to
prove evaluate in k + 1. Hence the proposition is true by mathematical induction. 6
Induction
Example 4.2. Let A be a set with n elements. Then the set P(A) of all subsets of A (the
set P(A) is called the power set of A) contains 2n elements.
Proof. We prove the theorem by induction over n. First we check the case in which A
contains 0 elements. In this case A is the empty set ∅ and his only subset is the empty set
itself. In other word P(A) = {∅} contains only one element and, since 20 = 1 this implies
that the theorem is true for n = 0.
Next we suppose that, if a set A contains n elements, then P(A) contains 2n elements
and, using this assumption, we show that, if A contains n + 1 elements, then P(A) contains
2n+1 elements.
Let us keep our assimption in mind and consider a set A with n + 1 element. To fix our
ideas we can think that A = {1, 2, . . . , n, n + 1}. There are two kind of subsets of A: those
which contain n + 1 and those which do not contain n + 1. All subsets which do not contain
n+1 can be considered as subset of A′ = {1, 2, 3, . . . , n}, hence, by our inductive assumption
there are 2n of them. All subsets which contain n + 1 are the union of a subset of A′ and the
set {n + 1}, hence there are 2n such subsets. Alltogether, we have 2n + 2n = 2n+1 subsets
of A. This complete the proof.
I conclude this note by remarking that what I presented here is just the most basic type
of induction you can find in mathematics. There exist many variant and sophistications you
can find. For istance:
• You can have different way to enumerate your propositions, so you might need to start
from a number other that 0 your argument.
• You can have your propositions enumerated as {Pk1 , Pk2 , Pk3 , . . . } with the function
mapping n to kn which is very difficult to compute. In this case a standard way to
procede is to assume that you can find a minimal counterexample to what you are
trying to prove and then show that you can (under this assumption) build a smaller
counterexample. (This technique is very common in the theory of finite groups.)
Exercises 4.3.
1. Let n be a positive integer. Prove that: 1 + 4 + 9 + · · · + n2 =
n(n+1)(2n+1)
.
6
2. Let n be a positive integer. Prove that 4 + 10 + 16 + · · · + (6n − 2) = n(3n + 1).
3. Prove that, for every x ∈ R \ {0, 1} and every n ∈ N, we have that: x0 + x1 + x2 +
n+1
· · · + xn = x a−1−1 .
4. Let n be a positive integer. Prove that n2 > n + 1 whenever n ≥ 2.
5. Let n be an integer number greater or equal to 4. Prove that n2 ≤ 2n .
6. Find the mistake in the following argument.
Theorem. All socks have the same colour.
Proof. We shall prove that all finite sets of socks are monochrome by induction on the
number of socks contained in the set. If the set contains only one sock, then the theorem
is trivially true. Now suppose that the proposition is true for sets containing n socks
and let A = {1, 2, 3, . . . , n, n + 1} be a set of n + 1 socks. The set A′ = {1, 2, 3, . . . , n}
7
Stefano Ferri
has n element, hence it is monochrome. Also the set A′′ = {2, 3, 4, . . . , n, n + 1} has
n element, hence, using again the inductive assumption, it is monocrome. Now, from
these two facts we deduce that n has the same colour has 1 because they both lie in A′
and also that n has the same colour as n + 1 because they both are in A′′ . Hence, all
socks 1, 2, 3, . . . , n, n + 1 have the same colour and the theorem is proved by induction.
5
Appendix (direct proofs of the previous examples)
Example 4.1 and 5.2 are propositions which are interesting in their own right. Both of them,
besides the proof I presented, have an easy direct proof.
Theorem 5.1. For every natural number n, we have that
1 + 2 + 3 + ··· + n =
n(n + 1)
.
2
Proof. In the following table there are 1 + 2 + 3 + · · · + n black dots.

•




• •
n row • • •


· · ·



• • • ... •
{z
}
|
n columns
Now we write picture twice the same table
using white dots):

• ◦




• •




• •
n row · ·


• •





• •


• •
|
(one is made using black dots and the other one
◦
◦
•
·
•
•
•
...
...
...
...
...
...
...
{z
◦
◦
◦
·
◦
•
•
n+1 columns
◦
◦
◦
·
◦
◦
•
◦
◦
◦
·
◦
◦
◦
}
The last lable contains n(n + 1) dots and it also contains twice 1 + 2 + 3 + · · · + n dots, from
.
which it follows that 1 + 2 + 3 + · · · + n = n(n+1)
2
Theorem 5.2. Let A be a set with n elements. Then the set P(A) of all subsets of A
contains 2n elements.
Proof. Counting the number of subsets of A is the same as counting the number of functions
between A and a set {0, 1}. Given a subset B of A we define a function χB : A −→ {0, 1}
by:
0 if x ∈ B
χB (x) :=
1 otherwise
(χB is called characteristic function of B). On the other hand, if we have a function
f : A −→ {0, 1} we can define a subset B of A by the following rule:
8
Induction
x ∈ B if and only if f (x) = 1.
Now, if we write A = {1, 2, 3 . . . , n}, and we have a function f : A −→ {0, 1}, we have
that f (1) is either 0 or 1, for each of the two possibility we can have that f (2) is 0 or 1, and,
carring on the process, for each of the four possibilities we have that f (3) can be either 0
or 1. Continuing this until the last element we have that we can build exactly 2 · 2 · 2 · · · · 2
(n times) functions or, which is the same, we have 2n functions, as requested.