Download Oh Yeah? Well, Prove It.

Oh Yeah? Well, Prove It.
MT 423A - Abstract Algebra
Fall 2009
A large part of mathematics consists of building up a theoretical framework that allows
us to solve problems. This theoretical framework is built upon a set of axioms. Axioms are
unproven assumptions that are the foundation of all mathematics. (Not everyone agrees
on what axioms should be used, but that’s a whole other story.) These basic axioms are
combined using the rules of logic into more complicated statements called theorems. If
the axioms are true and the logic is sound, then the theorems must also be true. New
theorems are created by combining axioms and known theorems. A proof of a theorem is
simply an explanation of why the theorem is true. Different people will require different
amounts of explanation. It will take some practice to get used to how much explanation
I expect.
In this class we will not attempt to build all of mathematics from a set of axioms (an
extremely difficult proposition!). Instead, we will start with some known mathematical
objects (like the integers), and see how some basic definitions, axioms and known theorems
can be built into a useful theoretical framework. In order to do this, we need to learn how
to prove new theorems.
One of the first definitions we will use is that of an even integer. I’m sure you all think
you know what an even integer is, but many of you will hesitate if asked whether zero is
even or odd. The following definition makes this obvious.
Definition: An integer is called even if it is divisible by 2, and called odd otherwise.
In other words, the even numbers are those that can be written as 2k for some integer k,
and the odd numbers are those that can be written as 2k + 1 for some integer k.
It should be clear upon a moment’s reflection that zero is an even number.
A statement that has not been proved is called a conjecture. Conjectures can usually
be written in the form “If P , then Q.” This can also be written as “P =⇒ Q”, which is
read “P implies Q”. P consists of one or more statements called the hypotheses and Q
is a statement called the conclusion.
Example 1: The statement “The sum of two even integers is even.” can be written as
the statement “If a and b are even integers, then a + b is an even integer.” The hypotheses
are “a and b are even integers” and the conclusion is “a + b is an even integer.”
Some conjectures are written in the form “P if and only if Q.” This is also written
as “P ⇐⇒ Q”, or “P iff Q”. In this case it is said that Q is a necessary and sufficient
condition for P . To prove an “if and only if” statement, we must prove two statements:
• If P , then Q
(Q is necessary for P ), and
• If Q, then P
(Q is sufficient for P )
One direction is frequently significantly easier to prove than the other.
Example 2: “The square of an integer n is even if and only if n is even.”
P is the statement: “The square of an integer n is even”, and
Q is the statement: “n is even.”
To prove this statement we must prove both
• If n2 is even, then n is even.
• If n is even, then n2 is even.
A proof that P =⇒ Q is a logically correct argument that shows that if the hypotheses
are true, then the conclusion must also be true. The argument may use axioms, definitions
and previously proved statements.
Once a statement has been proved, it is called a theorem. A theorem whose main
purpose is to prove other more important theorems is called a lemma. A theorem that
easily follows from a more important theorem is called a corollary.
There is no algorithm, or single list of instructions, for finding proofs, but here are
some tips to help you get started.
1. Understand what you are trying to prove. Look up definitions of all the terms. Even
though you may know what the terms mean, it is important to have the “official”
definitions in front of you, as these will frequently give you a clue about how you
might proceed.
2. Understand what the hypotheses are. Write down definitions of all the terms involved
in the hypotheses. This is almost always the first step in a proof.
3. It is often helpful to look at examples that illustrate the conjecture. This can give
you a better understanding of the problem. No collection of specific examples
constitues a general proof that a conjecture is true.
4. It is sometimes useful to work backward from the conclusion. Ask yourself, “What
would I have to know in order to know that the conclusion is true.” Then, “What
would I have to know in order to know that these statements are true.” Keep going
until you see a way to connect the hypotheses with the conclusion.
5. Once you have discovered your proof, write it up in a clear correct logical form. State
the main steps from the hypotheses to the conclusion with some justification for
each step. Avoid excessive detail, but provide enough detail so that your argument
is convincing.
There are several standard methods of proof. Here is a description of some of the most
common methods of proof, with examples of each.
• DIRECT PROOF: Begin with the hypotheses and follow a direct line of reasoning
to the conclusion.
Theorem: If a and b are even integers, then a + b is an even integer.
Proof: If a is an even integer, then a can be written as an integer multiple of 2 (this
is the definition of an even integer). Thus, we can write a = 2m for some integer m.
Similarly, since b is even, we can write b = 2n for some integer n (not necessarily
the same integer as we used to write a, of course).
Now look at a + b. We can write a + b as 2m + 2n = 2(m + n), and we see that a + b
is an integer multiple of 2. Therefore, a + b is an even integer. 2
It is important to mark the end of every proof. The box, 2, is commonly used for
this. Another common way to mark the end of a proof is with the letters QED. (In
fact, the box is often called a QED-box.) QED is an abbreviation of the latin phrase
“Quod Erat Demonstrandum” - “which was to be demonstrated”.
• COUNTER-EXAMPLE: To prove that a statement is false in general, it is enough
to find one specific case for which the statement is false. Such a statement is called
a counter-example.
Find a counter-example to the following conjecture:
Conjecture: If a, b and d are integers and d divides the product ab, then d divides
a or d divides b.
• CONTRADICTION: Assume the hypothesis is true and the conclusion is false, and
show that this contradicts something known to be true.
Theorem: There are infinitely many prime numbers.
Proof: This theorem is unusual in that there are no hypotheses explicitly stated!
We only have basic definitions and previous theorems to go on.
We will prove this by contradiction, using the Fundamental Theorem of Arithmetic
(Theorem 0.3 on page 6). Every integer greater than one is prime or a product of
primes.
In order to arrive at a contradiction, assume that there is a finite number of prime
numbers. If the number of prime numbers is finite (say there are n of them), then
we can list them.
p1 , p2 , . . . pn
Consider the (very large) number we get if we multiply all these primes together and
add 1.
V = p1 p2 . . . p n + 1
Notice that pk does not divide V for any k from 1 to n (since no primes divide 1). In
other words, none of the primes on the list divide V . The Fundamental Theorem of
Arithmetic, however, tells us that V can be written uniquely as a product of prime
numbers. In other words, some prime numbers have to divide V . But all the primes
are on the list, and none of them divide V .
This is a contradiction! Contradictions are sometimes denoted with the symbol →←.
This contradiction means that our initial assumption that there is a finite number
of prime numbers must be false. Therefore there must be infinitely many prime
numbers. 2
Some (uptight) mathematicians don’t like to use proof by contradiction. They insist
that any statement proved by contradiction can be proved directly. This may or
may not be true, but it is true that proof by contradiction is often a much simpler
and more elegant way to prove a statement that is difficult to prove directly. Always
try a direct proof first, but if you get stuck, a proof by contradiction will sometimes
work.
• CONTRAPOSITIVE: Sometimes the contrapositive of a statement is easier to prove
than the statement.
The contrapositive of the statement
“If P , then Q”
is the statement
“If not Q, then not P ”.
A statement and its contrapositive are logically equivalent. If one is true the other
is necessarily true, and vice versa.
For example, the statement “If I roll a 7, then I win.” is logically equivalent to the
statement “If I don’t win, then I don’t roll a 7”. Either they are both true or both
false.
Theorem: If n2 is even, then n is even.
Proof: We prove the contrapositive: If n is odd, then n2 is odd.
If n is odd, then n can be written as 2k + 1 for some integer k.
Then n2 = (2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1.
Since n2 is of the form 2x + 1 for the integer x = 2k 2 + 2k, n2 is an odd integer. 2
• INDUCTION: Induction is a technique for proving that a statement is true for all
positive integers. Proof by induction is like knocking over dominos - prove that you
can knock over the first domino, then prove that if one domino falls then it knocks
over the next domino. If you can prove both those things, then all the dominos will
fall.
More formally, if you wish use induction to prove a statement Sn true for all integers
n ≥ n0 , you must
– Prove a base case: prove the statement Sn is true for n = n0 .
– Prove the inductive step: assume the statement is true for case n, and use this
to prove the statement is true for case n + 1.
In other words prove that Sn =⇒ Sn+1
Sn is called the inductive hypothesis. You assume Sn is true, and use it to show
Sn+1 must also be true.
Theorem: The sum of the the first n positive integers is
we wish to prove the statement Sn :
n(n+1)
.
2
In other words,
n(n + 1)
2
for all n ≥ 1. (Note that Sn is simply a name for the above equation.)
1 + 2 + 3 + ... + n =
Proof: (By induction)
Base case: S1 is the statement 1 =
1(1+1)
.
2
This is true.
The inductive step: Assume Sn is true. That is,
1 + 2 + 3 + ... + n =
n(n + 1)
2
We wish to use this to prove Sn+1 :
1 + 2 + 3 + . . . + n + (n + 1) =
(n + 1)(n + 2)
2
To see this, we add (n + 1) to both sides of equation Sn and get
n(n + 1)
+ (n + 1)
2
Getting a common denominator and simplifying the right side produces
1 + 2 + 3 + . . . + n + (n + 1) =
n(n + 1)
n(n + 1) + 2(n + 1)
+ (n + 1) =
2
2
n2 + 3n + 2
2
(n + 1)(n + 2)
=
2
=
We see that if Sn is true then
(n + 1)(n + 2)
2
is also true, but this is statement Sn+1 . So Sn implies Sn+1 .
1 + 2 + 3 + . . . + n + (n + 1) =
Therefore, since we have shown a base case and proved the inductive step, the
statement Sn is true for all n ≥ 1.
In other words, the sum of the the first n positive integers is
n(n+1)
.
2
2
There you have it - a quick and dirty introduction to methods of proof. I hope you
find it helpful. Of course the only way to really learn how to write proofs is to grit your
teeth, try to write some, get stuck, get frustrated (but don’t stay frustrated!), ask for help,
make mistakes, learn from them, and repeat all of these steps as often as necessary. Many
of you will need to use office hours more than you are used to. I’ll be able to answer some
questions in class, but the best help will come one on one, or in small groups. Don’t be
afraid to ask for help. As long as you keep trying, I’ll be willing to help as best as I can.
EXERCISES:
1. Use the following outline to prove by induction that for all integers greater than or
equal to 1, the sum of the first n odd integers is n2 . Since the nth odd integer is
2n − 1, the symbolic form for this statement is
Sn : 1 + 3 + . . . + (2n − 1) = n2
(a) Check that the statement is true in the case n = 1.
(b) Write the inductive hypothesis, Sn . This is what you assume to be true.
(c) Write the expression Sn+1 . This is what you hope to show is true.
(d) What simple algebraic operation would you have to do to the left side of Sn to
get the left side of Sn+1 ? Of course, if you do this to the left side of equation
Sn , you must do it to the right side of Sn , as well. Do it!
(e) Does the left side of the expression from part (d) equal the left side of Sn+1 ?
(It better!)
(f) Do some simple algebra to see if the right side of part (d) equals the right side
of Sn+1 . If it does, you have shown that Sn =⇒ Sn+1 , and thus completed the
proof by induction.
2. Prove that if a, b and c are integers and a divides b and a divides c, then a divides
b + c and a divides b − c.
3. Prove that if a, b and c are integers and a divides b and b divides c, then a divides c.
4. Prove that the product of two even integers is divisible by 4.
5. Prove that the product of two odd integers has remainder 1 when divided by 4.
6. Prove that the square of any integer is congruent to 0 or 1 mod 4.
7. Prove that if a, b and c are integers such that a2 + b2 = c2 , then either a or b is even.
(This is an “inclusive or” - either a or b is even, or possibly both are even.)
8. Use induction to prove that 20 + 21 + 22 + 23 + . . . + 2n = 2n+1 − 1 for n ≥ 0.
9. The Fibonacci numbers are defined recursively by the formulas
F1 = 1
F2 = 1
Fn = Fn−1 + Fn−2
• calculate the first ten Fibonacci numbers.
• Use induction to prove that
F12 + F22 + F32 + . . . + Fn2 = Fn Fn+1
for all n ≥ 1.