Transcript Methods of Proof

With examples from Number Theory
(Rosen 1.5, 3.1, sections on methods of proving
theorems and fallacies)
Basic Definitions
Theorem - A statement that can be shown to be true.
Proof - A series of statements that form a valid argument.
• Start with your hypothesis or assumption
• Each statement in the series must be:
– Basic fact or definition
– Logical step (based on rules or basic logic)
– Previously proved theorem (lemma or corollary)
• Must end with what you are trying to prove
(conclusion).
The Proof Process
Assumptions
Logical Steps
-Definitions
-Already-proved statements
-logical steps (e.g., arithmetic
or algebraic)
Conclusion
(That which was to be proved)
Basic Number Theory Definitions
from Chapters 1.6, 2
•
•
•
•
•
Z = Set of all Integers
Z+ = Set of all Positive Integers
N = Set of Natural Numbers (Z+ and Zero)
R = Set of Real Numbers
Addition and multiplication on integers
produce integers. (a,b  Z)  [(a+b)  Z] 
[(ab)  Z]
 = “such that”
Number Theory Defs (cont.)
• n is even is defined as k  Z  n = 2k
• n is odd is defined as k  Z  n = 2k+1
• x is rational is defined as a,b  Z  x =
a/b, b0
• x is irrational is defined as a,b  Z  x
= a/b, b0 or a,b  Z, x  a/b, b0
• p  Z+ is prime means that the only
positive factors of p are p and 1. If p is not
prime we say it is composite.
Methods of Proof
p q (Example: if n is even, then n2 is even)
• Direct proof: Assume p is true and use a series of
previously proven statements to show that q is
true.
• Indirect proof: Show q p is true
(contrapositive), using any proof technique
(usually direct proof).
• Proof by contradiction: Assume negation of what
you are trying to prove (pq). Show that this
leads to a contradiction.
Direct Proof
Prove: nZ, if n is even, then n2 is even.
Tabular-style proof:
n is even
hypothesis
n=2k for some kZ definition of even
n2 = 4k2
algebra
n2 = 2(2k2) which is algebra and mult of
2*(an integer)
integers gives integers
n2 is even
definition of even
Same Direct Proof
Prove: nZ, if n is even, then n2 is even.
Sentence-style proof:
Assume that n is even. Thus, we know that n
= 2k for some integer k. It follows that n2
= 4k2 = 2(2k2). Therefore n2 is even since it
is 2 times 2k2, which is an integer.
Structure of a Direct Proof
Prove: nZ, if n is even, then n2 is even.
Proof:
Assume that n is even. Thus, we know that n
= 2k for some integer k. It follows that n2 =
4k2 = 2(2k2). Therefore n2 is even since it is
2 times 2k2 which is an integer.
Another Direct Proof
Prove: The sum of two rational numbers is
a rational number.
Proof: Let s and t be rational numbers. Then
s = a/b and t = c/d where a,b,c,d Z, b,d 0.
Then s+t = a/b + c/d = (ad+cb)/bd . But
since (ad+cb) Z and bd Z 0 (why?), then
(ad+cb)/bd is rational.
Structure of this Direct Proof
Prove: The sum of two rational numbers is a
rational number.
Proof: Let s and t be rational numbers.
Assumed
Then s = a/b and t = c/d where a,b,c,d Z , b,d 0. Def
Then s+t = a/b + c/d = (ad+cb)/bd .
But since (ad+cb) Z and bd Z 0, then
(ad+cb)/bd is rational.
Basic facts of
arithmetic
Conclusion from Def
Example of an Indirect Proof
Prove: If n3 is even, then n is even.
Proof: The contrapositive of “If n3 is even, then n is
even” is “If n is odd, then n3 is odd.” If the
contrapositive is true then the original statement must
be true.
Assume n is odd. Then kZ  n = 2k+1. It follows that
n3 = (2k+1)3 = 8k3+8k2+4k+1 = 2(4k3+4k2+2k)+1.
(4k3+4k2+2k) is an integer. Therefore n3 is 1 plus an
even integer. Therefore n3 is odd.
Assumption, Definition, Arithmetic, Conclusion
Discussion of Indirect Proof
Could we do a direct proof of If n3 is even,
then n is even?
Assume n3 is even . . . then what?
We don’t have a rule about how to take n3 apart!
Example: Proof by Contradiction
Prove: The sum of an irrational number and a
rational number is irrational.
Proof: Let q be an irrational number and r be a
rational number. Assume that their sum is
rational, i.e., q+r=s where s is a rational
number. Then q = s-r. But by our previous proof
the sum of two rational numbers must be rational,
so we have an irrational number on the left equal to
a rational number on the right. This is a
contradiction. Therefore q+r can’t be rational and
must be irrational.
Structure of Proof by Contradiction
• Basic idea is to assume that the opposite of what
you are trying to prove is true and show that it
results in a violation of one of your initial
assumptions.
• In the previous proof we showed that assuming
that the sum of a rational number and an irrational
number is rational and showed that it resulted in
the impossible conclusion that a number could be
rational and irrational at the same time. (It can be
put in a form that implies n  n is true, which is
a contradiction.)
2nd Proof by Contradiction
Prove: If 3n+2 is odd, then n is odd.
Proof: Assume 3n+2 is odd and n is even.
Since n is even, then n=2k for some integer k.
It follows that 3n+2 = 6k+2 = 2(3k+1).
Thus, 3n+2 is even. This contradicts the
assumption that 3n+2 is odd.
What Proof Approach?
• (n  Z  n3+5 is odd)  n is even indirect
• The sum of two odd integers is even direct
• Product of two irrational numbers is
irrational Is this true? Counterexample?
• The sum of two even integers is even direct
contradiction
• 2 is irrational
• If n  Z and 3n+2 is odd, then n is odd indirect
• If a2 is even, then a is even indirect
Using Cases
Prove: n Z, n3 + n is even.
Separate into cases based on whether n is even or
odd. Prove each separately using direct proof.
Proof: We can divide this problem into two
cases. n can be even or n can be odd.
Case 1: n is even. Then kZ  n = 2k.
n3+n = 8k3 + 2k = 2(4k3+k) which is even since
4k3+k must be an integer.
Cases (cont.)
Case 2: n is odd. Then kZ  n = 2k+1.
n3 + n = (8k3 +12k2 + 6k + 1) + (2k + 1) =
2(4k3 + 6k2 + 4k + 1) which is even since
4k3 + 6k2 + 4k + 1 must be an integer.
Therefore n Z, n3 + n is even
Even/Odd is a Special Case of Divisibility
We say that x is divisible by y if  k  Z  x=yk
• n is divisible by 2 if  k  Z  n = 2k (even)
• The other case is n = 2k+1(odd,remainder of 1)
• n is divisible by 3 if  k  Z  n = 3k
Other cases
• n = 3k + 1
• n = 3k + 2
This leads to modulo arithmetic
• n is divisible by 4 if  k  Z  n = 4k
Lemmas and Corollaries
• A lemma is a simple
theorem used in the
proof of other
theorems.
• A corollary is a
proposition that can be
established directly
from a theorem that
has already been
proved.
Remainder Lemma
Lemma: Let a=3k+1 where k is an integer.
Then the remainder when a2 is divided by 3
is 1.
Proof: Assume a =3k+1. Then
a2 = 9k2 + 6k + 1 = 3(3k2+2k) + 1.
Since 3(3k2+2k) is divisible by 3, the
remainder must be 1.
Divisibility Example
Prove: n2 - 2 is never divisible by 3 if n is an
integer.
Discussion: What does it mean for a number to be
divisible by 3? If a is divisible by 3 then  b  Z  a
= 3b. Remainder when n is divided by 3 is 0. Other
options are a remainder of 1 and 2.
So we need to show that the remainder when n2 - 2 is
divided by 3 is always 1 or 2 but never 0.
Divisibility Example (cont.)
Prove: n2 - 2 is never divisible by 3 if n is an
integer.
Let’s use cases!
There are three possible cases:
• Case 1: n = 3k
• Case 2: n = (3k+1)
• Case 3: n= (3k+2); kZ
n2-2 is never divisible by 3 if n  Z
Proof:
Case 1: n = 3k for kZ then
n2-2 = 9k2 - 2 = 3(3k2) - 2 =
3(3k2 - 1) + 1
The remainder when dividing by 3 is 1.
n2-2 is never divisible by 3 if nZ
Case 2: n = 3k+1 for kZ
n2-2 = (3k+1)2 - 2 = 9k2 + 6k +1-2 =
3(3k2 + 2k) - 1 = 3(3k2 + 2k -1) + 2
Thus the remainder when dividing by 3 is 2.
n2-2 is never divisible by 3 if nZ
Case 3: n = 3k+2 for kZ
n2-2 = (3k+2)2 - 2 = 9k2 + 12k +4 -2 =
3(3k2 + 4k) + 2
Thus the remainder when dividing by 3 is 2.
In each case the remainder when dividing n2-2
by 3 is nonzero. This proves the theorem.
More Complex Proof
Prove: 2 is irrational.
Direct proof is difficult.
Must show that there are no a,b, Z, b≠0 such
that a/b = 2 .
Try proof by contradiction.
More Complex Proof (cont.)
Proof by Contradiction of 2 is irrational:
Assume 2 is rational, i.e., 2 = a/b for some
a,b Z, b0.
Since any fraction can be reduced until there
are no common factors in the numerator and
denominator, we can further assume that:
2 = a/b for some a,b Z, b0 and a and b
have no common factors.
More Complex Proof (cont.)
(2)2 = (a/b)2 = a2/b2 = 2.
Now what do we want to do? Let’s show that
a2/b2 = 2 implies that both a and b are even!
Since a and b have no common factors, this is
a contradiction since both a and b even
implies that 2 is a common factor.
Clearly a2 is even (why?). Does that mean a
is even?
More Complex Proof (cont.)
Lemma 1: If a2 is even, then a is even.
Proof (indirect): If a is odd, then a2 is odd.
Assume a is odd. Then kZ  a = 2k+1.
a2 = (2k+1)2 = 4k2 + 4k + 1= 2(2k2+2k) + 1.
Therefore a is odd. So the Lemma must be
true.
More Complex Proof (cont.)
Back to the example!
So far we have shown that a2 is even. Then
by Lemma 1, a is even. Thus kZ  a =
2k.
Now, we will show that b is even.
From before, a2/b2 = 2  2b2 = a2 = (2k)2.
Dividing by 2 gives b2 = 2k2. Therefore b2 is
even and from Lemma 1, b is even.
More Complex Proof (cont.)
But, if a is even and b is even then they have a
common factor of 2. This contradicts our
assumption that our a/b has been reduced to
have no common factors.
Therefore 2  a/b for some a,b Z, b0.
Therefore 2 is irrational.
Fallacies
Incorrect reasoning occurs in the following
cases when the propositions are assumed to
be tautologies (since they are not).
• Fallacy of affirming the conclusion
• [(p  q)  q]  p
• Fallacy of denying the hypothesis
• [(p  q)  p]  q
• Fallacy of circular reasoning
• One or more steps in the proof are based on the
truth of the statement being proved.
Proof?
Prove if n3 is even then n is even.
Proof: Assume n3 is even.
Then kZ  n3 = 8k3 for some integer k. It
follows that n = 38k3 = 2k. Therefore n is
even.
Statement is true but argument is false.
Argument assumes that n is even in making
the claim n3=8k3, rather than n3 = 2k. This is
circular reasoning.