With examples from Number Theory

1
Methods of Proof

• With examples from Number Theory
• (Rosen 1.5, 3.1, sections on methods of proving
  theorems and fallacies)

2
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).

3
Basic Number Theory Definitionsfrom 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) ? (ab) ? Z ? (ab) ? Z

4
Number Theory Defs (cont.)
? such that

• n is even is defined as ?k ? Z ? n 2k
• n is odd is defined as ?k ? Z ? n 2k1
• 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.

5
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.

6
Direct Proof

• Prove ?n?Z, if n is even, then n2 is even.
• Tabular-style proof
• n is even hypothesis
• n2k 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

7
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.

8
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.

9
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
  st a/b c/d (adcb)/bd . But since (adcb)
  ?Z and bd ?Z ?0 (why?), then (adcb)/bd is
  rational.

10
Structure of this 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 st a/b c/d (adcb)/bd .
• But since (adcb) ?Z and bd ?Z ?0, then
  (adcb)/bd is rational.

Assumed
Def
Basic facts of arithmetic
Conclusion from Def
11
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 2k1. It
  follows that n3 (2k1)3 8k38k24k1
  2(4k34k22k)1. (4k34k22k) is an integer.
  Therefore n3 is 1 plus an even integer.
  Therefore n3 is odd.

Assumption, Definition, Arithmetic, Conclusion
12
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 dont have a rule about how to take n3 apart!
13
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., qrs 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 qr cant be rational
  and must be irrational.

14
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.)

15
2nd Proof by Contradiction
Prove If 3n2 is odd, then n is odd. Proof
Assume 3n2 is odd and n is even. Since n is
even, then n2k for some integer k. It follows
that 3n2 6k2 2(3k1). Thus, 3n2 is even.
This contradicts the assumption that 3n2 is odd.
16
What Proof Approach?

• (n ? Z ? n35 is odd) ? n is even
• The sum of two odd integers is even
• Product of two irrational numbers is irrational
• The sum of two even integers is even
• ?2 is irrational
• If n ? Z and 3n2 is odd, then n is odd
• If a2 is even, then a is even

indirect
direct
Is this true? Counterexample?
direct
contradiction
indirect
indirect
17
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. n3n 8k3 2k 2(4k3k) which is even
since 4k3k must be an integer.
18
Cases (cont.)

• Case 2 n is odd. Then ?k?Z ? n 2k1.
• 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

19
Even/Odd is a Special Case of Divisibility

• We say that x is divisible by y if ? k ? Z ?
  xyk
• n is divisible by 2 if ? k ? Z ? n 2k (even)
• The other case is n 2k1(odd,remainder of 1)
• n is divisible by 3 if ? k ? Z ? n 3k
• Other cases
• n 3k 1
• n 3k 2
• n is divisible by 4 if ? k ? Z ? n 4k

20
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.

21
Remainder Lemma

• Lemma Let a3k1 where k is an integer. Then
  the remainder when a2 is divided by 3 is 1.
• Proof Assume a 3k1. Then
• a2 9k2 6k 1 3(3k22k) 1.
• Since 3(3k22k) is divisible by 3, the remainder
  must be 1.

22
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.
23
Divisibility Example (cont.)

• Prove n2 - 2 is never divisible by 3 if n is an
  integer.
• Lets use cases!
• There are three possible cases
• Case 1 n 3k
• Case 2 n (3k1)
• Case 3 n (3k2) k?Z

24
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.

25
n2-2 is never divisible by 3 if n?Z

• Case 2 n 3k1 for k?Z
• n2-2 (3k1)2 - 2 9k2 6k 1-2
• 3(3k2 2k) - 1 3(3k2 2k -1) 2
• Thus the remainder when dividing by 3 is 2.

26
n2-2 is never divisible by 3 if n?Z

• Case 3 n 3k2 for k?Z
• n2-2 (3k2)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.

27
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.

28
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.

29
More Complex Proof (cont.)

• (?2)2 (a/b)2 a2/b2 2.
• Now what do we want to do? Lets 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?

30
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 2k1.
• a2 (2k1)2 4k2 4k 1 2(2k22k) 1.
• Therefore a is odd. So the Lemma must be true.

31
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.

32
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.

33
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.

34
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 n38k3, rather than n3 2k. This is
  circular reasoning.