Transcript Prove if n 3 is even then n is even. 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).
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) [(ab) Z] Z]
= “such that”
Number Theory Defs
(cont.) • n is
even
• n is
odd
is defined as k is defined as k Z Z n = 2k n = 2k+1 • x is
rational
a/b, b 0 is defined as a,b Z • x is
irrational
is defined as a,b = a/b, b 0 or a,b Z, x a/b, b 0 x = Z x • 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 n 2 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 n 2 is even.
Tabular-style proof:
hypothesis
n is even n=2k for some k Z n 2 = 4k 2 n 2 = 2(2k 2 ) which is 2*(an integer) n 2 is even
definition of even algebra algebra and mult of integers gives integers definition of even
Same Direct Proof
Prove:
n
Z, if n is even, then n 2 is even.
Sentence-style proof:
Assume that n is even. Thus, we know that n = 2k for some integer k. It follows that n 2 = 4k 2 = 2(2k 2 ). Therefore n 2 is even since it is 2 times 2k 2 , which is an integer.
Structure of a Direct Proof
Prove:
n
Z, if n is even, then n 2 is even.
Proof:
Assume that n is even. Thus, we know that n = 2k for some integer k. It follows that n 2 = 4k 2 = 2(2k 2 ). Therefore n 2 is even since it is 2 times 2k 2 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 n 3 is even, then n is even.
Proof:
The contrapositive of “If n even” is “If n is odd, then n 3 3 is even, then n is is odd.” If the contrapositive is true then the original statement must be true. Assume n is odd.
n 3 = (2k+1) 3 Then = 8k 3 +8k 2 k Z n = 2k+1.
+4k+1 = 2(4k 3 +4k It follows that 2 +2k)+1. (4k 3 +4k 2 +2k) is an integer. Therefore n 3 is 1 plus an even integer. Therefore n 3 is odd.
Assumption, Definition, Arithmetic, Conclusion
Discussion of Indirect Proof
Could we do a direct proof of
If n 3 is even, then n is even?
Assume n 3 is even . . . then what?
We don’t have a rule about how to take n 3 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.
• (n
What Proof Approach?
Z n 3 +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 2 is irrational contradiction • If n Z and 3n+2 is odd, then n is odd • If a 2 is even, then a is even indirect direct indirect
Using Cases
Prove:
n
Z, n 3 + 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 3 +n = 8k 3 + 2k = 2(4k 3 n = 2k.
+k) which is even since 4k 3 +k must be an integer.
Cases
(cont.) Case 2: n is odd. Then k Z n = 2k+1. n 3 + n = (8k 3 2(4k 3 + 6k 2 +12k 2 + 6k + 1) + (2k + 1) = + 4k + 1) which is even since 4k 3 + 6k 2 + 4k + 1 must be an integer. Therefore n Z, n 3 + n is even
Even/Odd is a Special Case of Divisibility
We say that x is divisible by y if
• n is divisible by 2 if
k
Z
k
Z
x=yk 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 This leads to modulo arithmetic • n = 3k + 1 • n = 3k + 2
• 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 a 2 is divided by 3 is 1.
Proof:
Assume a =3k+1. Then a 2 = 9k 2 + 6k + 1 = 3(3k 2 +2k) + 1. Since 3(3k 2 +2k) is divisible by 3, the remainder must be 1.
Divisibility Example
Prove: n 2 - 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 n 2 - 2 is divided by 3 is always 1 or 2 but never 0.
Divisibility Example
(cont.)
Prove: n 2 - 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
n
2
-2 is never divisible by 3 if n
Z
Proof:
Case 1: n = 3k for k Z then n 2 -2 = 9k 2 - 2 = 3(3k 2 ) - 2 = 3(3k 2 - 1) + 1 The remainder when dividing by 3 is 1.
n
2
-2 is never divisible by 3 if n
Z
Case 2: n = 3k+1 for k Z n 2 -2 = (3k+1) 2 - 2 = 9k 2 + 6k +1-2 = 3(3k 2 + 2k) - 1 = 3(3k 2 + 2k -1) + 2 Thus the remainder when dividing by 3 is 2.
n
2
-2 is never divisible by 3 if n
Z
Case 3: n = 3k+2 for k Z n 2 -2 = (3k+2) 2 - 2 = 9k 2 + 12k +4 -2 = 3(3k 2 + 4k) + 2 Thus the remainder when dividing by 3 is 2.
In each case the remainder when dividing n 2 -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 = a 2 /b 2 = 2.
Now what do we want to do? Let’s show that a 2 /b 2 = 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 a 2 is even?
is even (why?). Does that mean a
More Complex Proof (cont.)
Lemma 1:
If a 2 is even, then a is even.
Proof (indirect):
If a is odd, then a 2 Assume a is odd. Then k Z is odd.
a = 2k+1.
a 2 = (2k+1) 2 = 4k 2 + 4k + 1= 2(2k 2 +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 a 2 is even. Then by Lemma 1, a is even. Thus k Z a = 2k.
Now, we will show that b is even.
From before, a 2 /b 2 = 2 2b 2 = a 2 = (2k) 2 .
Dividing by 2 gives b 2 = 2k 2 . Therefore b 2 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 n 3 is even then n is even.
Proof:
Assume n 3 is even. Then k Z n 3 = 8k 3 for some integer k. It follows that n = 3 8k 3 = 2k. Therefore n is even.
Statement
is true but
argument
is false. Argument assumes that n is even in making the claim n 3 =8k 3 , rather than n 3 = 2k. This is circular reasoning.