The following definitions will be important in most of our discussions

The following definitions will be important in most

This preview shows page 27 - 31 out of 151 pages.

The following definitions will be important in most of our discussions: Definition 1.7.1. An integer n is even if there exists an integer k such that n = 2 k . So the set of even integers is E = { 2 k : k Z } = { . . . , - 4 , - 2 , 0 , 2 , 4 , . . . } . An integer n is odd if there exists an integer k such that n = 2 k + 1 . So the set of odd integers is O = { 2 k + 1 : k Z } = { . . . , - 3 , - 1 , 1 , 3 , 5 , . . . } . 1.7.1 Direct proof In order to prove a conditional statement P Q , we assume P and then deduce Q by using some of axioms, definitions, rules or logical inferences, previously proven theorems. Example 1.7.2. Use the direct proof to prove that: If n is an even integer then n 2 is also an even integer. Proof. In this statement we can let P : n is even integer 24
Image of page 27
Q : n 2 is even integer. We need to assume P and deduce Q . Assume n is even, i.e. n = 2 k for some k Z . Then n 2 = n · n = (2 k )(2 k ) = 4 k 2 = 2(2 k 2 ) = 2 m for m = 2 k 2 Z , hence n 2 is even. Example 1.7.3. Use the direct proof to prove that If n is even integer and m is odd integer, then n + m an odd integer. Proof. In this statement we can let P : n is even integer. Q : m is odd integer. R : n + m is odd integer. We need to assume P Q and deduce R , i.e. P Q R . Assume n is even integer and m is an odd integer, i.e. n = 2 k and m = 2 r + 1 for some k, r Z . Then n + m = (2 k ) + (2 r + 1) = 2( k + r ) + 1 = 2 q + 1 for q = k + r Z , hence n + m is odd integer. 1.7.2 Indirect proof Here we will look into three types of indirect proofs such as 25
Image of page 28
Proof by contradiction. Proof by contrapositive. Proof by counter-example. Proof by Contradiction In order to prove the statement P Q using contradiction, we prove that the statement is true by showing that it cannot possibly be false. That is, we assume a statement is false and then arrive at some contradiction. It is easy to show that ( P Q ) ≡ ¬ P Q. For us to prove using contradiction we assume ( P Q ) is false i.e. ¬ ( P Q ) is true. Also we can show that ¬ ( P Q ) P ∧ ¬ Q . So in order to prove that ( P Q ) using contradiction, we must assume P ∧¬ Q and therefore arrive at some contradiction. Example 1.7.4. Prove by contradiction that if x R and x 2 = 2, then x is irrational. Proof. We need to prove that P Q where P : x R and x 2 = 2. Q : x is irrational. Assume P ∧ ¬ Q i.e. Let x 2 = 2 and assume that x is a rational number, i.e. x = a b where a, b Z with b 6 = 0 and a and b do not have common divisor . x 2 = ( a b ) 2 = a 2 b 2 . Since x 2 = 2, then a 2 b 2 = 2 and then a 2 = 2 b 2 , hence a 2 is an even number. The only way this can happen is if a itself is even integer, i.e. a = 2 k for some k Z . Then a 2 = (2 k ) 2 = 4 k 2 . So 4 k 2 = 2 b 2 , hence 2 k 2 = b 2 . This implies that b 2 is even hence b is also even, i.e. b = 2 r for some r Z . But this contradicts a and 26
Image of page 29
b not having a common divisor since in this case we have both a and b divisible by 2. So our assumption was false, hence x is irrational number. Proof by contrapositive To prove the statement P Q by contrapositive, we must assume ¬ Q and deduce ¬ P , hence ¬ Q ⇒ ¬ P .
Image of page 30
Image of page 31

You've reached the end of your free preview.

Want to read all 151 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes