The following definitions will be important in most of our discussions

# The following definitions will be important in most

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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
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
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
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 .

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