Question

Prove the following statements. These exercises are cumulative, covering all techniques addressed in Chapters \(4-7\). Suppose \(x \in \mathbb{Z} .\) Then \(x\) is even if and only if \(3 x+5\) is odd.

Short answer

The statement 'if \(x\) is an even integer then \(3x+5\) is odd' and 'if \(3x+5\) is odd, then \(x\) is an even integer' have been proven true. So, the given statement is correct: \(x\) is even if and only if \(3x+5\) is odd.

Step-by-step solution

Proof for the forward condition

Assume \(x\) is an even integer. According to the definition of even numbers, \(x\) can be expressed as \(2n\) where \(n\) is an element of integers. So we replace \(x\) by \(2n\) in the expression \(3x+5\) which becomes \(3(2n) + 5 = 6n+5\). As \(6n\) is an even number, \(6n+5\) would be odd since the sum of an even number and an odd number (5 in this case) always results in an odd number. Thus we have proven that if \(x\) is even, then \(3x+5\) is odd.

Proof for the reverse condition

Assume \(3x+5\) is odd. Expressing it as \(3x+5 = 2m+1\) where \(m\) is an integer as per the definition of odd numbers. From this equation, we can isolate \(x\) to obtain \(x = (2m+1 - 5) / 3 = (2m-4) / 3 = 2(m -2) / 3 \). As seen here, \(x\) can be expressed as twice a quantity which means \(x\) is even. So, if \(3x + 5\) is odd, then \(x\) is even.