Question

Use the method of proof by contradiction to prove the following statements. (In each case, you should also think about how a direct or contrapositive proof would work. You will find in most cases that proof by contradiction is easier.) Suppose \(n \in \mathbb{Z} .\) If \(n\) is odd, then \(n^{2}\) is odd.

Short answer

The initial assumption that \(n^2\) is even while \(n\) is odd leads to a contradiction. Therefore, the original statement, 'if \(n\) is odd then \(n^2\) is odd' must be true.

Step-by-step solution

Define the Terms

An odd number can be defined as \(n = 2k+1\) where \(k\) is an integer. A square of a number is obtained by multiplying the number by itself.

Assume the Opposite

We are trying to prove: if \(n\) is odd then \(n^2\) is odd. Let's assume the opposite: that \(n\) is odd and \(n^2\) is even. An even number can be defined as \(m = 2l\), where \(m\) and \(l\) are integers.

Substitute and Simplify

Since we assumed \(n\) is odd, we know that \(n = 2k+1\), and since we assumed \(n^2\) is even, we know that \(n^2 = 2l\). Therefore, we can write the square as \((2k + 1)^2 = 2l\). Simplifying this gives us \(4k^2 + 4k + 1 = 2l\). Divide both sides by 2 to get \(2k^2 + 2k + 0.5 = l\).

Reach a Contradiction

We previously defined \(l\) as an integer. However, from our substitution and simplification, we get \(l = 2k^2 + 2k + 0.5\), which is not an integer because of the 0.5 term. This is a contradiction.