Paragraph

Let \(x\) and \(y\) be elements of the domain \(\Z\text{.}\) Assume \(f(x) = f(y)\text{.}\) If \(x\) and \(y\) are both even, then \(f(x) = x+1\) and \(f(y) = y+1\text{.}\) Since \(f(x) = f(y)\text{,}\) we have \(x + 1 = y + 1\) which implies that \(x = y\text{.}\) Similarly, if \(x\) and \(y\) are both odd, then \(x - 3 = y-3\) so again \(x = y\text{.}\) The only other possibility is that \(x\) is even an \(y\) is odd (or visa-versa). But then \(x + 1\) would be odd and \(y - 3\) would be even, so it cannot be that \(f(x) = f(y)\text{.}\) Therefore if \(f(x) = f(y)\) we then have \(x = y\text{,}\) which proves that \(f\) is injective.