\(|ab| = |a|\cdot|b|\) holds for all real numbers \(a\) and \(b\).


For the absolute value function definition and other properties see here.

Proof:

Given two real numbers \(a\) and \(b\). Consider the following cases:

  • \(a \geq 0, b \geq 0\):
    Since \(a \geq 0\), then \(|a| = a\). Similarly, since \(b \geq 0\) then \(|b| = b\). Therefore, \(|a||b| = ab\). But we also know that \(ab \geq 0\) since \(a \geq 0\) and \(b \geq 0\). Therefore \(|ab| = ab = |a||b|\).
  • \(a < 0, b < 0\):
    Since both \(a\) and \(b\) are less than zero then we know that their product must be positive. Therefore, \(|ab| = ab\). But since \(a < 0\) and \(b < 0\), then \(|a| = -a\) and \(|b| = -b\). So \(|a||b| = -a(-b) ab\). So \(|a||b| = ab = |ab|\) as required.
  • \(a < 0, b \geq 0\) or \(a \geq 0, b < 0\):
    Without the loss of generality suppose that \(a \geq 0\) and \(b < 0\). Since \(a \geq 0\), then \(|a| = a\) and since \(b < 0\), then \(|b| = -b\). Therefore, \(|a||b| = -ab\). But we know that the product of a positive number and a negative number is negative. So \(|ab| = -ab\). so \(|ab| = |a||b|\)

And we’re done! \(\blacksquare\)

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