Absolute Value of Product/Proof 1

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Theorem

Let $x, y \in \R$ be real numbers.


Then:

$\size {x y} = \size x \size y$

where $\size x$ denotes the absolute value of $x$.


Thus the absolute value function is completely multiplicative.


Proof

Let either $x = 0$ or $y = 0$, or both.

We have that $\size 0 = 0$ by definition of absolute value.

Hence:

$\size x \size y = 0 = x y = \size {x y}$


Let $x > 0$ and $y > 0$.

Then:

\(\ds x y\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {x y}\) \(=\) \(\ds x y\) Definition of Absolute Value

and:

\(\ds x\) \(=\) \(\ds \size x\) Definition of Absolute Value
\(\ds y\) \(=\) \(\ds \size y\)
\(\ds \leadsto \ \ \) \(\ds \size x \size y\) \(=\) \(\ds x y\)
\(\ds \) \(=\) \(\ds \size {x y}\)


Let $x < 0$ and $y < 0$.

Then:

\(\ds x y\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {x y}\) \(=\) \(\ds x y\) Definition of Absolute Value

and:

\(\ds -x\) \(=\) \(\ds \size x\) Definition of Absolute Value
\(\ds -y\) \(=\) \(\ds \size y\)
\(\ds \leadsto \ \ \) \(\ds \size x \size y\) \(=\) \(\ds \paren {-x} \paren {-y}\)
\(\ds \) \(=\) \(\ds xy\)
\(\ds \) \(=\) \(\ds \size {x y}\)


The final case is where one of $x$ and $y$ is positive, and one is negative.

Without loss of generality, let $x < 0$ and $y > 0$.

Then:

\(\ds x y\) \(<\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {x y}\) \(=\) \(\ds -\paren {x y}\) Definition of Absolute Value

and:

\(\ds -x\) \(=\) \(\ds \size x\) Definition of Absolute Value
\(\ds y\) \(=\) \(\ds \size y\)
\(\ds \leadsto \ \ \) \(\ds \size x \size y\) \(=\) \(\ds \paren {-x} y\)
\(\ds \) \(=\) \(\ds -\paren {x y}\)
\(\ds \) \(=\) \(\ds \size {x y}\)

The case where $x > 0$ and $y < 0$ is the same.

$\blacksquare$

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