Absolute Value Function is Completely Multiplicative/Proof 3

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

\(\ds \size {x y}\)

\(=\)

\(\ds \sqrt {\paren {x y}^2}\)

Definition 2 of Absolute Value

\(\ds \)

\(=\)

\(\ds \sqrt {x^2 y^2}\)

\(\ds \)

\(=\)

\(\ds \sqrt {x^2} \sqrt{y^2}\)

\(\ds \)

\(=\)

\(\ds \size x \cdot \size y\)

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: $\S 1.16$