Absolute Value Function is Completely Multiplicative

From ProofWiki
Jump to navigation Jump to search

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 1

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.

$\blacksquare$


Proof 2

Let $x$ and $y$ be considered as complex numbers which are wholly real.

That is:

$x = x + 0 i, y = y + 0 i$

From Complex Modulus of Real Number equals Absolute Value, the absolute value of $x$ and $y$ equal the complex moduli of $x + 0 i$ and $y + 0 i$.

Thus $\cmod x \cmod y$ can be interpreted as the complex modulus of $x$ multiplied by the complex modulus of $y$.

By Complex Modulus of Product of Complex Numbers:

$\cmod x \cmod y = \cmod {x y}$

As $x$ and $y$ are both real, so is $x y$.

Thus by Complex Modulus of Real Number equals Absolute Value, $\cmod {x y}$ can be interpreted as the absolute value of $x y$ as well as its complex modulus.

$\blacksquare$


Proof 3

\(\ds \size {x y}\) \(=\) \(\ds \sqrt {\paren {x y}^2}\) Definition 2 of Absolute Value
\(\ds \) \(=\) \(\ds \sqrt {x^2 y^2}\)
\(\ds \) \(=\) \(\ds \sqrt {x^2} \sqrt{y^2}\)
\(\ds \) \(=\) \(\ds \size x \cdot \size y\) Definition 2 of Absolute Value

$\blacksquare$


Proof 4

Follows directly from:

Real Numbers form Ordered Integral Domain
Product of Absolute Values on Ordered Integral Domain.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Absolute_Value_Function_is_Completely_Multiplicative&oldid=736907"