Absolute Value of Product

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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 $x$.


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:

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

and:

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


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

Then:

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

and:

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


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:

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

and:

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

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.

$\blacksquare$


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