Prove: $|xy|=|x|\cdot |y|$

Question

Can you help me to prove this equality? $$|xy|=|x|\cdot |y|$$

I don't know how to start proving the equality.

Answer 1

Using the fact that $\lvert a \rvert = \sqrt{a^2}$ for all $a \in \mathbb{R}$, we obtain $$ \begin{array}{rcl} \lvert xy \rvert &=& \sqrt{ (xy)^2 } \\ &=& \sqrt{ x^2 \cdot y^2 } \\ &=& \sqrt{ x^2 } \cdot \sqrt{y^2} \\ &=& \lvert x \rvert \cdot \lvert y \rvert \end{array} $$ for all $x,y \in \mathbb{R}$.

Answer 2

If $x$ and $y$ are both positive it is obviously true, and the case of one or both being zero is equally trivial.

If one of them is negative (say $y$), you have $$|xy|=x(-y)=x|y|=|x||y|.$$

If both are negative, you have $$|xy|=(-x)(-y)=|x||y|,$$

which exhausts the possibilities.

Answer 3

There are three cases to consider depending whether each of $x$ and $y$ is negative, zero, or positive. First, if either $x=0$ or $y=0$, we have $$ |xy| = 0 = |x||y| \text{.} $$

Then, if both $x$ and $y$ have the same sign, $xy > 0$ and either \begin{align} x &< 0, y < 0, \text{ so } & |xy| &= xy = (-x)(-y) = |x||y| \text{ or} \\ x &> 0, y > 0, \text{ so } & |xy| &= xy = (x)(y) = |x||y| \text{.} \end{align}

Finally, if $x$ and $y$ have opposite signs, $xy < 0$ and either \begin{align} x &< 0, y > 0, \text{ so } & |xy| &= -(xy) = (-x)(y) = |x||y| \text{ or} \\ x &> 0, y < 0, \text{ so } & |xy| &= -(xy) = (x)(-y) = |x||y| \text{.} \end{align}