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\title{Latex Example 2}
\author{Karen Mortensen and revised by GF 28dec09}
\maketitle
\textbf{ Theorem.}\textit{ For all real numbers $x$ and $y$, \[|xy|=|x||y|\].}

\textbf{ Proof.} To prove this, first suppose that $x \ge 0$ and $y \ge 0$.
Then $xy \ge 0$.
By definition of absolute value, $|xy| = xy$, $|x| = x$ and $|y|=y$.
Therefore $|xy| = |x| |y|$.

Next suppose that $x \ge 0$ and $y < 0$.  Then $xy \le 0$.  By definition,
$|xy| = -(xy)$, $|x| = x$ and $|y| = -y$.  Since $ -(xy) = x(-y)$, we conclude
that $|xy| = |x| |y|$.

Next suppose that $x < 0$ and $y \ge 0$. The argument from the previous
paragraph, with the roles of $x$ and $y$ reversed, shows that $|xy| = |x| |y|$.

Finally, suppose that $x < 0$ and $y < 0$.  Then $xy >0$.  By definition,
$|xy| = xy$, $|x| = -x$ and $|y| = -y$.  Since $xy = (-x) (-y)$, we conclude
that $|xy| = |x| |y|$.

Having considered all possible cases for the signs of $x$ and $y$, we have
proved that $|xy| = |x| |y|$.
\hfill q.e.d.

This proof is of importance because it establishes that absolute value
commutes with multiplication, a commonly used property of absolute value.
The proof illustrates the strategy of dividing a mathematical statement into
several cases and proving each case separately.  Once the statement was broken
tiny bit of algebra.  This demonstrates the importance of going back to
definitions when constructing proofs.

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