# Real Multiplication Distributes over Addition

## Theorem

The operation of multiplication on the set of real numbers $\R$ is distributive over the operation of addition:

$\forall x, y, z \in \R:$
$x \times \paren {y + z} = x \times y + x \times z$
$\paren {y + z} \times x = y \times x + z \times x$

## Algebraic Proof

From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.

Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}, z = \eqclass {\sequence {z_n} } {}$, where $\eqclass {\sequence {x_n} } {}$, $\eqclass {\sequence {y_n} } {}$ and $\eqclass {\sequence {z_n} } {}$ are such equivalence classes.

From the definition of real multiplication, $x \times y$ is defined as:

$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$

From the definition of real addition, $x + y$ is defined as:

$\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$

Thus:

 $\ds x \times \paren {y + z}$ $=$ $\ds \eqclass {\sequence {x_n} } {} \times \paren {\eqclass {\sequence {y_n} } {} + \eqclass {\sequence {z_n} } {} }$ $\ds$ $=$ $\ds \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n + z_n} } {}$ $\ds$ $=$ $\ds \eqclass {\sequence {x_n \times \paren {y_n + z_n} } } {}$ $\ds$ $=$ $\ds \eqclass {\sequence {\paren {x_n \times y_n} + \paren {x_n \times z_n} } } {}$ Rational Multiplication Distributes over Addition $\ds$ $=$ $\ds \eqclass {\sequence {x_n \times y_n} } {} + \eqclass {\sequence {x_n \times z_n} } {}$ $\ds$ $=$ $\ds \paren {\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} } + \paren {\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {z_n} } {} }$ $\ds$ $=$ $\ds \paren {x \times y} + \paren {x \times z}$

By Real Addition is Commutative and Real Multiplication is Commutative, it follows that:

$\paren {y + z} \times x = \paren {y \times x} + \paren {z \times x}$

$\blacksquare$

## Geometric Proof

In the words of Euclid:

If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments.

Let $A$ and $BC$ be two straight lines.

Let $BC$ be cut at random at points $D$ and $E$.

Then the rectangle contained by $A$ and $BC$ is equal to the sum of the rectangles contained by $A$ and $BD$, by $A$ and $DE$, and by $A$ and $EC$, as follows:

Construct $BF$ perpendicular to $BC$.

Then we have that:

$\Box BCHG = \Box BDKG + \Box DELK + \Box ECHL$

Now $\Box BCHG$ is the rectangle contained by $A$ and $BC$, because it is contained by $BC$ and $BG$, and $BG = A$.

Similarly, from Opposite Sides and Angles of Parallelogram are Equal:

$\Box BDKG$ is the rectangle contained by $A$ and $BD$, because it is contained by $BD$ and $BG = A$
$\Box DEKL$ is the rectangle contained by $A$ and $DE$, because it is contained by $DE$ and $DK$, and $DK = A$
$\Box ECHL$ is the rectangle contained by $A$ and $EC$, because it is contained by $EC$ and $EL$, and $EL = A$.

Hence the result.

$\blacksquare$