Multiplicative Inverse

Two numbers are multiplicative inverses when their multiplication results in $1$.

For example:

${\Large {1 \over 2}}$ and $2$ are multiplicative inverses because ${\Large 2 \cdot {1 \over 2}=1}$

More examples:

The multiplicative inverse of $5$ is ${\Large {1 \over 5}}$
${\Large 5 \cdot {1 \over 5}=1}$

The multiplicative inverse of $3$ is ${\Large {1 \over 3}}$
${\Large 3 \cdot {1 \over 3}=1}$

The multiplicative inverse of ${\Large {5 \over 7}}$ is ${\Large {7 \over 5}}$
${\Large {7 \over 5} \cdot {5 \over 7}=1}$

The multiplicative inverse of ${\Large {9 \over 23}}$ is ${\Large {23 \over 9}}$
${\Large {23 \over 9} \cdot {9 \over 23}=1}$

The multiplicative inverse of $0.5$ is $2$

${\Large 2 \cdot 0.5=1}$

The multiplicative inverse of $0.25$ is $4$

${\Large 4 \cdot 0.25=1}$

Examples with negative numbers:

The multiplicative inverse of $-3$ is ${\Large -{1 \over 3}}$ because ${\Large -3 \cdot \left(-{1 \over 3}\right)=1}$

The multiplicative inverse of ${\Large -{2 \over 5}}$ is${\Large -{5 \over 2}}$ because ${\Large -{5 \over 2} \cdot \left(-{2 \over 5}\right)=1}$

Whenever a is different from $0$, it happens that ${\Large a\cdot{1 \over a} = 1}$

Why can't zero have a multiplicative inverse?

Zero $0 )$ has no multiplicative inverse because ${\Large {1 \over 0}}$ is undefined. No number multiplied by $0 )$ can equal $1 )$, since any number times $0 )$ always equals $0 )$.

Alternative notation: The multiplicative inverse can also be written using exponent notation: $a^{-1} = {\Large {1 \over a}}$

Division is equivalent to multiplication by the multiplicative inverse,

that is:  ${\Large {{2 \over {1 \over 3}} = 2 \cdot 3 = 6}}$

This is because $3$ is the multiplicative inverse of  ${\Large {1 \over 3}}$

In general:  ${\Large a /{1 \over b} = a \cdot b}$

Why does this work?

Dividing by a fraction is the same as multiplying by its reciprocal. When you divide by ${\Large {1 \over b}}$, you're essentially asking "how many ${\Large {1 \over b}}$'s fit into $a$?" The answer is $a \cdot b$.

Solve the following exercises

• ${\Large {5+{{4 \over 7} \over 2} =}}$
• ${\Large {{6 \over 0.75} - 2 \cdot 3 =}}$
• ${\Large {{3{1 \over 2}-{{1 \over 3} \over {1 \over 6}}}=}}$
• ${\Large {{{{10 \over 7} \over 2 } + {2 \over {7 \over 8} }}=}}$
• ${\Large {{{3 \over 5} \over {9 \over 10}} + {{7 \over 9} \over {1 \over 3}}=}}$