# Fractions

If $p$ and $q$ are algebraic expressions, the quotient, or ratio, $p/q$ is called a fractional expression (or simply a fraction) with numerator $p$ and denominator $q$. Always remember that the denominator of a fraction cannot be zero. If $q = 0$, the expression $\frac{p}{q}$ has simply no meaning. Therefore, whenever we use a factional expression, we shall automatically assume that the variables involved are restricted to numerical values that will give a nonzero denominator.

A fractional expression in which both the numerator and the denominator are polynomials is called a rational expression. Examples of rational expression are $\frac{{2x}}{1}$,   $\frac{3}{y}$,   $\frac{{5x + 3}}{{17}}$, $\frac{{2st – {t^3}}}{{3{s^4} – {t^4}}}$,   and   $\frac{{{x^2} – 1}}{{{y^2} – 1}}$.

Notice that a fraction such as $3x/\sqrt y$ is not a rational expression because its denominator is not polynomial. We say that a rational is reduced to lower terms or simplified if its numerator and denominator have no common factors (other than $1$or$– 1$). Thus, to simplify a rational expression, we factor both the numerator and the denominator into prime factors and then cancel common factors by using the following property.

The Cancellation Property for Fractions

If $q \ne 0$ and $k \ne 0$, then $\frac{{pk}}{{qk}} = \frac{p}{q}$
cancellation is usually indicated by slanted lines drawn through the canceled factors; for instance,
$\frac{{{x^2} – 1}}{{{x^2} – 3x + 2}} = \frac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)}} = \frac{{x + 1}}{{x – 2}}$

If one fraction can be obtained from another by canceling common factors or by multiplying numerator and denominator by the same nonzero expression, then the two fractions are said to be equivalent. Thus, the calculation above shows that $\frac{{{x^2} – 1}}{{{x^2} – 3x + 2}}$   and   $\frac{{x – 1}}{{x – 2}}$ are equivalent fractions.

Example:

Reduce each fraction to its lowest terms.

(a) $\frac{{14{x^7}y}}{{56{x^3}{y^2}}}$
(b) $\frac{{cd – {c^2}}}{{{c^2} – {d^2}}}$
(c) $\frac{{5{x^2} – 14x – 3}}{{2{x^2} + x – 21}}$
(d) $\frac{{16x – 32}}{{8y\left( {x – 2} \right) – 4\left( {x – 2} \right)}}$

Solution:

(a) $\frac{{14{x^7}y}}{{56{x^3}{y^2}}} = \frac{{14{x^3}y \cdot {x^4}}}{{14{x^3}y \cdot 4y}} = \frac{{{x^4}}}{{4y}}$

(b) First, we factor the numerator and denominator, and then we use the fact that $d – c = – \left( {c – d} \right)$:
$\frac{{cd – {c^2}}}{{{c^2} – {d^2}}} = \frac{{c\left( {d – c} \right)}}{{\left( {c – d} \right)\left( {c + d} \right)}} = \frac{{ – c\left( {c – d} \right)}}{{\left( {c – d} \right)\left( {c + d} \right)}} = \frac{{ – c}}{{c + d}}$

(c) $\frac{{5{x^2} – 14x – 3}}{{2{x^2} + x – 21}} = \frac{{\left( {5x + 1} \right)\left( {x – 3} \right)}}{{\left( {2x + 7} \right)\left( {x – 3} \right)}} = \frac{{5x + 1}}{{2x + 7}}$

(d) $\frac{{16x – 32}}{{8y\left( {x – 2} \right) – 4\left( {x – 2} \right)}} = \frac{{16\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {8y – 4} \right)}}$
$= \frac{{\left( 4 \right)\left( 4 \right)\left( {x – 2} \right)}}{{4\left( {x – 2} \right)\left( {2y – 1} \right)}} = \frac{4}{{2y – 1}}$