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

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