Examples: \[ \begin{array}{rcl}{2/3}&=&\frac{2}{3}\\ -6/4&=&-\frac{6}{4}=\frac{-6}{4}\\7&=&7/1=\frac{7}{1}\end {array}\]

Because division by #0# is impossible, the denominator should not be #0#.

In formulas: If #a#, #b#, #c#, and #d# are real numbers with #b\ne0# and #d\ne0#, then

1. #\frac{a}{b}=\frac{a\cdot d}{b\cdot d}#
2. #\frac{a}{b}=\frac{c}{d}# if and only if #a\cdot d = b\cdot c#

Examples: \[ \begin{array}{rcl}\frac{15}{-10}&=&\frac{-3\cdot(-5)}{2\cdot(-5)}=\frac{-3}{2}\\ \frac{15}{-10}&=&\frac{-9}{6}\end {array}\]

The fraction \(\frac{-4}{-6}\) for example, can be simplified to \(\frac{2}{3}\) by dividing the numerator and denominator by #-2#.

It goes without saying that two rational numbers with the same fraction are the same.

Conversely: assume #\frac{a}{b}# and #\frac{c}{d}# to be two simplified fractions, where #a#, #b#, #c# and #d# are numbers with #b# and #d# unequal to #0#, and assume that they are the same: #\frac{a}{b}=\frac{c}{d}#. We show that numerator and denominator are also equal in that case; meaning: #a=c# and #b=d#.

The property of fractions shows that #a\cdot d= b\cdot c#. Since #\frac{a}{b}# and #\frac{c}{d}# are simplified, we have #\gcd(a,b)=\gcd(c,d)#. So no divisor of #a# is a divisor of #b#. Because #a# is a divisor of #b\cdot c# (after all, #a\cdot d= b\cdot c#), #a# must be a divisor of #c#. Similarly, #c# must be a divisor of #a#. It follows that #a=c# or #a=-c#.

If #a=0#, then #c=0#. In other words, the numerators are #0#. From the definition of simplified fraction, it follows that the corresponding denominators are equal to #1#, so #b=d=1#. We conclude that #a=c# and #b=d#, which is what we needed to prove.

Now suppose #a\ne0#. Then #c\ne0# also holds. If #a=-c#, then from #a\cdot d= b\cdot c# it follows that #b=-d#. But this contradicts the requirement that the denominators #b# and #d# of simplified fractions are both positive. Consequently, #a=c#. But #a\cdot d= b\cdot c# also implies #b=d#, which is what we needed to prove.