### Theory:

Unlike fractions are fractions with different denominators.

Let us see how we can order the unlike fractions with the same numerator and compare the unlike fractions with different denominator.

**Ordering of unlike fractions with the same numerator**:

$\frac{6}{13},\phantom{\rule{0.147em}{0ex}}\frac{6}{14},\phantom{\rule{0.147em}{0ex}}\frac{6}{15}$ are unlike fractions with the same numerator.

The fractions can be ordered as, $\frac{6}{15}<\frac{6}{14}<\frac{6}{13}$ because the unlike fraction with the highest denominator will be the smallest among unlike fractions with the same numerator.

**$\frac{4}{5},\frac{2}{3},\frac{6}{7}$. To order unlike fractions with different numerators, we have to convert all the unlike fractions to like fractions. To do that follow the steps below:**

**Comparison of unlike fractions with different numerator:**

**i)**Take LCM (Least Common Multiple) of all the denominators.

**ii)**Make the denominators of all the fractions into LCM.

**iii)**Compare the numerators of the like fractions, the fraction with the greatest numerator will be the greatest fraction.

Example:

Find the greatest fraction among $\frac{4}{5},\frac{2}{3},\frac{6}{7}$

**i)**LCM of all the denominators \(5, 3, 7\) is \(105\).

**ii)**$\frac{4\times 21}{5\times 21},\frac{2\times 35}{3\times 35},\frac{6\times 15}{7\times 15}$ \(=\) $\frac{84}{105},\frac{70}{105},\frac{90}{105}$

- To convert \(5\) into \(105\), \(105 ÷ 5 = 21\), multiply \(4/5\) with \(21\) in both numerator and denominator to get \(105\) as the denominator.
- To convert \(3\) into \(105\), \(105 ÷ 3 = 35\), multiply \(2/3\) with \(35\) in both numerator and denominator to get \(105\) as the denominator.
- To convert \(7\) into \(105\), \(105 ÷ 7 = 15\), multiply \(6/7\) with \(15\) in both numerator and denominator to get \(105\) as the denominator.

**iii)**Compare the numerators of like fractions, \(84\), \(70\) and \(90\). Where \(90 > 84 > 70\) so $\frac{90}{105}>\frac{84}{105}>\frac{70}{105}$.

The corresponding fractions are, $\frac{6}{7}>\frac{4}{5}>\frac{2}{3}$.

**Hence, the greatest fraction among**$\frac{4}{5},\frac{2}{3},\frac{6}{7}$

**is**$\frac{6}{7}$.