## The operation of sorting fractions in ascending order:

^{- 25}/_{52}, ^{- 28}/_{42}, ^{- 40}/_{53}

### Analyze the fractions to be compared and ordered, by category:

#### negative proper fractions: - ^{25}/_{52}, - ^{28}/_{42}, - ^{40}/_{53};

### Reduce (simplify) fractions to their lowest terms equivalents:

#### - ^{25}/_{52} already reduced to the lowest terms;

the numerator and denominator have no common prime factors:

25 = 5^{2};

52 = 2^{2} × 13;

#### - ^{28}/_{42} = - ^{(22 × 7)}/_{(2 × 3 × 7)} = - ^{((22 × 7) ÷ (2 × 7))}/_{((2 × 3 × 7) ÷ (2 × 7))} = - ^{2}/_{3}

#### - ^{40}/_{53} already reduced to the lowest terms;

the numerator and denominator have no common prime factors:

40 = 2^{3} × 5;

53 is a prime number;

## To sort fractions, build them up to the same numerator.

### Calculate LCM, the least common multiple of the fractions' numerators

#### LCM will be the common numerator of the compared fractions.

#### The prime factorization of the numerators:

#### 25 = 5^{2}

#### 2 is a prime number

#### 40 = 2^{3} × 5

#### Multiply all the unique prime factors, by the largest exponents:

#### LCM (25, 2, 40) = 2^{3} × 5^{2} = 200

### Calculate the expanding number of each fraction

#### Divide LCM by the numerator of each fraction:

#### For fraction: - ^{25}/_{52} is 200 ÷ 25 = (2^{3} × 5^{2}) ÷ 5^{2} = 8

#### For fraction: - ^{2}/_{3} is 200 ÷ 2 = (2^{3} × 5^{2}) ÷ 2 = 100

#### For fraction: - ^{40}/_{53} is 200 ÷ 40 = (2^{3} × 5^{2}) ÷ (2^{3} × 5) = 5

### Expand the fractions

#### Build up all the fractions to the same numerator (which is LCM).

Multiply the numerators and denominators by their expanding number:

#### - ^{25}/_{52} = - ^{(8 × 25)}/_{(8 × 52)} = - ^{200}/_{416}

#### - ^{2}/_{3} = - ^{(100 × 2)}/_{(100 × 3)} = - ^{200}/_{300}

#### - ^{40}/_{53} = - ^{(5 × 40)}/_{(5 × 53)} = - ^{200}/_{265}

### The fractions have the same numerator, compare their denominators.

#### The larger the denominator the larger the negative fraction.

## ::: Comparing operation :::

The final answer: