Compare two fractions with different denominators

### 1.1. Knowledge to remember

**For example :** Compare two fractions \(\frac{2}{3}\) and \(\frac{3}{4}\).

a) Take two equal tapes. Divide the first paper tape into 3 equal parts, take 2 parts, i.e. take \(\frac{2}{3}\) paper tape. Divide the second paper tape into 4 equal parts, take 3 parts, i.e. take \(\frac{3}{4}\) paper tape.

Looking at the picture we see:

b) We can compare two fractions \(\frac{2}{3}\) and \(\frac{3}{4}\) as follows:

- Reduce the denominators of the two fractions \(\frac{2}{3}\) and \(\frac{3}{4}\) :

\(\frac{2}{3} = \frac{{2 \times 4}}{{3 \times 4}} = \frac{8}{{12}};\,\,\,\,\ ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{4} = \frac {{3 \times 3}}{{4 \times 3}} = \frac{9}{{12}}\)

- Compare two fractions with the same denominator:

\(\frac{8}{{12}} < \frac{9}{{12}}\) (because 8 < 9)

- Conclusion: \(\frac{2}{3} < \frac{3}{4}\).

*To compare two fractions with different denominators, we can reduce the denominators of the two fractions, and then compare the numerators of the two new fractions.*

### 1.2. Textbook exercise solution page 122

**Lesson 1: **Compare two fractions

a) \(\frac{3}{4}\) and \(\frac{4}{5}\) b) \(\frac{5}{6}\) and \(\frac{7}{ 8}\) c) \(\frac{2}{5}\) and \(\frac{3}{{10}}\).

__Solution guide:__

- To compare two fractions with different denominators, we can reduce the denominators of the two fractions, and then compare the numerators of the two new fractions.

a) Reducing the denominator of two fractions \(\frac{3}{4}\) and \(\frac{4}{5}\) :

\(\frac{3}{4} = \frac{{3 \times 5}}{{4 \times 4}} = \frac{{15}}{{20}};\,\,\,\ ,\,\,\frac{4}{5} = \frac{{4 \times 4}}{{5 \times 4}} = \frac{{16}}{{20}}\)

Since \(\frac{{15}}{{20}} < \frac{{16}}{{20}}\) \(\frac{3}{4} < \frac{4}{5} \).

b) Denominator of two fractions \(\frac{5}{6}\) and \(\frac{7}{8}\):

\(\frac{5}{6} = \frac{{5 \times 8}}{{6 \times 8}} = \frac{{40}}{{48}};\,\,\,\ ,\,\,\,\,\frac{7}{8} = \frac{{7 \times 6}}{{8 \times 6}} = \frac{{42}}{{48}}\ )

Since \(\frac{{40}}{{48}} < \frac{{42}}{{48}}\) \(\frac{5}{6} < \frac{7}{8} \).

c) Reduce the denominator to the fraction \(\frac{2}{5}\) and keep the fraction \(\frac{3}{{10}}\):

\(\frac{2}{5} = \frac{{2 \times 2}}{{5 \times 2}} = \frac{4}{{10}}\)

Since \(\frac{4}{{10}} > \frac{3}{{10}}\) \(\frac{2}{5} > \frac{3}{{10}}\) .

**Lesson 2:** Simplify and compare two fractions

\(\frac{6}{{10}}\) and \(\frac{4}{5}\) b) \(\frac{3}{4}\) and \(\frac{6}{ {twelfth}}\)

__Solution guide:__

- Reduce the given fractions to the simplest fraction (if possible).
- To compare two fractions with different denominators, we can reduce the denominators of the two fractions, and then compare the numerators of the two new fractions.

a) Reduce the fraction \(\frac{6}{{10}}\) and keep the fraction \(\frac{4}{5}\) :

\(\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5}\)

Since \(\frac{3}{5} < \frac{4}{5}\) \(\frac{6}{{10}} < \frac{4}{5}\) .

b) Reduce the fraction \(\frac{6}{{12}}\) and keep the fraction \(\frac{3}{4}\) :

\(\frac{6}{{12}} = \frac{{6:3}}{{12:3}} = \frac{2}{4}\)

Since \(\frac{3}{4} > \frac{2}{4}\) \(\frac{3}{4} > \frac{6}{{12}}\).

**Lesson 3: **Mai eats \(\frac{3}{8}\) the cake, Hoa eats \(\frac{2}{5}\) the cake. Who ate more cake?

__Solution guide:__

Denominator of two fractions:

\(\frac{3}{8} = \frac{{3 \times 5}}{{8 \times 5}} = \frac{{15}}{{40}};\,\,\,\ ,\,\,\,\,\frac{2}{5} = \frac{{2 \times 8}}{{5 \times 8}} = \frac{{16}}{{40}}. \)

Since \(\frac{{16}}{{40}} > \frac{{15}}{{40}}\) \(\frac{2}{5} > \frac{3}{8} \).

So Hoa is the one who eats more cakes.

### 1.3. Solve the exercises Textbook Practice page 122

**Lesson 1: **Compare two fractions

a) \(\frac{5}{8}\) and \(\frac{7}{8}\) b) \(\frac{{15}}{{25}}\) and \(\frac {4}{5}\) c) \(\frac{9}{7}\) and \(\frac{9}{8}\) d) \(\frac{{11}}{{20} }\) and \(\frac{6}{{10}}\)

__Solution guide:__

- Of two fractions with the same denominator, the fraction with the smaller numerator is the smaller fraction.
- Of two fractions with the same numerator, the fraction with the smaller denominator is the larger fraction.

a) Since 5 < 8 \(\frac{5}{8} < \frac{7}{8}\)

b) Simplify fractions: \(\frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5}\)

Since \(\frac{3}{5} < \frac{4}{5}\) \(\frac{{15}}{{25}} < \frac{4}{5}\).

c) Denominator of two fractions \(\frac{9}{7}\) and \(\frac{9}{8}\):

\(\frac{9}{7} = \frac{{9 \times 8}}{{7 \times 8}} = \frac{{72}}{{56}};\,\,\,\ ,\,\,\,\,\,\frac{9}{8} = \frac{{9 \times 7}}{{8 \times 7}} = \frac{{63}}{{56} }\).

Since \(\frac{{72}}{{56}} > \frac{{63}}{{56}}\) \(\frac{9}{7} > \frac{9}{8} \).

d) Denominator of two fractions \(\frac{{11}}{{20}}\) and \(\frac{6}{{10}}\) :

\(\frac{6}{{10}} = \frac{{6 \times 2}}{{10 \times 2}} = \frac{{12}}{{20}}\) ; keep the fraction \(\frac{{11}}{{20}}\)

Since \(\frac{{11}}{{20}} < \frac{{12}}{{20}}\) \(\frac{{12}}{{20}} < \frac{6) }{{ten}}\).

**Lesson 2: **Compare two fractions in two different ways

a) \(\frac{8}{7}\) and \(\frac{7}{8}\) b) \(\frac{9}{5}\) and \(\frac{5}{ 8}\) c) \(\frac{{12}}{{16}}\) and \(\frac{{28}}{{21}}\)

__Solution guide:__

- Method 1: Reduce the denominator of two fractions and then compare the two fractions after converging.
- Method 2: Compare two given fractions with 11.

a) Method 1: Reduce the denominator of two fractions \(\frac{8}{7}\) and \(\frac{7}{8}\)

\(\frac{8}{7} = \frac{{8 \times 8}}{{7 \times 8}} = \frac{{64}}{{56}};\,\,\,\ ,\,\frac{7}{8} = \frac{{7 \times 7}}{{8 \times 7}} = \frac{{49}}{{56}}\)

Since \(\frac{{64}}{{56}} > \frac{{49}}{{56}}\) \(\frac{8}{7} > \frac{7}{8} \).

Method 2: We have \(\frac{8}{7} > 1;\,\,\,\frac{7}{8} < 1\).

Therefore, \(\frac{8}{7} > \frac{7}{8}\).

b) Method 1: Reduce the denominator of two fractions \(\frac{9}{5}\) and \(\frac{5}{8}\)

\(\frac{9}{5} = \frac{{9 \times 8}}{{5 \times 8}} = \frac{{72}}{{40}};\,\,\,\ ,\,\,\frac{5}{8} = \frac{{5 \times 5}}{{8 \times 5}} = \frac{{25}}{{40}}\)

Since \(\frac{{72}}{{40}} > \frac{{25}}{{40}}\) \(\frac{9}{5} > \frac{5}{8} \).

Method 2: \(\frac{9}{5} > 1;\,\,\,\,\frac{5}{8} < 1\)

Therefore, \(\frac{9}{5} > \frac{5}{8}\)

c) Method 1: Reducing two fractions \(\frac{{12}}{{16}}\) and \(\frac{{28}}{{21}}\) we have

\(\frac{{12}}{{16}} = \frac{{12:4}}{{16:4}} = \frac{3}{4};\,\,\,\,\ ,\,\,\frac{{28}}{{21}} = \frac{{28:7}}{{21:7}} = \frac{4}{3}\)

Using the denominators of the two fractions \(\frac{3}{4}\) and \(\frac{4}{3}\) we have :

\(\frac{3}{4} = \frac{{3 \times 3}}{{4 \times 3}} = \frac{9}{{12}};\,\,\,\,\ ,\,\,\,\,\,\,\frac{4}{3} = \frac{{4 \times 4}}{{3 \times 4}} = \frac{{16}}{{ twelfth}}\)

Since \(\frac{9}{{12}} < \frac{{16}}{{12}}\) \(\frac{3}{4} < \frac{4}{3}\) .

Hence \(\frac{{12}}{{16}} < \frac{{28}}{{21}}\).

Method 2: \(\frac{{12}}{{16}} < 1;\,\,\,\frac{{28}}{{21}} > 1\)

Therefore, \(\frac{{12}}{{16}} > \frac{{28}}{{21}}\)

**Lesson 3: **Compare two fractions with the same numerator

a) Example: Compare \(\frac{4}{5}\) and \(\frac{4}{7}\)

We have: \(\frac{4}{5} = \frac{{4 \times 7}}{{5 \times 7}} = \frac{{28}}{{35}};\,\, \,\,\,\,\,\,\,\,\frac{4}{7} = \frac{{4 \times 5}}{{7 \times 5}} = \frac{{20} }{{35}}\)

Since 28 > 20, \(\frac{4}{5} > \frac{4}{7}\)

*Comment :*

Of two fractions with the same numerator, the fraction with the smaller denominator is the larger fraction.

b) Compare two fractions : \(\frac{9}{{11}}\) and \(\frac{9}{{14}}\) ; \(\frac{8}{9}\) and \(\frac{8}{{11}}\)

__Solution guide:__

- Of two fractions with the same numerator, the fraction with the smaller denominator is the larger fraction.

We have : 11 < 14 so \(\frac{9}{{11}} > \frac{9}{{14}}\);

9 < 11 so \(\frac{8}{9} > \frac{8}{{11}}\).

**Lesson 4: **Write fractions in order from smallest to largest

a) \(\frac{6}{7};\frac{4}{7};\frac{5}{7}\) b) \(\frac{2}{3};\frac{5} {6};\frac{3}{4}\)

__Solution guide:__

- Compare the given fractions and arrange them in order from smallest to largest.

a) We have: 4 < 5 < 6 so \(\frac{4}{7} < \frac{5}{7} > \frac{6}{7}\)

So the given fractions in order from smallest to largest are: 47;57;6747;57;67

b) Reducing the denominator of three fractions \(\frac{2}{3};\frac{5}{6};\frac{3}{4}\), choosing a common denominator of 12.

\(\begin{array}{l}

\frac{2}{3} = \frac{{2 \times 4}}{{3 \times 4}} = \frac{8}{{12}};\,\,\,\,\,\ ,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{6} = \frac{{5 \times 2}}{{6 \times 2}} = \frac{{10}}{{12}}\\

\frac{3}{4} = \frac{{3 \times 3}}{{4 \times 3}} = \frac{9}{{12}}.

\end{array}\)

Since \(\frac{8}{{12}} < \frac{9}{{12}} < \frac{{10}}{{12}}\) \(\frac{2}{3} < \frac{3}{4} < \frac{5}{6}\).

So the given fractions in order from smallest to largest are \(\frac{2}{3};\frac{3}{4};\frac{5}{6}\).

.

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