Question

verify that the sum of three rational numbers 1/2, 2/3 and 5/4 remains the same even after changing the grouping.​

Answer 1

$\underline{\underline{\underline{\underline{\mathfrak{\blue{ \:\:\:Solution:-\:\:\:}}}}}}$

$\sf {\bigg(\dfrac{1}{2} + \dfrac{2}{3}\bigg) + \dfrac{5}{4} \: \leftarrow \bigg(First\:add\: \dfrac{1}{2}\: and\: \dfrac{2}{3} \bigg)}$

$\longrightarrow \sf {\bigg(\dfrac{1 \times 3}{2 \times 3} + \dfrac{2 \times 2}{3 \times 2}\bigg) + \dfrac{5}{4}}$

$\longrightarrow \sf {\bigg(\dfrac{3}{6} + \dfrac{4}{6}\bigg) + \dfrac{5}{4}}$

$\longrightarrow \sf {\bigg(\dfrac{3 + 4}{6} \bigg) + \dfrac{5}{4}}$

$\longrightarrow \sf {\dfrac{7}{6} + \dfrac{5}{4}}$

$\longrightarrow \sf {\dfrac{7 \times 2}{6 \times 2} + \dfrac{5 \times 3}{4 \times 3}}$

$\longrightarrow \sf {\dfrac{14}{12} + \dfrac{15}{12}}$

$\longrightarrow \sf {\dfrac{14 + 15 }{12} }$

$\longrightarrow \sf {\dfrac{29}{12} }$

_____________________

$\sf {\dfrac{1}{2} + \bigg(\dfrac{2}{3} + \dfrac{5}{4}\bigg) \: \leftarrow (Grouping\:is\: changed)}$

$\longrightarrow \sf {\dfrac{1}{2} + \bigg(\dfrac{2 \times 4}{3 \times 4} + \dfrac{5 \times 3}{4 \times 3}\bigg) }$

$\longrightarrow \sf {\dfrac{1}{2} + \bigg(\dfrac{8}{12} + \dfrac{15}{12}\bigg) }$

$\longrightarrow \sf {\dfrac{1}{2} + \bigg(\dfrac{8 + 15}{12} \bigg) }$

$\longrightarrow \sf {\dfrac{1}{2} + \dfrac{23}{12}}$

$\longrightarrow \sf {\dfrac{1 \times 6}{2 \times 6} + \dfrac{23}{12}}$

$\longrightarrow \sf {\dfrac{6}{12} + \dfrac{23}{12}}$

$\longrightarrow \sf {\dfrac{6 + 23}{12} }$

$\longrightarrow \sf {\dfrac{29}{12} }$

• Hence Verified :)

Answer 2

$\huge\purple{ \:\:\:\:\:\:\underline{\underline{ \blue{✰ᗩᑎՏᗯᗴᖇ✰}}}}$

Solution:−

\begin{gathered}\sf {\bigg(\dfrac{1}{2} + \dfrac{2}{3}\bigg) + \dfrac{5}{4} \: \leftarrow \bigg(First\:add\: \dfrac{1}{2}\: and\: \dfrac{2}{3} \bigg)}\\\\\end{gathered}

(

2

1

+

3

2

)+

4

5

←(Firstadd

2

1

and

3

2

)

\begin{gathered}\longrightarrow \sf {\bigg(\dfrac{1 \times 3}{2 \times 3} + \dfrac{2 \times 2}{3 \times 2}\bigg) + \dfrac{5}{4}}\\\\\end{gathered}

⟶(

2×3

1×3

+

3×2

2×2

)+

4

5

\begin{gathered}\longrightarrow \sf {\bigg(\dfrac{3}{6} + \dfrac{4}{6}\bigg) + \dfrac{5}{4}}\\\\\end{gathered}

⟶(

6

3

+

6

4

)+

4

5

\begin{gathered}\longrightarrow \sf {\bigg(\dfrac{3 + 4}{6} \bigg) + \dfrac{5}{4}}\\\\\end{gathered}

⟶(

6

3+4

)+

4

5

\begin{gathered}\longrightarrow \sf {\dfrac{7}{6} + \dfrac{5}{4}}\\\\\end{gathered}

⟶

6

7

+

4

5

\begin{gathered}\longrightarrow \sf {\dfrac{7 \times 2}{6 \times 2} + \dfrac{5 \times 3}{4 \times 3}}\\\\\end{gathered}

⟶

6×2

7×2

+

4×3

5×3

\begin{gathered}\longrightarrow \sf {\dfrac{14}{12} + \dfrac{15}{12}}\\\\\end{gathered}

⟶

12

14

+

12

15

\begin{gathered}\longrightarrow \sf {\dfrac{14 + 15 }{12} }\\\\\end{gathered}

⟶

12

14+15

\begin{gathered}\longrightarrow \sf {\dfrac{29}{12} }\\\end{gathered}

⟶

12

29

_____________________

\begin{gathered}\sf {\dfrac{1}{2} + \bigg(\dfrac{2}{3} + \dfrac{5}{4}\bigg) \: \leftarrow (Grouping\:is\: changed)}\\\\\end{gathered}

2

1

+(

3

2

+

4

5

)←(Groupingischanged)

\begin{gathered}\longrightarrow \sf {\dfrac{1}{2} + \bigg(\dfrac{2 \times 4}{3 \times 4} + \dfrac{5 \times 3}{4 \times 3}\bigg) }\\\\\end{gathered}

⟶

2

1

+(

3×4

2×4

+

4×3

5×3

)

\begin{gathered}\longrightarrow \sf {\dfrac{1}{2} + \bigg(\dfrac{8}{12} + \dfrac{15}{12}\bigg) }\\\\\end{gathered}

⟶

2

1

+(

12

8

+

12

15

)

\begin{gathered}\longrightarrow \sf {\dfrac{1}{2} + \bigg(\dfrac{8 + 15}{12} \bigg) }\\\\\end{gathered}

⟶

2

1

+(

12

8+15

)

\begin{gathered}\longrightarrow \sf {\dfrac{1}{2} + \dfrac{23}{12}}\\\\\end{gathered}

⟶

2

1

+

12

23

\begin{gathered}\longrightarrow \sf {\dfrac{1 \times 6}{2 \times 6} + \dfrac{23}{12}}\\\\\end{gathered}

⟶

2×6

1×6

+

12

23

\begin{gathered}\longrightarrow \sf {\dfrac{6}{12} + \dfrac{23}{12}}\\\\\end{gathered}

⟶

12

6

+

12

23

\begin{gathered}\longrightarrow \sf {\dfrac{6 + 23}{12} }\\\\\end{gathered}

⟶

12

6+23

\begin{gathered}\longrightarrow \sf {\dfrac{29}{12} }\\\end{gathered}

⟶

12

29

Hence Verified :)