verify that the sum of three rational numbers 1/2, 2/3 and 5/4 remains the same even after changing the grouping.
Answers
_____________________
Hence Verified :)
Answer:
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 :)
Step-by-step explanation: