for x is equal to 1 by 10 y is equal to minus 3 by 5 z is equal to 7 by 20 find the value of expensive( x minus y) minus z and x minus( y minus z )are they equal
Answers
Given values of x, y and z are
\begin{gathered}x=\frac{1}{10},\\ \\y=\frac{-3}{5}\\ \\\& \ z= \frac{7}{20}\end{gathered}
x=
10
1
,
y=
5
−3
& z=
20
7
We have to find the value of the expressions (x-y)-z\ \& \ x-(y-z)(x−y)−z & x−(y−z) and check whether they are equal or not.
In short, we have to check whether subtraction of rational numbers is associative or not, using the given values.
So, lets find the values of expressions first,
\because (x-y)-z = (\frac{1}{10}-[\frac{-3}{5}])- \frac{7}{20}∵(x−y)−z=(
10
1
−[
5
−3
])−
20
7
\begin{gathered}\Rightarrow (x-y)-z = (\frac{1}{10}+\frac{3}{5})- \frac{7}{20}\\ \\\Rightarrow (x-y)-z = (\frac{1+6}{10})- \frac{7}{20}\\\\\Rightarrow (x-y)-z = \frac{7}{10}- \frac{7}{20}\\ \\\Rightarrow (x-y)-z = \frac{14-7}{20} \\\\\Rightarrow (x-y)-z = \frac{7}{20}\end{gathered}
⇒(x−y)−z=(
10
1
+
5
3
)−
20
7
⇒(x−y)−z=(
10
1+6
)−
20
7
⇒(x−y)−z=
10
7
−
20
7
⇒(x−y)−z=
20
14−7
⇒(x−y)−z=
20
7
Now, x-(y-z) = \frac{1}{10}-(\frac{-3}{5}- \frac{7}{20})x−(y−z)=
10
1
−(
5
−3
−
20
7
)
\begin{gathered}\Rightarrow x-(y-z) = \frac{1}{10}+(\frac{-12-7}{20})\\ \\\Rightarrow x-(y-z) = \frac{1}{10}-(- \frac{19}{20})\\\\\Rightarrow x-(y-z) = \frac{1}{10}+ \frac{19}{20}\\ \\\Rightarrow x-(y-z) = \frac{2+19}{20} \\\\\Rightarrow x-(y-z) = \frac{21}{20}\end{gathered}
⇒x−(y−z)=
10
1
+(
20
−12−7
)
⇒x−(y−z)=
10
1
−(−
20
19
)
⇒x−(y−z)=
10
1
+
20
19
⇒x−(y−z)=
20
2+19
⇒x−(y−z)=
20
21
It is clear that,
\begin{gathered}(x-y)-z \neq x-(y-z)\\ \\As \ \frac{7}{20} \neq \frac{21}{20}\end{gathered}
(x−y)−z
=x−(y−z)
As
20
7
=
20
21
i.e. the given expressions are not equal.
This example also proves that subtraction of rational numbers is not associative.