Question

Verify the property "reciprocal of $( x \times y)$ = reciprocal of x $ \times$ reciprocal of y" for the given values.
A) \[x = {\text{ }}\dfrac{7}{{11}}{\text{ }}and{\text{ }}y = {\text{ }}\dfrac{3}{2}\]
B) \[x = {\text{ }}\dfrac{5}{7}{\text{ }}and{\text{ }}y = {\text{ }}\dfrac{6}{5}\]
C) \[x = {\text{ }}1{\text{ }}and{\text{ }}y = {\text{ 3}}\]
D) \[x = {\text{ }}\dfrac{3}{7}{\text{ }}and{\text{ }}y = {\text{ }}\dfrac{0}{5}\]

Answer 1

Hint: A fraction is part of a whole number. It has two parts – a numerator and a denominator.
The reciprocal of a fraction is just switching the numerator (top number) and the denominator (bottom number). The negative reciprocal takes the negative of that number.
We need to verify the property "reciprocal of $( x \times y)$ = reciprocal of x $ \times$ reciprocal of y". So, in every scenario we will replace the value of x and y in the property LHS and RHS and prove the property.

Complete step-by-step answer: A) \[x = {\text{ }}\dfrac{7}{{11}}{\text{ }}and{\text{ }}y = {\text{ }}\dfrac{3}{2}\]
reciprocal of $( x \times y)$ = reciprocal of x $ \times$ reciprocal of y
LHS of equation = reciprocal of $( x \times y)$=$reciprocal\left( {\dfrac{7}{{11}}{\text{ }} \times \quad \dfrac{3}{2}} \right)$= $reciprocal\left( {\dfrac{{21}}{{22}}} \right)$=$\dfrac{{22}}{{21}}$
RHS of equation = reciprocal of x $ \times$ reciprocal of y
=\[reciprocal{\text{ }}\left( {\dfrac{7}{{11}}} \right){\text{ }} \times {\text{ }}reciprocal{\text{ }}\left( {\dfrac{3}{2}} \right)\]=$\left( {\dfrac{{11}}{7}{\text{ }} \times \quad \dfrac{2}{3}} \right)$=$\dfrac{{22}}{{21}}$
So, hence proved LHS = RHS.
B) \[x = {\text{ }}\dfrac{5}{7}{\text{ }}and{\text{ }}y = {\text{ }}\dfrac{6}{5}\]
reciprocal of $( x \times y)$ = reciprocal of x $ \times$ reciprocal of y
LHS of equation = reciprocal of $( x \times y)$=$reciprocal\left( {\dfrac{5}{7}{\text{ }} \times \quad \dfrac{6}{5}} \right)$= $reciprocal\left( {\dfrac{6}{7}} \right)$=$\dfrac{7}{6}$
RHS of equation = reciprocal of x $ \times$ reciprocal of y
=\[reciprocal{\text{ }}\left( {\dfrac{5}{7}} \right){\text{ }} \times {\text{ }}reciprocal{\text{ }}\left( {\dfrac{6}{5}} \right)\]=$\left( {\dfrac{7}{5}{\text{ }} \times \quad \dfrac{5}{6}} \right)$=$\dfrac{7}{6}$
So, hence proved LHS = RHS.
C) \[x = {\text{ }}1{\text{ }}and{\text{ }}y = {\text{ 3}}\]
reciprocal of $( x \times y)$ = reciprocal of x $ \times$ reciprocal of y
LHS of equation = reciprocal of $( x \times y)$=$reciprocal\left( {{\text{1 }} \times \;3} \right)$= $reciprocal\left( 3 \right)$=$\dfrac{1}{3}$
RHS of equation = reciprocal of x $ \times$ reciprocal of y
=\[reciprocal{\text{ }}\left( 1 \right){\text{ }} \times {\text{ }}reciprocal{\text{ }}\left( 3 \right)\]=$\left( {\dfrac{1}{1}{\text{ }} \times \quad \dfrac{1}{3}} \right)$=$\dfrac{1}{3}$
So, hence proved LHS = RHS.
D) \[x = {\text{ }}\dfrac{3}{7}{\text{ }}and{\text{ }}y = {\text{ }}\dfrac{0}{5}\]
reciprocal of $( x \times y)$ = reciprocal of x $ \times$ reciprocal of y
LHS of equation = reciprocal of $( x \times y)$=$reciprocal\left( {\dfrac{3}{7}{\text{ }} \times \quad \dfrac{0}{5}} \right)$= $reciprocal\left( {\dfrac{0}{5}} \right)$=$\infty $
RHS of equation = reciprocal of x $ \times$ reciprocal of y
=\[reciprocal{\text{ }}\left( {\dfrac{3}{7}} \right){\text{ }} \times {\text{ }}reciprocal{\text{ }}\left( {\dfrac{0}{5}} \right)\]=$\left( {\dfrac{3}{7}{\text{ }} \times \quad \infty } \right)$=$\infty $
So, hence proved LHS = RHS.

Note: Every number has a reciprocal except for 0. There is nothing you can multiply by 0 to create a product of 1, so it has no reciprocal.
The reciprocal of a number is 1 divided by the number.
The reciprocal of a number is also called its multiplicative inverse.
The product of a number and its reciprocal is 1.
All numbers except 0 have a reciprocal.
The reciprocal of a fraction is found by flipping its numerator and denominator.