Question

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

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) $$x={\displaystyle \frac{7}{11}}andy={\displaystyle \frac{3}{2}}$$
reciprocal of $(x\times y)$ = reciprocal of x $\times$ reciprocal of y
LHS of equation = reciprocal of $(x\times y)$=$reciprocal\left({\displaystyle \frac{7}{11}}\times {\displaystyle \frac{3}{2}}\right)$= $reciprocal\left({\displaystyle \frac{21}{22}}\right)$=${\displaystyle \frac{22}{21}}$
RHS of equation = reciprocal of x $\times$ reciprocal of y
=$$reciprocal\left({\displaystyle \frac{7}{11}}\right)\times reciprocal\left({\displaystyle \frac{3}{2}}\right)$$=$\left({\displaystyle \frac{11}{7}}\times {\displaystyle \frac{2}{3}}\right)$=${\displaystyle \frac{22}{21}}$
So, hence proved LHS = RHS.
B) $$x={\displaystyle \frac{5}{7}}andy={\displaystyle \frac{6}{5}}$$
reciprocal of $(x\times y)$ = reciprocal of x $\times$ reciprocal of y
LHS of equation = reciprocal of $(x\times y)$=$reciprocal\left({\displaystyle \frac{5}{7}}\times {\displaystyle \frac{6}{5}}\right)$= $reciprocal\left({\displaystyle \frac{6}{7}}\right)$=${\displaystyle \frac{7}{6}}$
RHS of equation = reciprocal of x $\times$ reciprocal of y
=$$reciprocal\left({\displaystyle \frac{5}{7}}\right)\times reciprocal\left({\displaystyle \frac{6}{5}}\right)$$=$\left({\displaystyle \frac{7}{5}}\times {\displaystyle \frac{5}{6}}\right)$=${\displaystyle \frac{7}{6}}$
So, hence proved LHS = RHS.
C) $$x=1andy=\text{ 3}$$
reciprocal of $(x\times y)$ = reciprocal of x $\times$ reciprocal of y
LHS of equation = reciprocal of $(x\times y)$=$reciprocal\left(\text{1 }\times 3\right)$= $reciprocal\left(3\right)$=${\displaystyle \frac{1}{3}}$
RHS of equation = reciprocal of x $\times$ reciprocal of y
=$$reciprocal\left(1\right)\times reciprocal\left(3\right)$$=$\left({\displaystyle \frac{1}{1}}\times {\displaystyle \frac{1}{3}}\right)$=${\displaystyle \frac{1}{3}}$
So, hence proved LHS = RHS.
D) $$x={\displaystyle \frac{3}{7}}andy={\displaystyle \frac{0}{5}}$$
reciprocal of $(x\times y)$ = reciprocal of x $\times$ reciprocal of y
LHS of equation = reciprocal of $(x\times y)$=$reciprocal\left({\displaystyle \frac{3}{7}}\times {\displaystyle \frac{0}{5}}\right)$= $reciprocal\left({\displaystyle \frac{0}{5}}\right)$=$\mathrm{\infty }$
RHS of equation = reciprocal of x $\times$ reciprocal of y
=$$reciprocal\left({\displaystyle \frac{3}{7}}\right)\times reciprocal\left({\displaystyle \frac{0}{5}}\right)$$=$\left({\displaystyle \frac{3}{7}}\times \mathrm{\infty }\right)$=$\mathrm{\infty }$
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.