# Let ‘O’ be a binary operation on the set ${Q_0}$ of all non zero rational numbers defined by $a \circ b = \dfrac{{ab}}{2}$ for all $a,b \in {Q_0}$. Find the invertible elements of ${Q_0}$.

Last updated date: 29th Mar 2023

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Answer

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Hint: We have to use the fundamental theorem of binary operation and find identity elements w.r.t the binary operation and find inverse of any element.

Complete step-by-step answer:

Let e be the identity element

Then according to the definition of the identity elements we get

$a \circ e = e \circ a = a,a \in {Q_0}$

Then we have,

$\dfrac{{ae}}{2} = a$

Or e = 2 (since $a \ne 0$ )

So the identity element is 2

Given ‘O’ be a binary operation on the set ${Q_0}$ of all non-zero rational numbers defined by

$a \circ b = \dfrac{{ab}}{2},a,b \in {Q_0}$

Now the identity element of the set is 2 with respect to the binary operation O

To find the inverse of any element

Let $a \in {Q_0}$ again also let $x \in {Q_0}$ be the inverse of a

Then according to the definition of the inverse element we get

$x \circ a = a \circ x = 2$ (identity element)

Or, $\dfrac{{xa}}{2} = 2$

$ \Rightarrow x = \dfrac{4}{a}$ (since $a \in {Q_0}$)

So, inverse of a is $\dfrac{4}{a}$

Note: For such problems use the fundamental theorem of binary operation and try to find the identity element so with the help of the identity element you can easily find the inverse element.

Complete step-by-step answer:

Let e be the identity element

Then according to the definition of the identity elements we get

$a \circ e = e \circ a = a,a \in {Q_0}$

Then we have,

$\dfrac{{ae}}{2} = a$

Or e = 2 (since $a \ne 0$ )

So the identity element is 2

Given ‘O’ be a binary operation on the set ${Q_0}$ of all non-zero rational numbers defined by

$a \circ b = \dfrac{{ab}}{2},a,b \in {Q_0}$

Now the identity element of the set is 2 with respect to the binary operation O

To find the inverse of any element

Let $a \in {Q_0}$ again also let $x \in {Q_0}$ be the inverse of a

Then according to the definition of the inverse element we get

$x \circ a = a \circ x = 2$ (identity element)

Or, $\dfrac{{xa}}{2} = 2$

$ \Rightarrow x = \dfrac{4}{a}$ (since $a \in {Q_0}$)

So, inverse of a is $\dfrac{4}{a}$

Note: For such problems use the fundamental theorem of binary operation and try to find the identity element so with the help of the identity element you can easily find the inverse element.

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