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Question: Let ‘O’ be a binary operation on the set \({Q_0}\) of all non zero rational numbers defined by \(a \...

Let ‘O’ be a binary operation on the set Q0{Q_0} of all non zero rational numbers defined by ab=ab2a \circ b = \dfrac{{ab}}{2} for all a,bQ0a,b \in {Q_0}. Find the invertible elements of Q0{Q_0}.

Explanation

Solution

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
ae=ea=a,aQ0a \circ e = e \circ a = a,a \in {Q_0}
Then we have,
ae2=a\dfrac{{ae}}{2} = a
Or e = 2 (since a0a \ne 0 )
So the identity element is 2
Given ‘O’ be a binary operation on the set Q0{Q_0} of all non-zero rational numbers defined by
ab=ab2,a,bQ0a \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 aQ0a \in {Q_0} again also let xQ0x \in {Q_0} be the inverse of a
Then according to the definition of the inverse element we get
xa=ax=2x \circ a = a \circ x = 2 (identity element)
Or, xa2=2\dfrac{{xa}}{2} = 2
x=4a\Rightarrow x = \dfrac{4}{a} (since aQ0a \in {Q_0})
So, inverse of a is 4a\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.