
On the set of positive rational, a binary operation \[*\] is defined by \[a*b = \dfrac{{2ab}}{5}\]. If \[2*x = {3^{ - 1}}\] then find the value of $x$.
Answer
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Hint: Here we have a set of positive rational numbers under the binary operation.
Also it is defined by some condition.
We use this condition, and then we will get the value.
Complete step-by-step answer:
It is given set of all positive rational number and is denoted by \[{Q^ + }\]
It is form \[{Q^ + }\]=\[\dfrac{a}{b},b \ne 0,a,b\] is always positive, \[a\underline > 0,b > 0\]
Also given that binary operation is ‘\[*\]’ is defined by \[a*b = \dfrac{{2ab}}{5}\] , where\[a,b \in {Q^ + }\]
Again it is given that \[2*x = {3^{ - 1}}\], where \[x \in {Q^ + }\]
Also, \[2\] belongs to \[{Q^ + }\]
Because \[2\] can be written as \[\dfrac{2}{1}\], where \[\dfrac{2}{1} \in {Q^ + }\]
Hence, \[2\]\[ \in {Q^ + }\]
Here \[2*x\]can be written by using \[a*b\]
Also, it is defined by \[a*b = \dfrac{{2ab}}{5}\]
Here \[a = 2\]and\[b = x\], we get
\[2*x\]=\[\dfrac{{2(2x)}}{5}\]
Substitute \[2*x = - 3\]we get.
\[\dfrac{{2(2x)}}{5}\]=\[{3^{ - 1}}\]
\[\dfrac{{2(2x)}}{5}\]=\[\dfrac{1}{3}\]
Multiply \[2\]by\[2x\], we get,
\[\dfrac{{4x}}{5}\]=\[\dfrac{1}{3}\]
Except\[x\], take all the values to the right hand side, and reverse it.
\[x = \dfrac{5}{4} \times \dfrac{1}{3}\]
Denominator \[4\] multiply with \[3\], numerator \[5\] multiplied by \[1\]
\[x = \dfrac{5}{{12}}\]
Therefore, \[x = \dfrac{5}{{12}}\] belongs to a set of positive rational numbers.
That is, \[\dfrac{5}{{12}} \in {Q^ + }\]
Note: Here, we state the rational numbers are represented in \[\dfrac{a}{b}\] form where \[b\] is not equal to zero.
We can also say, \[0\] is rational number, as we can represent in many forms such as \[\dfrac{0}{1},\dfrac{0}{2},\dfrac{0}{3},....\] but \[\dfrac{1}{0},\dfrac{2}{0},\dfrac{3}{0},...\] are not rational, since they give us infinite values.
From these, we can say, a number is rational if we can write it as a fraction, where both denominators and numerator are integers and denominator is a non-zero number.
A binary operator is an operator that operates on two operands and manipulates them to return a result.
Also it is defined by some condition.
We use this condition, and then we will get the value.
Complete step-by-step answer:
It is given set of all positive rational number and is denoted by \[{Q^ + }\]
It is form \[{Q^ + }\]=\[\dfrac{a}{b},b \ne 0,a,b\] is always positive, \[a\underline > 0,b > 0\]
Also given that binary operation is ‘\[*\]’ is defined by \[a*b = \dfrac{{2ab}}{5}\] , where\[a,b \in {Q^ + }\]
Again it is given that \[2*x = {3^{ - 1}}\], where \[x \in {Q^ + }\]
Also, \[2\] belongs to \[{Q^ + }\]
Because \[2\] can be written as \[\dfrac{2}{1}\], where \[\dfrac{2}{1} \in {Q^ + }\]
Hence, \[2\]\[ \in {Q^ + }\]
Here \[2*x\]can be written by using \[a*b\]
Also, it is defined by \[a*b = \dfrac{{2ab}}{5}\]
Here \[a = 2\]and\[b = x\], we get
\[2*x\]=\[\dfrac{{2(2x)}}{5}\]
Substitute \[2*x = - 3\]we get.
\[\dfrac{{2(2x)}}{5}\]=\[{3^{ - 1}}\]
\[\dfrac{{2(2x)}}{5}\]=\[\dfrac{1}{3}\]
Multiply \[2\]by\[2x\], we get,
\[\dfrac{{4x}}{5}\]=\[\dfrac{1}{3}\]
Except\[x\], take all the values to the right hand side, and reverse it.
\[x = \dfrac{5}{4} \times \dfrac{1}{3}\]
Denominator \[4\] multiply with \[3\], numerator \[5\] multiplied by \[1\]
\[x = \dfrac{5}{{12}}\]
Therefore, \[x = \dfrac{5}{{12}}\] belongs to a set of positive rational numbers.
That is, \[\dfrac{5}{{12}} \in {Q^ + }\]
Note: Here, we state the rational numbers are represented in \[\dfrac{a}{b}\] form where \[b\] is not equal to zero.
We can also say, \[0\] is rational number, as we can represent in many forms such as \[\dfrac{0}{1},\dfrac{0}{2},\dfrac{0}{3},....\] but \[\dfrac{1}{0},\dfrac{2}{0},\dfrac{3}{0},...\] are not rational, since they give us infinite values.
From these, we can say, a number is rational if we can write it as a fraction, where both denominators and numerator are integers and denominator is a non-zero number.
A binary operator is an operator that operates on two operands and manipulates them to return a result.
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