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How do you find the additive and multiplicative inverse of \[{1}/{3}\;\]?
Answer
455.1k+ views
Hint: For a number a, it's multiplicative inverse b is such that \[a.b=1\] which is the multiplicative identity. For a number a, it's additive inverse c would be such that \[a+c=0\] where 0 is additive identity.
Complete step by step answer:
As per the given question, we have to find the additive and multiplicative inverse of \[{1}/{3}\;\].
Let \[{a=1}/{3}\;\].
Additive inverse of a :
Let the additive inverse of a is b. As per the given condition to find the additive inverse the value of c should be such that \[a+c=0\].
We know that \[{a=1}/{3}\;\],b is the additive inverse. Now, substituting these values in the equation \[a+c=0\]. we get
\[\Rightarrow \dfrac{1}{3}+b=0\]
Now, we subtract \[{1}/{3}\;\] from both sides of the equation. then we get
\[\Rightarrow \dfrac{1}{3}+b-\dfrac{1}{3}=0-\dfrac{1}{3}\]
Now on solving above equation we get
\[\begin{align}
& \Rightarrow b+0=-\dfrac{1}{3} \\
& \Rightarrow b=-\dfrac{1}{3} \\
\end{align}\]
From the above equation, we can say that the additive inverse b is \[-\dfrac{1}{3}\].
Therefore, the additive inverse of \[{1}/{3}\;\] is \[-\dfrac{1}{3}\].
Multiplicative inverse of a :
Let multiplicative inverse of a be c. As per the given condition to find the multiplicative inverse the value of b should be such that \[a.b=1\].
We know that \[{a=1}/{3}\;\],c is the multiplicative inverse. Now, substituting these values in the equation \[a.b=1\]. we get
\[\Rightarrow \dfrac{1}{3}\times c=1\]
Now, we multiply with 3 on both sides of the equation. Then we get
\[\Rightarrow \dfrac{1}{3}\times c\times 3=1\times 3\]
Now on solving above equation we get
\[\begin{align}
& \Rightarrow 1\times c=3 \\
& \Rightarrow c=3 \\
\end{align}\]
From the above equation, we can say that the multiplicative inverse c is 3.
Therefore, the multiplicative inverse of \[{1}/{3}\;\] is 3.
Note: In order to solve these types of problems, we must have enough knowledge about the definitions and conditions of additive and multiplicative inverse. We must avoid calculation mistakes to get the correct results.
Complete step by step answer:
As per the given question, we have to find the additive and multiplicative inverse of \[{1}/{3}\;\].
Let \[{a=1}/{3}\;\].
Additive inverse of a :
Let the additive inverse of a is b. As per the given condition to find the additive inverse the value of c should be such that \[a+c=0\].
We know that \[{a=1}/{3}\;\],b is the additive inverse. Now, substituting these values in the equation \[a+c=0\]. we get
\[\Rightarrow \dfrac{1}{3}+b=0\]
Now, we subtract \[{1}/{3}\;\] from both sides of the equation. then we get
\[\Rightarrow \dfrac{1}{3}+b-\dfrac{1}{3}=0-\dfrac{1}{3}\]
Now on solving above equation we get
\[\begin{align}
& \Rightarrow b+0=-\dfrac{1}{3} \\
& \Rightarrow b=-\dfrac{1}{3} \\
\end{align}\]
From the above equation, we can say that the additive inverse b is \[-\dfrac{1}{3}\].
Therefore, the additive inverse of \[{1}/{3}\;\] is \[-\dfrac{1}{3}\].
Multiplicative inverse of a :
Let multiplicative inverse of a be c. As per the given condition to find the multiplicative inverse the value of b should be such that \[a.b=1\].
We know that \[{a=1}/{3}\;\],c is the multiplicative inverse. Now, substituting these values in the equation \[a.b=1\]. we get
\[\Rightarrow \dfrac{1}{3}\times c=1\]
Now, we multiply with 3 on both sides of the equation. Then we get
\[\Rightarrow \dfrac{1}{3}\times c\times 3=1\times 3\]
Now on solving above equation we get
\[\begin{align}
& \Rightarrow 1\times c=3 \\
& \Rightarrow c=3 \\
\end{align}\]
From the above equation, we can say that the multiplicative inverse c is 3.
Therefore, the multiplicative inverse of \[{1}/{3}\;\] is 3.
Note: In order to solve these types of problems, we must have enough knowledge about the definitions and conditions of additive and multiplicative inverse. We must avoid calculation mistakes to get the correct results.
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