
How do you add or subtract a negative fraction?
Answer
511.5k+ views
Hint: We first explain the mathematical notation of a negative fraction. We explain the conditions of the terms in the fraction. Then we take a random number to add or subtract the negative fraction. We perform the binary operation and get the solution of the problem.
Complete step-by-step solution:
We first define a negative fraction as $-\dfrac{p}{q}$ where $p,q\in {{\mathbb{Z}}^{+}}$.
Now we take any number x where $x\in \mathbb{R}$.
We need to add or subtract $-\dfrac{p}{q}$ to x.
At the time of adding we get $x+\left( -\dfrac{p}{q} \right)$. We take the L.C.M of the denominators 1 and q which gives q as a result.
Solving we get
$\begin{align}
& x+\left( -\dfrac{p}{q} \right) \\
& =x-\dfrac{p}{q} \\
& =\dfrac{qx-p}{q} \\
\end{align}$
At the time of subtraction, we get $x-\left( -\dfrac{p}{q} \right)$. We take the L.C.M of the denominators 1 and q which gives q as a result.
Solving we get
$\begin{align}
& x-\left( -\dfrac{p}{q} \right) \\
& =x+\dfrac{p}{q} \\
& =\dfrac{qx+p}{q} \\
\end{align}$
This means the addition or subtraction of a negative fraction $-\dfrac{p}{q}$ to a number x will be
$x\pm \left( -\dfrac{p}{q} \right)=\dfrac{qx\mp p}{q}$ respectively. For the addition we get ‘minus’ sign in between and for the addition we get ‘plus’ sign.
We take an example where we are adding 5 with $-\dfrac{2}{3}$. The L.C.M of the denominators is 3.
We get \[5+\left( -\dfrac{2}{3} \right)=5-\dfrac{2}{3}=\dfrac{15-2}{3}=\dfrac{13}{3}\].
Now we subtract $-\dfrac{2}{3}$ from 5. The L.C.M of the denominators is 3.
We get \[5-\left( -\dfrac{2}{3} \right)=5+\dfrac{2}{3}=\dfrac{15+2}{3}=\dfrac{17}{3}\].
Note: The value of q can be 1 in $-\dfrac{p}{q}$. Any integer is considered to be a fraction where the value of the value of the denominator is 1. The fractional of 6 will be $\dfrac{6}{1}$.
Complete step-by-step solution:
We first define a negative fraction as $-\dfrac{p}{q}$ where $p,q\in {{\mathbb{Z}}^{+}}$.
Now we take any number x where $x\in \mathbb{R}$.
We need to add or subtract $-\dfrac{p}{q}$ to x.
At the time of adding we get $x+\left( -\dfrac{p}{q} \right)$. We take the L.C.M of the denominators 1 and q which gives q as a result.
Solving we get
$\begin{align}
& x+\left( -\dfrac{p}{q} \right) \\
& =x-\dfrac{p}{q} \\
& =\dfrac{qx-p}{q} \\
\end{align}$
At the time of subtraction, we get $x-\left( -\dfrac{p}{q} \right)$. We take the L.C.M of the denominators 1 and q which gives q as a result.
Solving we get
$\begin{align}
& x-\left( -\dfrac{p}{q} \right) \\
& =x+\dfrac{p}{q} \\
& =\dfrac{qx+p}{q} \\
\end{align}$
This means the addition or subtraction of a negative fraction $-\dfrac{p}{q}$ to a number x will be
$x\pm \left( -\dfrac{p}{q} \right)=\dfrac{qx\mp p}{q}$ respectively. For the addition we get ‘minus’ sign in between and for the addition we get ‘plus’ sign.
We take an example where we are adding 5 with $-\dfrac{2}{3}$. The L.C.M of the denominators is 3.
We get \[5+\left( -\dfrac{2}{3} \right)=5-\dfrac{2}{3}=\dfrac{15-2}{3}=\dfrac{13}{3}\].
Now we subtract $-\dfrac{2}{3}$ from 5. The L.C.M of the denominators is 3.
We get \[5-\left( -\dfrac{2}{3} \right)=5+\dfrac{2}{3}=\dfrac{15+2}{3}=\dfrac{17}{3}\].
Note: The value of q can be 1 in $-\dfrac{p}{q}$. Any integer is considered to be a fraction where the value of the value of the denominator is 1. The fractional of 6 will be $\dfrac{6}{1}$.
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