Answer

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**Hint:**Take the l.c.m. of ‘2’ and ‘3’. Divide ‘2’ with the l.c.m. and multiply the result with ‘1’. Again divide ‘3’ with the l.c.m. and multiply the result with ‘1’. Subtracting both of them as numerator and the l.c.m. as the denominator obtains the solution.

**Complete step-by-step solution:**

Fraction subtraction: For subtracting two fractions first we have to take the l.c.m. of the denominators of the two fractions. The first part of the numerator is obtained by multiplying a factor (which is obtained by dividing the l.c.m. with the first part of the denominator) with the existing denominator. Similarly, the second part of the numerator is obtained by multiplying a factor (which is obtained by dividing the l.c.m. with the second part of the denominator) with the existing denominator. Subtracting both the parts as numerator and the l.c.m. as a denominator together is the solution.

Coming to our expression $\dfrac{1}{2}-\dfrac{1}{3}$

As we know, the l.c.m. of ‘2’ and ‘3’ is $2\times 3=6$

So, the denominator of the result is ‘6’

For the numerator part

Dividing ‘2’ by the l.c.m. ‘6’ we get $6\div 2=3$

So the first part of the numerator of the result$=1\times 3=3$

Again, dividing ‘3’ by the l.c.m. ‘6’ we get $6\div 3=2$

So the second part of the numerator of the result$=1\times 2=2$

Hence the numerator of the result$=3-2=1$

Considering both numerator and denominator together we get the result as $\dfrac{1}{6}$

Which can be written as

$\begin{align}

& \dfrac{1}{2}-\dfrac{1}{3} \\

& \Rightarrow \dfrac{1\cdot 3-1\cdot 2}{6} \\

& \Rightarrow \dfrac{3-2}{6} \\

& \Rightarrow \dfrac{1}{6} \\

\end{align}$

This is the required solution.

**Note:**We obtained the result of $\dfrac{1}{2}-\dfrac{1}{3}$ as $\dfrac{1}{6}$ by fraction subtraction and as we know $\dfrac{1}{2}\times \dfrac{1}{3}=\dfrac{1\times 1}{2\times 3}=\dfrac{1}{6}$. So we can conclude that $\dfrac{1}{2}$ and $\dfrac{1}{3}$ are such fractions whose subtraction and multiplication is the same i.e. $\dfrac{1}{6}$.

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