Question

Check the commutative property of addition for $\frac{102}{201},\frac{3}{4}$

Answer 1

We know that the Commutative Property of Addition states that changing the order of addends does not change the sum. i.e., if $a$ and $b$ are two real numbers, then $a+b=b+a$.

Let $a=\frac{102}{201}$ and$b=\frac{3}{4}$.

Let us first find $a+b$ as follows:

$a+b=\frac{102}{201}+\frac{3}{4}=\frac{102\times 4}{201\times 4}+\frac{3\times 201}{4\times 201}=\frac{408}{804}+\frac{603}{804}=\frac{1011}{804}......\left(1\right)$

Now, we find $b+a$ as follows:

$b+a=\frac{3}{4}+\frac{102}{201}=\frac{3\times 201}{4\times 201}+\frac{102\times 4}{201\times 4}=\frac{603}{804}+\frac{408}{804}=\frac{1011}{804}......\left(2\right)$

Since equation $\left(1\right)$ is equal to equation $\left(2\right)$,

$\therefore a+b=b+a$ i.e. $\frac{102}{201}+\frac{3}{4}=\frac{3}{4}+\frac{102}{201}$

Hence, the given numbers $\frac{102}{201}$ and $\frac{3}{4}$ satisfies the commutative property of addition.