Question

$\text{Name the multiplication property of rational numbers shown below:}\left(i\right)\frac{3}{5}\times \frac{-8}{9}=\frac{-8}{9}\times \frac{3}{5}\left(ii\right)\frac{-3}{4}\times (\frac{5}{7}\times \frac{-8}{15})=(\frac{-3}{5}\times \frac{5}{7})\times \frac{-8}{15}\left(iii\right)\frac{4}{5}\times (\frac{3}{-8}+\frac{-4}{7})=\frac{4}{5}\times \frac{3}{-8}+\frac{4}{5}\times \frac{-4}{7}\left(iv\right)\frac{-7}{5}\times \frac{5}{-7}=1\left(v\right)\frac{8}{-9}\times 1=1\times \frac{8}{-9}=\frac{8}{-9}\left(vi\right)\frac{-3}{4}\times 0=0$

Answer 1

(i)

$\frac{3}{5}\times \frac{-8}{9}=\frac{-8}{9}\times \frac{3}{5}$

Since, it is in the form of "$a\times b=b\times a$",

It is a commutativity property of rational numbers under multiplication.
(ii)

$\frac{-3}{4}\times (\frac{5}{7}\times \frac{-8}{15})=(\frac{-3}{5}\times \frac{5}{7})\times \frac{-8}{15}$

Since, it is in the form of "$a\times (b\times c)=(a\times b)\times c$"

It is an associative property of rational numbers under multiplication.

(iii)

$\frac{4}{5}\times (\frac{3}{-8}+\frac{-4}{7})=\frac{4}{5}\times \frac{3}{-8}+\frac{4}{5}\times \frac{-4}{7}$

Since, it is in the form of " $a\times (b+c)=(a\times b)+(a\times c)$" .

It is a distributive property of rational numbers over addition.

(iv)

$\frac{-7}{5}\times \frac{5}{-7}=1$

Since, the multiplicative inverse of $\frac{a}{b}=\frac{b}{a},a,b\ne 0$

i.e., $\frac{a}{b}\times \frac{b}{a}=1$

So, $\frac{-7}{5}\times \frac{5}{-7}=1$ has the existence of multiplicative inverse.

(v)

$\frac{8}{-9}\times 1=1\times \frac{8}{-9}=\frac{8}{-9}$

Since, it is in the form of "$a\times 1=1\times a=a$".

and "$1$" is called the multiplicative identity.

So $\frac{8}{-9}\times 1=1\times \frac{8}{-9}=\frac{8}{-9}$ has the existence of multiplicative identity.

(vi)

$\frac{-3}{4}\times 0=0$

One of zero's unique rules is called the multiplication property of zero.

The multiplication property of zero states that the product of any number and zero is zero.

Hence, It is a "multiplication property of zero"