'Rational numbers are commutative under addition but not commutative under subtraction.' Justify the statement with an example.
Answer
Let $\frac{1}{2}$ and $\frac{1}{4}$ be two rational numbers. Now, $\frac{1}{2}+\frac{1}{4}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$ which implies that rational numbers are commutative under addition. Now, $\quad \frac{1}{2}-\frac{1}{4}=\frac{2-1}{4}=\frac{1}{4}$ But $\quad \frac{1}{4}-\frac{1}{2}=\frac{1-2}{4}=-\frac{1}{4}$ Thus, $\quad \frac{1}{2}-\frac{1}{4} \neq \frac{1}{4}-\frac{1}{2}$ So, rational numbers are not commutative under subtraction.
Write the multiplicative and additive identity for rational numbers?
Answer
1 is the multiplicative identity of a rational number $\because$ For all rational numbers let $x, x \times 1=x=1 \times x$ and 0 is the additive identity for a rational numbers. $\because$ For all rational numbers let $x, x+0=x=0+x$
Give the reason why multiplicative inverse of 0 does not exist?
Answer
The multiplicative inverse for any rational number $x$ is $\frac{1}{x}$. Here, multiplicative inverse of 0 is $\frac{1}{0}$. But $\frac{1}{0}$ is not defined. So, multiplicative inverse of 0 is not defined.
$\frac{4}{5}+\frac{9}{2}+\left(\frac{-4}{5}\right)=\frac{4}{5}+\left(\frac{-4}{5}\right)+\frac{9}{2}\quad$ [by commutative property of addition] $=0+\frac{9}{2}$ $[\because x+(-x)=0$, for all $x$ being a rational number$]$ $=\frac{9}{2}\quad$ [by additive identity]
Give an example to show that subtraction is not associative for rational numbers.
Answer
For rational numbers $\frac{5}{4}, \frac{3}{4}$ and $\frac{1}{4}$, we see that $\frac{5}{4}-\left(\frac{3}{4}-\frac{1}{4}\right)=\frac{5}{4}-\frac{2}{4}=\frac{3}{4}$ and $\left(\frac{5}{4}-\frac{3}{4}\right)-\frac{1}{4}=\frac{2}{4}-\frac{1}{4}=\frac{1}{4}$ Thus, $\quad \frac{5}{4}-\left(\frac{3}{4}-\frac{1}{4}\right) \neq\left(\frac{5}{4}-\frac{3}{4}\right)-\frac{1}{4}$ So, we can say that subtraction is not associative for rational numbers.
What is the sum of additive identity of whole numbers and rational numbers?
Answer
The additive identity of whole numbers is 0 and the additive identity of rational numbers is also 0. Thus, the sum of additive identity of whole numbers and rational numbers is 0.
For two integers p and q, $\frac{\text{p}}{\text{q}}$ is a rational number, then write condition for q, $\frac{\text{p}}{\text{q}}$ to be a rational number.
Answer
$\text{q}$ will be a non-zero integer i.e. $\text{q} \neq 0$ is the condition for being $\frac{\text{p}}{\text{q}}$ a rational number.