Multiplicative inverse of $\frac{a}{b}$ is $\frac{c}{d}$ if $\frac{a}{b}\times \frac{c}{d}=1$.
Now, $\frac{-15}{51}\times \frac{-17}{5}=\frac{3}{3}=1$

Reciprocal of $\frac{11}{66}=\frac{66}{11}$
$\Rightarrow \frac{7}{6}\times \frac{66}{11}=7\times \frac{11}{11}=7$

Zero is a whole number as well as the rational number. Zero can be expressed in the form of $\frac{p}{q}$  as $\frac{0}{1}$. Hence zero is a rational number since it satisfies the definition of rational number where p = 0 and q $\ne$ 0.

When we multiply and divide a rational number by the same number, the value of rational number remains unchanged. Therefore,
$\frac{9}{4}\times \frac{7}{7}=\frac{63}{28}\frac{9}{4}\times \frac{4}{4}=\frac{36}{16}$

L.C.M of 7 and 8 is 56.
$\frac{3}{7}$ and $\frac{4}{8}$ can be written as $\frac{24}{56}$ and $\frac{28}{56}$.
Check for numbers which lie between $\frac{24}{56}$ and $\frac{28}{56}$.
Now, $\frac{13}{28}$ = $\frac{26}{56}$ lies between $\frac{3}{7}$ and $\frac{4}{8}$.

A number which can be represented in the form of $\frac{p}{q},q\ne 0$ where p and q are integers is called a rational number.
For example, $\frac{2}{3},5,$ $-4$ etc are all rational numbers as they can be expressed in $\frac{p}{q}$ form with $p$ and $q$ as integers and $q\ne 0$.

If 'a' and 'b' are any two whole numbers, their product will also be a whole number. We know that the multiplication of any two numbers can be expressed as repeated addition of one number. So, if we are adding any whole number to itself again and again, then we will surely get a whole number.
Let's assume $a=2,b=3$.
$a\times b=2\times 3$ which is same as adding '2' three times to itself, which will result in a whole number.

Rational numbers can be compared by making their denominators equal.
Here, we need to compare all given rational numbers $\frac{25}{4},\frac{22}{3},\frac{13}{2},\frac{15}{2},\frac{17}{2}$ and $\frac{8}{1}$.
We know LCM of 1, 2, 3 and 4, i.e., the denominators of the above fractions, is 12.
Converting all the above fractions into like fractions, we get
$\frac{25}{4}=\frac{25\times 3}{4\times 3}=\frac{75}{12}$
$\frac{22}{3}=\frac{22\times 4}{3\times 4}=\frac{88}{12}$
$\frac{13}{2}=\frac{13\times 6}{2\times 6}=\frac{78}{12}$
$\frac{15}{2}=\frac{15\times 6}{2\times 6}=\frac{90}{12}$
$\frac{17}{2}=\frac{17\times 6}{2\times 6}=\frac{102}{12}$
$8=\frac{8}{1}=\frac{8\times 12}{1\times 12}=\frac{96}{12}$
Arranging these fractions in an ascending order, we have
$\frac{75}{12},\frac{78}{12},\frac{88}{12},\frac{90}{12},\frac{96}{12},\frac{102}{12}$
Thus, $\frac{13}{2}$ lies between $\frac{25}{4}$ and $\frac{22}{3}$.

Reciprocal of $\frac{5}{11}$ is $\frac{11}{5}$.

$\frac{11}{5}÷\frac{-55}{18}$
$=\frac{11\times 18}{5\times (-55)}$

$=\frac{-18}{25}$

An operation is commutative if changing the order of operands does not change the result.
Addition and multiplication are commutative for whole numbers.
$\text{For example,}$ $2+3=5,$  $3+2=5$.
$\text{Also,}$ $3\times 4=12,$  $4\times 3=12$