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Fill in the blanks:
(i) The product of two positive rational numbers is always______.
(ii) The product of a positive rational number and a negative rational number is always________.
(iii) The product of two negative rational numbers is always________.
(iv) The reciprocal of a positive rational number is________.
(v) The reciprocal of a negative rational number is________.
(vi) Zero has reciprocal. The product of a rational number and its reciprocal is______.
(viii) The numbers and are their own reciprocals______.
(ix) If a is reciprocal of b, then the reciprocal of b is______.
(x) The number 0 is the reciprocal of any number______.
(xi) Reciprocal of $\frac{1}{a}$, $a≠0$ is______.
(xii) $(17 \times 12)^{-1} = 17^{-1} \times$________ .
To do:
We have to fill in the given blanks.
Solution:
(i) The product of two positive rational numbers is always positive.
For example,
$5\times4=20$
(ii) The product of a positive rational number and a negative rational number is always negative.
For example,
$5\times(-4)=-20$
(iii) The product of two negative rational numbers is always positive.
For example,
$(-5)\times(-4)=20$ 
(iv) The reciprocal of a positive rational number is positive.
For example,
Reciprocal of 5 is $\frac{1}{5}$ . 
(v) The reciprocal of a negative rational number is negative.
For example,
Reciprocal of $-5$ is $\frac{-1}{5}$ .  
(vi) Any number divided by zero is not defined.
Therefore,
Zero has no reciprocal.
(vii) The product of a rational number and its reciprocal is 1.
For example,
Reciprocal of 5 is $\frac{1}{5}$
$5\times\frac{1}{5}=1$
(viii) The numbers that are their own reciprocals are $1$ and $-1$.
For example,
Reciprocal of 1 is $\frac{1}{1}=1$
Reciprocal of $-1$ is $\frac{1}{-1}=\frac{-1}{1}=-1$
(ix) If $a$ is reciprocal of $b$, then the reciprocal of $b$ is $a$.
For example,
Reciprocal of 5 is $\frac{1}{5}$
Reciprocal of $\frac{1}{5}=\frac{1}{\frac{1}{5}}$
$=\frac{5}{1}$
$=5$
(x) Any number divided by zero is not defined.
Therefore,
The number 0 is not the reciprocal of any number.
(xi) Reciprocal of $a$ where $a≠0$ is $\frac{1}{a}$.
Therefore,
Reciprocal of $\frac{1}{a}$, $a≠0$ is $a$.
(xii) $(a \times b)^{-1}=(a)^{-1}\times(b)^{-1}$ where $a, b≠0$.
This implies,
$(17 \times 12)^{-1} = 17^{-1} \times12^{-1}$
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