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Rational Numbers : Exercise 1.1 (Mathematics NCERT Class 8th)


Q.1 Using appropriate properties find.
(i)  - {2 \over 3} \times {3 \over 5} + {5 \over 2} - {3 \over 5} \times {1 \over 6}
(ii)
{2 \over 5} \times \left( { - {3 \over 7}} \right) - {1 \over 6} \times {3 \over 2} + {1 \over {14}} \times {2 \over 5}
Sol. (i) Given,  - {2 \over 3} \times {3 \over 5} + {5 \over 2} - {3 \over 5} \times {1 \over 6}
=  - {2 \over 3} \times {3 \over 5} - {3 \over 5} \times {1 \over 6} + {5 \over 2} (Using Commutative property)
= {3 \over 5}\left( { - {2 \over 3} - {1 \over 6}} \right) + {5 \over 2} (Using Distributive property)
= {3 \over 5}\left( {{{ - 4 - 1} \over 6}} \right) + {5 \over 2}
= {3 \over 5}\left( {{{ - 5} \over 6}} \right) + {5 \over 2}
= {3 \over 5} \times {{ - 5} \over 6} + {5 \over 2}
= {{ - 3} \over 6} + {5 \over 2}
= {{ - 3 + 15} \over 6}
= {{12} \over 6}
= 2

(ii) Given, {2 \over 5} \times \left( { - {3 \over 7}} \right) - {1 \over 6} \times {3 \over 2} + {1 \over {14}} \times {2 \over 5}
= {2 \over 5} \times \left( { - {3 \over 7}} \right) + {1 \over {14}} \times {2 \over 5} - {1 \over 6} \times {3 \over 2} (Using Commutative property)
= {2 \over 5}\left( { - {3 \over 7} + {1 \over {14}}} \right) - {1 \over 6} \times {3 \over 2} (Using Distributive property)
= {2 \over 5}\left( {{{ - 6 + 1} \over {14}}} \right) - {1 \over 6} \times {3 \over 2}
= {2 \over 5}\left( {{{ - 5} \over {14}}} \right) - {1 \over 6} \times {3 \over 2}
= {2 \over 5} \times {{ - 5} \over {14}} - {1 \over 4}
= {{ - 1} \over 7} - {1 \over 4}
= {{ - 4 - 7} \over {28}}
= {{ - 11} \over {28}}

Q.2 Write the additive inverse of each of the following.
(i){2 \over 8}
(ii) {{ - 5} \over 9}
(iii)
{{ - 6} \over { - 5}}
(iv){2 \over { - 9}} (v) {{19} \over { - 6}}
Sol. (i) {2 \over 8}
We know, {2 \over 8} + \left( { - {2 \over 8}} \right) = {2 \over 8} - {2 \over 8} = 0
Hence, the additive inverse of {2 \over 8} is \left( { - {2 \over 8}} \right).

(ii) {{ - 5} \over 9}
We know, {{ - 5} \over 9} + {5 \over 9} = {{ - 5 + 5} \over 9} = 0
Hence, the additive inverse of {{ - 5} \over 9} is {5 \over 9}.

(iii) {{ - 6} \over { - 5}}
We know, {{ - 6} \over { - 5}} = {6 \over 5}
Now, {6 \over 5} + \left( {{{ - 6} \over 5}} \right) = {{6 - 6} \over 5} = 0
Hence, the additive inverse of {{ - 6} \over { - 5}}is {{ - 6} \over 5}.

(iv) {2 \over { - 9}}
We know, {2 \over { - 9}} + {2 \over 9} = {{ - 2 + 2} \over 9} = 0
Hence, the additive inverse of {2 \over { - 9}}is {2 \over 9}.

(v) {{19} \over { - 6}}
We know, {{19} \over { - 6}} + {{19} \over 6} = {{ - 19 + 19} \over 6} = 0
Hence, the additive inverse of {{19} \over { - 6}}is {{19} \over 6}.

Q.3 Verify that - (-x) = x for.
(i) x = {{11} \over {15}}
(ii)
x = - {{13} \over {17}}

Sol. (i) x = {{11} \over {15}}
The additive inverse of x = {{11} \over {15}} is  - x = - {{11} \over {15}}
Thus, {{11} \over {15}} + \left( { - {{11} \over {15}}} \right) = 0
Now, the additive inverse of  - {{11} \over {15}} is {{11} \over {15}}
Thus,  - \left( { - {{11} \over {15}}} \right) = {{11} \over {15}}
Hence, proved that - ( - x) = x.

(ii) x = - {{13} \over {17}}
The additive inverse of x = - {{13} \over {17}}is  - x = {{13} \over {17}}
Thus,  - {{13} \over {17}} + {{13} \over 7} = 0
The additive inverse of {{13} \over {17}}is  - {{13} \over {17}}
Hence, proved that  - ( - x) = x.

Q.4 Find the multiplicative inverse of the following.
(i) -13
(ii){{ - 13} \over {19}}
(iii) {1 \over 5}
(iv)  - {5 \over 8} \times {{ - 3} \over 7}
(v)  - 1 \times {{ - 2} \over 5}
(vi)1
Sol. The multiplicative inverse is defined as the reciprocal of the given number.
(i) -13
Hence, the multiplicative inverse of -13 is equal to {{ - 1} \over {13}}

(ii) {{ - 13} \over {19}}
Hence, the multiplicative inverse of {{ - 13} \over {19}}is equal to {{19} \over { - 13}}

(iii) {1 \over 5}
Hence, the multiplicative inverse of {1 \over 5} is equal to {5 \over 1} or 5.

(iv)  - {5 \over 8} \times {{ - 3} \over 7}
We know that,  - {5 \over 8} \times {{ - 3} \over 7} ={{( - 5) \times ( - 3)} \over {8 \times 7}} = {{15} \over {56}}
Hence, multiplicative inverse of {{15} \over {56}} is {{56} \over {15}}

(v)  - 1 \times {{ - 2} \over 5}
We know that,  - 1 \times {{ - 2} \over 5} = {2 \over 5}
Hence, multiplicative inverse of {{2} \over {5}} is {{5} \over {2}}

(vi)1
We know that, -1 is equal to {1 \over { - 1}} = -1
Hence, multiplicative inverse of -1 is -1.

Q.5 Name the property under multiplication used in each of the following.
(i) {{ - 4} \over 5} \times 1 = 1 \times {{ - 4} \over 5} = {{ - 4} \over 5}
(ii)
 - {{13} \over {17}} \times {{ - 2} \over 7} = {{ - 2} \over 7} \times - {{13} \over 7}
(iii)
 - {{19} \over {29}} \times {{29} \over { - 19}} = 1

Sol. (i) {{ - 4} \over 5} \times 1 = 1 \times {{ - 4} \over 5} = {{ - 4} \over 5}
We know that, 1 is the multiplicative identity for rational numbers.
Hence, the property of multiplicative identity is used here.

(ii)  - {{13} \over {17}} \times {{ - 2} \over 7} = {{ - 2} \over 7} \times - {{13} \over 7}
When rational numbers are swapped between one operators and still their result does not change, then we say that the numbers follow the commutative property for that operation.
Hence, commutative property is used here.

(iii)  - {{19} \over {29}} \times {{29} \over { - 19}} = 1
The reciprocal of is {{29} \over { - 19}}
Thus, multiplicative inverse property is used here.

Q.6 Multiply {6 \over {13}}by the reciprocal of {{ - 7} \over {16}}.
Sol. We know that, the reciprocal of {{ - 7} \over {16}} is {{16} \over { - 7}}
So,{6 \over {13}} \times {{16} \over { - 7}} = {{6 \times 16} \over {13 \times ( - 7)}}
 = {{96} \over { - 91}}

Q.7 Tell what property allows you to compute {1 \over 3} \times \left( {6 \times {4 \over 3}} \right) as \left( {{1 \over 3} \times 6} \right) \times {4 \over 3}
Sol. When rational numbers are rearranged between one or more same operations and still their result does not change then we say that they follow the associative property for that operation.
Thus, given equation follows the associative property.

Q.8 Is  the multiplicative inverse of  - 1{1 \over 8} ? Why or why not?
Sol. We can write,  - 1{1 \over 8} = {{ - 7} \over 8}
Now, multiplying both numbers we get, {8 \over 9} \times {{ - 7} \over 8} = {{ - 7} \over 9} \ne 1
The result is not equal to 1.
Hence,  - 1{1 \over 8} is not the multiplicative inverse of {8 \over 9}.

Q.9 Is 0.3 the multiplicative inverse of {1 \over 3}? Why or why not?
Sol. We know that, 0.3 = {3 \over {10}}
The multiplicative inverse of {3 \over {10}} is {{10} \over 3} 
Again, we know that {{10} \over 3} = 3{1 \over 3}
Hence, 3{1 \over 3} is the multiplicative inverse of 0.3

Q.10 Write.
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Sol. (i) Zero (0) is the rational number that does not have a reciprocal.
(ii) 1 and – 1 are the rational numbers that are equal to their reciprocals.
(iii) Zero (0) is the rational number that is equal to its negative.

Q.11 Fill in the blanks.
(i) Zero has __________ reciprocal.
(ii) The numbers ________ and ________ are their own reciprocals.
(iii) The reciprocal of – 5 is _____________.
(iv) Reciprocal of 1/x, where x ≠ 0is ______________.
(v) The product of two rational numbers is always a _____________.
(vi) The reciprocal of a positive rational number is ____________.
Sol. (i) No.
(ii) 1 and – 1.
(iii) -1/5.
(iv) x.
(v) Rational Number.
(vi) Positive.

 



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