Any number which can be expressed in the form of p/q, or number in the form of `p/q`; where q ≠ 0 is called rational number.

For example: `1/2`, `2/3`, `-3/4`, etc.

Since, q can be equal to 1, thus, all integers can be expressed in the form of p/q, hence, all integers are also rational number.

As, 0 (zero) is also an integer, hence, 0 (zero) is also rational number.

Question 1: Using appropriate properties find;

(i) `-2/3xx3/5+5/2-3/5xx1/6`

**Solution:** Given, `-2/3xx3/5+5/2-3/5xx1/6`

`=-2/3xx3/5-3/5xx1/6+5/2`

(Using commutativity)

`=3/5(-2/3-1/6)+5/2`

(Using distributivity)

`=3/5((-4-1)/(6))+5/2`

`=3/5(-5/6)+5/2`

`=3/5xx(-5)/(6)+5/2`

`=-3/6+5/2=(-3+15)/(6)`

`=(12)/(6)=2`

(ii) `2/5xx(-3)/(7)-1/6xx3/2+(1)/(14)xx2/5`

**Solution:** Given, `2/5xx(-3)/(7)-1/6xx3/2+(1)/(14)xx2/5`

Using commutative property, we get

`=2/5xx(-3)/(7)+(1)/(14)xx2/5-1/6xx3/2`

Using distributive property, we get

`=2/5(-3/7+(1)/(14))-1/6xx3/2`

`=2/5((-6+1)/(14))-1/6xx3/2`

`=2/5xx(-5)/(14)-1/6xx3/2`

`=-1/7-1/4`

`=(-4-7)/(28)=-(11)/(28)`

Question 2: Write the additive inverse of each of the following.

(i) `2/8`

**Solution:** Since, `2/8+(-2/8)`

`=2/8-2/8=0`

So, additive inverse of `2/8` is `-2/8`

(ii) `-5/8`

**Solution:** Since, `-5/9+5/9=0`

So, additive inverse of `-5/9` is `5/9`

(iii) `(-6)/(-5)`

**Solution:** `(-6)/(-5)=6/5`

Since, `6/5+(-6/5)=0`

So, additive inverse of `6/5` is `-6/5`

(iv) `(2)/(-9)`

**Solution:** Since, `(2)/(-9)+2/9=0`

So, additive inverse of `(2)/(-9)` is `2/9`

(v) `(19)/(-6)`

**Solution:** Since, `(19)/(-6)+(19)/(6)=0`

So, additive inverse of `(19)/(-6)` is `(19)/(6)`

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