Write the additive inverse of :

(i) 5

(ii) -7

(iii) $\dfrac{5}{9}$

(iv) $\dfrac{-3}{17}$

(v) 0

(vi) $11\dfrac{5}{17}$

(vii) $-5\dfrac{3}{8}$

(viii) -37

(ix) 1

The additive inverse of a number is the number which, when added to the original number, results in zero.

(i) Let x be the additive inverse of 5, then :

⇒ 5 + x = 0

⇒ x = -5.

Hence, additive inverse of 5 = -5.

(ii) Let x be the additive inverse of -7, then :

⇒ -7 + x = 0

⇒ x = 7.

Hence, additive inverse of -7 = 7.

(iii) Let x be the additive inverse of $\dfrac{5}{9}$, then :

⇒ $\dfrac{5}{9}$ + x = 0

⇒ x = $-\dfrac{5}{9}$.

Hence,additive inverse of $\dfrac{5}{9} = -\dfrac{5}{9}$.

(iv) Let x be the additive inverse of $-\dfrac{3}{17}$, then :

⇒ $-\dfrac{3}{17}$ + x = 0

⇒ x = $\dfrac{3}{17}$.

Hence,additive inverse of $-\dfrac{3}{17} = \dfrac{3}{17}$.

(v) Let x be the additive inverse of 0, then :

⇒ 0 + x = 0

⇒ x = 0.

Hence, additive inverse of 0 is 0.

(vi) Let x be the additive inverse of $11\dfrac{5}{17}$, then :

⇒ $11\dfrac{5}{17}$ + x = 0

⇒ $\dfrac{192}{17}$ + x = 0

⇒ x = $-\dfrac{192}{17}$.

Hence, additive inverse of $11\dfrac{5}{17} = -\dfrac{192}{17}$.

(vii) Let x be the additive inverse of $-5\dfrac{3}{8}$, then :

⇒ $-5\dfrac{3}{8}$ + x = 0

⇒ $-\dfrac{43}{8}$ + x = 0

⇒ x = $\dfrac{43}{8}$.

Hence, additive inverse of $-5\dfrac{3}{8} = \dfrac{43}{8}$.

(viii) Let x be the additive inverse of -37, then :

⇒ -37 + x = 0

⇒ x = 37.

Hence, additive inverse of -37 = 37.

(ix) Let x be the additive inverse of 1, then :

⇒ 1 + x = 0

⇒ x = -1.

Hence, additive inverse of 1 = -1.