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Find the additive inverse of each of the following:

(i) \(\frac{1}{3}\)

(ii) \(\frac{23}{9}\)

(iii) \(-18\)

(iv) \(\frac{-17}{8}\)

(v) \(\frac{15}{-4}\) 

(vi) \(\frac{-16}{-5}\)

(vii) \(\frac{-3}{11}\)

(viii) 0

(ix) \(\frac{19}{-6}\)

(x) \(\frac{-8}{-7}\)

1 Answer

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Best answer

Additive inverse of a number \(\frac{a}{b}\) is the number \(-\frac{a}{b}\) such that, \(\frac{a}{b}\) + \((\frac{-a}{b})= 0\)

Therefore,

(i) Additive inverse of \(\frac{a}{b}\) is \(-\frac{a}{b}\) 

(ii) Additive inverse of \(\frac{23}{9}\) is \(\frac{-23}{9}\)  

(iii) Additive inverse of -18 is 18 

(iv) Additive inverse of \(\frac{-17}{8}\) is \(\frac{17}{8}\)

(v) 

\(\frac{15}{-4}= \frac{15\times-1}{-4\times-1}= \frac{-15}{4}\)

Therefore, 

Additive inverse of \(\frac{-15}{4}\)is \(\frac{15}{4}\)

(vi) 

\(\frac{-16}{-5}= \frac{-16\times-1}{-5\times-1}= \frac{16}{5}\)

Additive inverse of \(\frac{16}{5}\) is \(\frac{-16}{5}\) 

(vii) Additive inverse of \(\frac{-3}{11}\) is \(\frac{3}{11}\) 

(viii) Additive inverse of 0 is 0

(ix)

\(\frac{19}{-6}= \frac{19\times-1}{-6\times-1}= \frac{-19}{6}\) 

Therefore, 

Additive inverse of \(\frac{-19}{6}\) is \(\frac{19}{6}\) 

(x) 

\(\frac{-8}{-6}= \frac{-8\times-1}{-7\times-1}= \frac{8}{7}\)

Additive inverse of \(\frac{8}{7}\) is \(\frac{-8}{7}\)

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