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Question Number 45794 by Tawa1 last updated on 16/Oct/18

Answered by MJS last updated on 16/Oct/18

a∗c=b ⇒ b∗c^(−1) =a ⇒ c^(−1) =b  b∗c=d ⇒ d∗c^(−1) =b ⇒ c^(−1) =b  c∗c=a ⇒ a∗c^(−1) =c ⇒ c^(−1) =b  d∗c=c ⇒ c∗c^(−1) =d ⇒ c^(−1) =b  ⇒ c^(−1) =b

$${a}\ast{c}={b}\:\Rightarrow\:{b}\ast{c}^{−\mathrm{1}} ={a}\:\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$$${b}\ast{c}={d}\:\Rightarrow\:{d}\ast{c}^{−\mathrm{1}} ={b}\:\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$$${c}\ast{c}={a}\:\Rightarrow\:{a}\ast{c}^{−\mathrm{1}} ={c}\:\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$$${d}\ast{c}={c}\:\Rightarrow\:{c}\ast{c}^{−\mathrm{1}} ={d}\:\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$$$\Rightarrow\:{c}^{−\mathrm{1}} ={b} \\ $$

Commented by Tawa1 last updated on 16/Oct/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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