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Question Number 190241 by astridmei last updated on 30/Mar/23
show that a⊛b=a+ab+b is a monoid when G=Z
$${show}\:{that}\:{a}\circledast{b}={a}+{ab}+{b}\:{is}\:{a}\:{monoid}\:{when}\:{G}={Z} \\ $$
Answered by mehdee42 last updated on 30/Mar/23
∗:Z×Z→Z  ; a∗b=a+b+ab  (Z,∗) is monoid whenever  1)∀a,b∈Z →a∗b∈Z  2)∀a,b,c∈Z→(a∗b)∗c=a∗(b∗c)  3)∃e∈Z ;∀a∈Z →a∗ e=e∗a=a  ∴  1)a∗b=a+b+ab∈Z  ✓  2)(a∗b)∗c=(a+b+ab)∗c=a+b+ab+c+(a+b+ab)c=a+b+c+ab+ac+bc+abc  a∗(b∗c)=a∗(b+c+bc)=a+b+c+bc+a(b+c+bc)=a+b+c+ab+ac+bc+abc        ⇒(2)✓  3) ∀ a∈Z →a∗0=o∗a=a ✓
$$\ast:\mathbb{Z}×\mathbb{Z}\rightarrow\mathbb{Z}\:\:;\:{a}\ast{b}={a}+{b}+{ab} \\ $$$$\left(\mathbb{Z},\ast\right)\:{is}\:{monoid}\:{whenever} \\ $$$$\left.\mathrm{1}\right)\forall{a},{b}\in\mathbb{Z}\:\rightarrow{a}\ast{b}\in\mathbb{Z} \\ $$$$\left.\mathrm{2}\right)\forall{a},{b},{c}\in\mathbb{Z}\rightarrow\left({a}\ast{b}\right)\ast{c}={a}\ast\left({b}\ast{c}\right) \\ $$$$\left.\mathrm{3}\right)\exists{e}\in\mathbb{Z}\:;\forall{a}\in\mathbb{Z}\:\rightarrow{a}\ast\:{e}={e}\ast{a}={a} \\ $$$$\therefore \\ $$$$\left.\mathrm{1}\right){a}\ast{b}={a}+{b}+{ab}\in\mathbb{Z}\:\:\checkmark \\ $$$$\left.\mathrm{2}\right)\left({a}\ast{b}\right)\ast{c}=\left({a}+{b}+{ab}\right)\ast{c}={a}+{b}+{ab}+{c}+\left({a}+{b}+{ab}\right){c}={a}+{b}+{c}+{ab}+{ac}+{bc}+{abc} \\ $$$${a}\ast\left({b}\ast{c}\right)={a}\ast\left({b}+{c}+{bc}\right)={a}+{b}+{c}+{bc}+{a}\left({b}+{c}+{bc}\right)={a}+{b}+{c}+{ab}+{ac}+{bc}+{abc}\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{2}\right)\checkmark \\ $$$$\left.\mathrm{3}\right)\:\forall\:{a}\in\mathbb{Z}\:\rightarrow{a}\ast\mathrm{0}={o}\ast{a}={a}\:\checkmark \\ $$$$ \\ $$

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