Answer to Question #271521 in Discrete Mathematics for No one

Question

Given that 𝐺 = {𝑎 ∈ ℝ|𝑎 ≠ −1} and 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 + 𝑎𝑏, show that (𝐺, ∗) is indeed a group.

Answer 1

group properties:

associativity:

$(a*b)*c=(𝑎 + 𝑏 + 𝑎𝑏)*c=𝑎 + 𝑏 + 𝑎𝑏+c+(𝑎 + 𝑏 + 𝑎𝑏)c=$

$=𝑎 + 𝑏 + 𝑎𝑏+c+ac+bc+abc$

$a*(b*c)=a*(b+c+bc)=a+b+c+bc+a (b+c+bc)=$

$=a+b+c+bc+ab+ac+abc$

$(a*b)*c=a*(b*c)$

unit element e:

$e*a=e+a+ae=a$

$e=0$

inverse element b=a^-1:

$a*b=a+b+ab=e=0$

$b=-\frac{a}{a+1}$

so, (𝐺, ∗) is a group