Show that * is associative

Question

Could you show me how to prove the following to be associative? Please take me through the process step by step.

$$a*b=a+b+2ab$$

Where $*$ is a binary operation and $a$ and $b$ are real numbers.

I have got this far:

$$(a*b)*c=a*(b*c)$$ So $$(a+b+2ab)*c$$ And $$a*(b+c+2bc)$$

I am not sure what the next step would be.

Many thanks, Conor

Answer 1

The operation $*$ will be associative iff $$a*(b*c)=(a*b)*c$$

So let's compute both sides of equation:

$$a*(b*c)=a*(b+c+2bc)=a+b+c+2bc+2ab+2ac+4abc$$

$$(a*b)*c=(a+b+2ab)*c=a+b+2ab+c+2ac+2bc+4abc$$

Since they are the same, the operation is associative.

Answer 2

Given

$$ a * b = a + b + 2 a b $$

Then

$$ a * (b * c) = a * \big( b + c + 2 b c \big) = a + \big( b + c + 2 b c \big) + 2 a \big( b + c + 2 b c \big)\\ = a + b + c + 2ab + 2ac + 2bc + 4abc $$

and

$$ (a * b) * c = \big( a + b + 2 a b \big) * c = \big( a + b + 2 a b \big) + c + 2 \big( a + b + 2 a b \big) c\\ = a + b + c + 2ab + 2ac + 2bc + 4abc $$

so

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

Answer 3

Let $X=\mathbb R \setminus \{-1/2\}$ and consider the bijection $f\colon X \to \mathbb R \setminus \{0\}$ given by $f(x)=1+2x$. Then $$ a*b=a+b+2ab=(2a+2b+4ab)/2=(1+2a+2b+4ab-1)/2\\=f^{-1}((1+2a)(1+2b))=f^{-1}(f(a)\cdot f(b)) $$

So, the operation $a*b$ is the pullback of ordinary multiplication via $f$. By construction, all properties of $a*b$ are reflection of the properties of ordinary multiplication, and $f$ is an isomorphism of groups.

In particular, $a*b$ is associative, has $0=f^{-1}(1)$ as neutral element, and $f^{-1}(1/f(a))=a/(1+2a)$ as the inverse of $a$.

Answer 4

The operation $*$ will be associative if this condition is true $$a*(b*c)=(a*b)*c$$

So let's compute both sides of this equation:

$$f1=a*(b*c)=a*(b+c+2bc)$$

$$f1=a*(b+c+2bc)=a+(b+c+2bc)+2a(b+c+2bc)$$

$$f1=a+(b+c+2bc)+2a(b+c+2bc)=a+b+c+2bc+2ab+2ac+4abc$$

$$f2=(a*b)*c=(a+b+2ab)*c$$

$$f2=(a+b+2ab)*c=(a+b+2ab)+c+2c(a+b+2ab)$$

$$f2=(a+b+2ab)+c+2c(a+b+2ab)=a+b+c+2bc+2ab+2ac+4abc$$

As you can see $f1=f2$

So they are the same and the operation is associative.

Answer 5

A quicker way is to remark that $$a*(b*c)=a*(b+c+2bc)=a+b+c+2ab+2ac+2bc+4abc$$ is invariant under permutation of $a,b,c$ and $*$ is commutative (i.e. $a*b=b*a$). Therefore, $$a*(b*c)=c*(a*b)=(a*b)*c$$ i.e. $*$ is associative.