Show that * is associative
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
$\endgroup$ 55 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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 $a$.
$\endgroup$ 4 $\begingroup$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
$$ a*(b*c) = (a*b)*c $$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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