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

Ok. Can you write down the right- and left-hand side and compare? — Peter Franek, Jun 19 '14 at 12:30
At the place where you are stuck, $a(b+c+2bc)$, the expression $(b+c+2bc)$ is one unit. Let's temporarily rename that unit $B$. Then $a(b+c+2bc)=a*B = (a+B+2aB)$. And now you re-substitute $B=(b+c+2bc)$ to get $(a+(b+c+2bc+)+2a(b+c+2bc)$ and the rest is algebra. Recognizing these "units" , and correctly applying this kind of substitution, is a basic and important mathematical skill. — Lee Mosher, Jun 19 '14 at 14:09

score 6 · Accepted Answer · answered Jun 19 '14 at 12:34

6

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.

answered Jun 19 '14 at 12:34

ajotatxe

65,084

Thanks but can you explain the last step were you merge the a and then the c with what is in the brackets? – CMB Jun 19 '14 at 12:39
@ConorMacBride: just set $x=(a+b+2ab)$ then $x*c=x+c+2xc$ then put $x$ in the last equation. – mesel Jun 19 '14 at 12:48

score 2 · Answer 2 · answered Jun 19 '14 at 12:56

2

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 $a$.

answered Jun 19 '14 at 12:56

lhf

216,483

See also http://math.stackexchange.com/a/373743/589 and http://math.stackexchange.com/a/513054/589. – lhf Jun 19 '14 at 12:57
This is a special case of transport of structure. – Bill Dubuque Jun 19 '14 at 13:50
@BillDubuque, that term is new to me, it makes sense, of course. Thanks! – lhf Jun 19 '14 at 13:53
1

Sometimes that name is only used in less trivial cases, e.g. in Gauss's Disq. Arith. the proof of associativity of composition of binary quadratic forms comprises many pages of unilluminating abstruse calculations, whereas nowadays this can be done simply by transporting the class group structure from ideals to primitive binary quadratic forms. And +1 for the general viewpoint. – Bill Dubuque Jun 19 '14 at 13:56

score 1 · Answer 3 · answered Mar 02 '20 at 23:15

1

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.

answered Mar 02 '20 at 23:15

Maxence1402

590

score 1 · Answer 4 · answered Jun 19 '14 at 12:44

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

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

score 0 · Answer 5 · answered Jul 09 '14 at 13:52

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.

Show that * is associative

5 Answers5