Associative Property

Question

What I learn as a basic theory of associative is that $$(a \times b) \times c = a \times (b \times c) \text{ and } (a+b)+c = a+(b+c).$$

However when doing my exercises on this topic, I came across questions with only 2 variables. Eg. $a^2+b^2$. Is this associative and why?

The binary operation is given as $a * b = a^2 + b^2$.

Answer 1

HINTS

If $a*b := a^2 + b^2$, what is $(a*b)*c$ and what is $a*(b*c)$? Can you check if they are the same?

Answer 2

"$+$" and "$\times$" are binary operations; the statement that they are associative of course uses three variables.

The question is: Given a binary operation $*$ defined by $a * b = a^2 + b^2$, is it associative? That is, is it true that $$ a * (b * c) = (a * b) * c? $$ Hint: no. Expand the left- and right-hand sides, using the definition of $*$, and find $a,b,c$ for which the two expressions are unequal.

Answer 3

It's not associative. We have $$(a*b)*c=(a^2+b^2)*c=(a^2+b^2)^2+c^2,$$ while $$a*(b*c)=a*(b^2+c^2)=a^2+(b^2+c^2)^2.$$ If you take for example $a=1$, $b=2$ and $c=3$ you get $$(a*b)*c=34$$ and $$a*(b*c)=170.$$

Answer 4

The definition of a binary operation only needs two variables to define it. The operation however needs be performable upon all the elements of a group. Associativity needs to be shown on three variables but most hold true for all elements.

So although the $a*b := a^2 + b^2$ is described with two variables it can be performed on all.

So to show * is associative you must show: $a*(b*c) = (a*b)*c$ or in other words does $a^2 + (b^2 + c^2)^2$ always equal $(a^2 + b^2)^2 + c^2$.

Answer 5

The property of a binary operation $a*b$ be associative allows us that the expression $a*b*c$ (and $a_1*a_2*.....a_n$ for a finite number n, in particular $\Sigma a_i$ and $\prod a_i$) be well defined. When this is not the case, you have in general $(a*b)*c\ne a*(b*c)$ so $a*b*c$ has no an univocal sense which happen with your operation $a*b=a^2+b^2$

Answer 6

For $a\ast b$ to be associative it should be $$(a\ast b)\ast c=a\ast (b\ast c),\text{ for all }a,b,c\in \Bbb R,\tag{1}$$ since I assume we are discussing a sum of squares of real numbers. This is, it should be $$(a^2+b^2)^2+c^2=a^2+(b^2+c^2)^2,\ \forall a,b,c\in \Bbb R.$$ But now it's not difficult to find a counterexample to the above equality: $$(0^2+1^2)^2+2^2=(0+1)^2+2^2=1^2+2^2=1+4=5$$ while $$0^2+(1^2+2^2)^2=0+(1+4)^2=0+5^2=0+25=25.$$ Since $(0\ast 1)\ast 2\ne 0\ast (1\ast 2)$, $(1)$ does not hold, this is $$a\ast b:=a^2+b^2$$ is not associative over the real numbers (and it is not over any subset of reals containing $0$, $1$ and $2$, like $\Bbb Q$ or $\Bbb N$).