How to prove that given a binary operator $*$, $a * b = a^2 - ab + b^2$ is associative?

Question

No matter how hard I try I cannot seem to prove that, given the binary operator $*$, the operation $a * b = a^2 - ab + b^2$ is associative. This is what I have tried:

$(a * b) * c = a * (b * c) [Associativity];

$(a^2 - ab + b^2) * c = a * (b^2 - bc + c^2)$;

I have tried using $(a - b)^2$: $(a - b)^2 * c = a * (b - c)^2$;

Like other associativity problems, I could prove that but reaching an equality after expanding both sides, but that doesn't seems to be the case here. What am I doing wrong? Is this even an associative operation?

Answer 1

$$1 * (2 * 3)=1 * (2^2-2*3+3^2)=1 * 7=1^2-1*7+7^2=43$$ $$(1 * 2) * 3=(1^2-1*2+2^2) * 3=3 * 3=3^2-3*3+3^2=9$$ Therefore, $*$ is not associative. Always remember to guess and check values of $a,b,c$ to see if an operation is associative before trying to prove it!

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Now, to show you why this is not associative, let's expand this out. In the following equations, the quantities are not actually equal, but we are going to show this by expanding each side out separately.

As you said, we have: $$(a^2-ab+b^2) * c=a * (b^2-bc+c^2)$$ Substitute into the formula: $$(a^2-ab+b^2)^2-c(a^2-ab+b^2)+c^2=a^2-a(b^2-bc+c^2)+(b^2-bc+c^2)^2$$ Distribute the $-c$ on the left side and the $-a$ on the right sode: $$(a^2-ab+b^2)^2-ca^2+abc-cb^2+c^2=a^2-ab^2+abc-ac^2+(b^2-bc+c^2)^2$$ Use the $(u+v+w)^2=u^2+v^2+w^2+2uv+2uw+2vw$ identity to square the trionmials: $$a^4+a^2b^2+b^4-2a^3b+2a^2b^2-2ab^3-ca^2+abc-cb^2+c^2 \\ =a^2-ab^2+abc-ac^2+b^4+b^2c^2+c^4-2b^3c+2b^2c^2-2bc^3$$ Add like terms and rearrange terms: $$abc+a^4-2a^3b+3a^2b^2-2ab^3+b^4-ca^2-cb^2+c^2 \\ =abc+b^4-2b^3c+3b^2c^2-2bc^3+c^4-ab^2-ac^2+a^2$$ Hopefully, you see now that wherever there is $a$ on the left side, there is $b$ on the right side, wherever there is $b$ on the left side, there is $c$ on the right side, and wherever there is $a$ on the right side. Therefore, the right side is the left side with those substitutions, or $(b * c) * a$. However, originally, the right side was $a* (b * c)$, so: $$a * (b * c)=(b * c) * a$$ However, this just shows commutativity, not associativity, so this operation isn't associative since the left and right side in the above equations aren't the same.

Answer 2

No. $(a,b,c) = (4,2,1).$

$$4 * 2 = 16 - 8 + 4 = 12.$$ $$ 12 * 1 = 144 - 12 + 1 = 133 $$ $$ 2 * 1 = 4 - 2 + 1 = 3 $$ $$ 4 * 3 = 16 - 12 + 9 = 13 $$

Answer 3

Hint $\, $ Let $\,d(f(a))\,$ be the degree of $\,a\,$ in the polynomial $f(a)$.

Then $\,d(a*b) = 2\, $ so $\,d((a*b)*c) = 4\,$ but $\,d(a*(b*c)) = 2\,$ by $\,d(b*c) = 0\,$

Answer 4

Just do it:

$(a*b)*c = (a^2 - ab + b^2)*c = (a^2 - ab + b^2)^2-a^2c +abc-b^2c+c^2=a^4 -2a^3b + 3a^2b^2 -2ab^3+b^4-a^2c+abc-b^2c+c^2$

$a*(b*c) = a^2-a (b^2-bc+c^2) +(b^2 - bc + c^2)^2$

It's pretty clear these need not be equal. One is "dominated" by $a^4$ and $b^4$ while the other is dominated by $c^4$ and $b^4$.

Let $a=100$, $b=c=0$. Then $(a*b)*c = (a*0)^2 -(a*0)0+0^2=(a*0)^2=(100^2-0*100+0)^2=100^4$. Whereas $a*(b*c)=100^2 -100 (0*0)+(0*)^2=100^2-100 (0^2-0*0+0^2)+(0^2-0*0+0^2)^2=100^2$