$S = \mathbb{Z}, a * b = a+b^2$

Commutative: $a*b = b*a$

$a*b = a + b^2$ and $b*a = b+a^2$ and they aren't the same at all.

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

$(a*b)*c = (a+b^2)* c = a+b^2+c^2$ and $a*(b*c) = a + (b+c^2)^2$ and they aren't the same at all. therefore it's not a binary operation on the set of integers. But the book says it is binary operation, I don't know where my mistake is in.


1 Answer 1


A binary operation $f$ on $\mathbb{S}$ is defined as an operation that takes $2$ values $a, b\in \mathbb{S}$ and returns a single value $c\in\mathbb{S}$, regardless of commutativity or associativity. For example, subtraction is a binary operation on $\mathbb{Z}$ because it always returns a value in $\mathbb{Z}$ even though it isn't commutative or associative. Applying these criteria, $*$ is a binary operation on $\mathbb{Z}$

  • $\begingroup$ I understand that, but commutative and associativity aren't the same. If we plug in numbers LHS $\neq$ RHS at all. $\endgroup$
    – EM4
    Oct 18, 2020 at 16:20
  • $\begingroup$ "$c = f(a,b)$" is an equation, not an operation. If you fix that one huge error, your answer is good. $\endgroup$
    – user21820
    Oct 18, 2020 at 19:14
  • $\begingroup$ I edited to fix your last edit. "$f$" is not the same as "$f(a,b)$". When $f$ is a function, "$f(a,b)$" is an expression given $(a,b)$ in the domain of $f$. It is not a good idea to propagate the common abuse of notation that confuses these. $\endgroup$
    – user21820
    Oct 20, 2020 at 15:10
  • $\begingroup$ @user1993: If you understand this answer now, you should mark it as accepted. $\endgroup$
    – user21820
    Oct 20, 2020 at 15:12
  • $\begingroup$ @user21820, yes I did. $\endgroup$
    – EM4
    Oct 20, 2020 at 16:08

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