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Since there's no $value$ such that $b *$ $value$ = $b$ there's no identity element for *. Therefore, there's no inverse either.

Since $a * b = a$ and $b * a = a$, we have $a * b = b * a$, so the operation is commutative. I am not sure, though.

Since $(a * b)(a * b) = a * (b * ( a * b)) = (a * (a * b)) * b = a$, * is associative. Still not sure.

Is any of the above correct?

{a, b} has 16 operations defined on it. How do we know that?

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  • $\begingroup$ It's certainly associative, since no matter how you write the parentheses, the result is always $a$, and therefore the way you write the parentheses doesn't matter, so the operation is associative. $\endgroup$ – MJD Jun 21 '14 at 23:51
  • $\begingroup$ Try writing down the operation table for every one of the 16 possible operations and I think you'll soon see why it is 16. $\endgroup$ – MJD Jun 21 '14 at 23:53
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For commutativity, you have to check that $x * y = y * x$ for all $x,y \in \{a,b\}$. This includes $(x,y) = (a,a)$ and $(b,b)$.

For associativity, you have to check that $(x*y)*z = x*(y*z)$ for all $x,y,z \in \{a,b\}$, not just for some particular values of $x,y,z$.

Of course, both of these are easy since $x*y = a = y*x$ and $(x*y)*z = a = x*(y*z)$.

I can define a binary operation on $\{a,b\}$ by arbitrarily defining $a*a$, $a*b$, $b*a$, and $b*b$. For each one, I have two choices $a$ or $b$. So, how many binary operations on $\{a,b\}$ can I define?

Your reasoning for why there is no identity/inverse element is correct.

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  • $\begingroup$ $a * a = a * a = a$ and $b * b = b * b = a$. So * is commutative. Is at least this part correct? $\endgroup$ – DirtySkank Jun 21 '14 at 23:57
  • $\begingroup$ Yes, although you don't need to test it case by case. Since $x*y = a$ and $y*x = a$, we have $x*y = y*x$ for all $x,y \in \{a,b\}$. $\endgroup$ – JimmyK4542 Jun 21 '14 at 23:59
  • $\begingroup$ You don't have to check $(a,a)$ and $(b,b)$ for commutativity. How would those pairs ever fail? $\endgroup$ – Ben Millwood Jun 22 '14 at 0:09
  • $\begingroup$ $a * (b * c) = (a * b) * c?$ Is $b * c$ defined, though? $\endgroup$ – DirtySkank Jun 22 '14 at 0:14
  • $\begingroup$ I think you are mixing up notation. In the definition of commutativity: "$\forall a,b,c \in S: a*(b*c) = (a*b)*c$", $a,b,c$ are arbitrary elements of the set $S$. In this problem, our set is $\{a,b\}$, so $a$ and $b$ are specific elements of the set. $\endgroup$ – JimmyK4542 Jun 22 '14 at 0:19
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Answer is always the same for the whole set, so as MJD comments, it must be associative and I think it must be commutative but it can't have an inverse or identity element

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