Discuss whether or not the following binary operations are commutative, associative, ...

Question

Discuss whether or not the following binary operations are commutative, associtive, have neutral elements and for which elements there are inverse elements. In between are what I have said, but if it doesn't seem right please let me know.

$1. (R, \circ ) \text{ with } x \circ y := |x \cdot y|;$

1: Commutative, Associative, no neutral element (unsure about this), and inverses for R+

$2. (N_0 ,\circ ) \text{ with } x \circ y := |x - y|; $

2: Commutative, Not Associative, neutral element = 0, and elements are own inverses

$3. (R, \circ ) \text{ with } x \circ y := 2 \cdot x - 3 \cdot y ;$

3: Not Commutative, Not Associative, no neutral element, and no inverses

$4. (R, \circ ) \text{ with } x \circ y := x + y^2;$

4: Not Commutative, Not Associative, no neutral element, and no inverses

$5. (R_{2 \times 2} , \circ ) \text{ with } A \circ B := A . B - B . A , (A \cdot B \text{ is matrix multiplication});$

5: Not Commutative, Not Associative, no neutral element, and no inverses

$6. (R, \circ ) \text{ with } x y := x + y - 2x^2y^2;$

6: Commutative, Not Associative, neutral element = 0, not sure about inverse.

$7. (\{\pm 1\} \times Z, \circ ) \text{ with } (\epsilon, a) \circ (\delta, b) := (\epsilon \cdot \delta, a \cdot\delta + b).$

I am unsure about this one.

Answer 1

1. This is indeed commutative and associative, and it has no neutral element, since there is no $x\in\Bbb R$ such that $x\circ-1=-1$, for instance. Since it has no neutral element, the question of inverses doesn’t even arise: inverses are defined in terms of a neutral element.

2. You’re fine here, though it would be good to provide a counterexample to associativity; $(3\circ 3)\circ 4=0\circ 4=4\ne 2=3\circ 1=3\circ(3\circ 4)$ is one.

3. Looks good.

4. This is also correct, though you might want to notice that the operation does have a one-sided neutral element on the right: $x\circ 0=x$ for all $\in\Bbb R$.

5. And again correct, though a counterexample to associativity would be a good idea.

6. For inverses fix $a\in\Bbb R$; to find $a^{-1}$, you must solve $a+x+2a^2x^2=0$ for $x$. That’s just a quadratic, so you know that $$x=\frac{-1\pm\sqrt{1-8a^3}}{4a^2}\;;$$ for which $a$ does this have a real solution?

7. Compare $\langle -1,0\rangle\circ\langle -1,1\rangle$ with $\langle -1,1\rangle\circ\langle -1,0\rangle$. Then multiply out and compare $$\big(\langle\epsilon_1,a_1\rangle\circ\langle\epsilon_2,a_2\rangle\big)\circ\langle\epsilon_3,a_3\rangle$$ and $$\langle\epsilon_1,a_1\rangle\circ\big(\langle\epsilon_2,a_2\rangle\circ\langle\epsilon_3,a_3\rangle\big)\;.$$ For a neutral element you want $\langle\epsilon_0,a_0\rangle$ such that $\langle\epsilon_0,a_0\rangle\circ\langle\epsilon,a\rangle$ for each $\langle\epsilon,a\rangle$, i.e., $\langle\epsilon_0\epsilon,\epsilon a_0+a\rangle=\langle\epsilon,a\rangle$. It’s clear that $\epsilon_0$ has to be $1$, not $-1$; is there an $a_0\in\Bbb Z$ that works?