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.

  • 5
    $\begingroup$ Tell us what you've tried. $\endgroup$ – Alex Provost May 11 '13 at 22:46
  • 2
    $\begingroup$ What did you get for each of them? $\endgroup$ – Alex Becker May 11 '13 at 22:49
  • $\begingroup$ @tedg, welcome to Math.SE. Thank you for your questions. Normally here it is considered polite to ask just one question, and to share what you've tried so that we don't duplicate your efforts. $\endgroup$ – vadim123 May 11 '13 at 23:08
  • $\begingroup$ Note that when there is no neutral element, you can't even ask whether an element has an inverse. The very definition of inverse is predicated on the existence of a neutral element. $\endgroup$ – Gerry Myerson May 11 '13 at 23:27
  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?

  • $\begingroup$ Thanks for the help, I now understand the few i was unclear on. Sorry my attempts at the questions were vague, i just wanted to double check my answers. $\endgroup$ – tedg May 12 '13 at 11:13
  • $\begingroup$ @tedg: You’re welcome. It looks to me as if you mostly had the right idea. $\endgroup$ – Brian M. Scott May 12 '13 at 15:17

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