Relations of functions

Question

I am doing an assignment and I want to make sure I understood my definitions can someone check my table and if I went wrong please tell me where and why.

Original question Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, irreflecive, and/or transitive where $(x,y) \in R$ if and only if $$(a) x+y=0\\ (b)x=\pm y \\ (c)x-y \in \Bbb Q\\ (d)x=2y $$

$$ \begin{matrix} & reflexive & symmetric & antisymmetric & irreflexive & transitive\\ x+y=0 & & x & & x & \\ x=\pm y & x & x & & & \\ x-y \in \Bbb Q & x & x & & & x\\ x=2y & & & x & x\\ \end{matrix} $$

What is the underlying set of the relations? $\mathbb{R}$? $\mathbb{C}$? — Clive Newstead, Apr 04 '14 at 02:17

score 2 · Accepted Answer · answered Apr 04 '14 at 02:25

2

The definition of irreflexive I've always seen is that the relation NEVER holds on an object with itself. Then the first and fourth are not irreflexive, since $0=2(0)$ and $0+0=0$. If your book/class uses a different definition, then you're fine, however.

answered Apr 04 '14 at 02:25

Hayden

16,737

so none of these are reflexive and a and d remain not reflexive – wolfcall Apr 04 '14 at 02:28
3

The second and third are reflexive, but the first and fourth are neither reflexive nor irreflexive. – Hayden Apr 04 '14 at 02:32
thanks I didn't think of the $(0,0)$ tuple – wolfcall Apr 04 '14 at 02:34
Yeah, a lot of people forget to check the possibility of zero. Similarly, people forget to check the possibility of the empty set when dealing with sets. Now you know! – Hayden Apr 04 '14 at 02:36
3

Your second relation is also transitive: if $x = \pm y$ and $y = \pm z$ then $x = \pm z$. (This is an equivalence relation, whose equivalence classes look like ${ x, -x }$.) – Clive Newstead Apr 04 '14 at 02:40
Thanks both Clive and Hayden for the help – wolfcall Apr 04 '14 at 11:16

Relations of functions

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