For each of the following relations on the set of all integers, determine whether the relation is reflexive, symmetric, and/or transitive:

a. (π‘₯,𝑦)βˆˆπ‘… if and only if π‘₯<𝑦.

b. (π‘₯,𝑦)βˆˆπ‘† if and only π‘₯𝑦β‰₯1.

c. (π‘₯,𝑦)βˆˆπ‘‡ if and only π‘₯=βˆ’π‘¦.

d. (π‘₯,𝑦)βˆˆπ‘ˆ if and only π‘₯=|𝑦|

I've made some attempts at solving these of course, but have only yielded fitting answers for b and d, which I know are symmetric/transitive, and solely symmetric, respectively (unless I made some errors in deducing this which is definitely not out of the question). I even tried looking at post like the one below to determine whether or not what I was doing was correct, but it didn't provide much I could really use.
I appreciate your help and assistance

Determine whether the relations are symmetric, antisymmetric, or reflexive.

  • $\begingroup$ Welcome to Mathematics Stack Exchange. For reflexive, for all $x$, is (a) $x<x$? (b) $x^2\ge1$? (c) $x=-x$? (d) $x=|x|$? $\endgroup$ – J. W. Tanner May 3 '19 at 19:12

In this type of exercise, you need to apply the definition of these types of relations and you have to ask yourself the right questions:

  • Reflexivity : for all $x$, is it true that $$ x < x \text{ (a). }\\ x^2 \ge 1 \text{ (b). }\\ x = -x \text{ (c). }\\ x = |x| \text{ (d). } $$

  • Symmetry : for all $x$ and $y$, is it true that $$ x<y \implies y<x \text{ (a). }\\ xy \ge 1 \implies yx \ge 1 \text{ (b). }\\ x = -y \implies y = -x \text{ (c). }\\ x = |y| \implies y = |x| \text{ (d). } $$

  • Transitivity : for all $x,y,z$, is it true that $$ x<y \text{ and } y<z \implies x<z \text{ (a). }\\ xy \ge 1 \text{ and } yz \ge 1 \implies xz \ge 1 \text{ (b). }\\ x = -y \text{ and } y = -z \implies x = -z \text{ (c). }\\ x = |y| \text{ and } y = |z| \implies x = |z| \text{ (d). } $$

Can you finish? If you have a question don't hesitate.

  • $\begingroup$ Symmetry: for all $x$ and $\mathbf y$ ... $\endgroup$ – J. W. Tanner May 3 '19 at 19:37
  • $\begingroup$ Sorry about that. $\endgroup$ – Monadologie May 3 '19 at 19:39
  • $\begingroup$ Not a big deal; thanks for correcting. I always remember that R, S, and T have 1, 2, and 3 variables, respectively $\endgroup$ – J. W. Tanner May 3 '19 at 19:45
  • $\begingroup$ I'm looking through it now since I had to step out. I understand it for the most part $\endgroup$ – ThePooberteen May 3 '19 at 20:03

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