I am doing an exercise where I need to find which one of the statements under are reflexive, symmetric and transitive. I will write what I think for each of the statements, could someone see if I am thinking right
Our domain is all real numbers:
a) x + y = 0
This one is reflexive only for (0,0) but not for all real numbers, it is symmetric for (0,0) but not for all real numbers, but I dont know how to see if it is symmetrical.
b) x = +- y
This is not reflexive for domain R, because (1,1) not true
I dont know how to see if its not symmetrical or not.
c) x = 2y
Not reflexive, (1,1) not true
to find symmetric do we do it like this? x = 2y -> x/2 = y means not symmetric
dont know how to find transitive.
d) xy >= 0
Not reflexive, (1,1) not treu
symmetric because xy, y*x
transitive?
e) x*y = 0
same as d)
f) x = 1 or y = 1
Not sure what to do
Kinda long post, but would love to get some help. Thank you
Best Answer
a) It's not reflexive but symmetric because addition is commutative: $x+y=0\iff y+x=0$. It's not transitive: for that it's enough to show up a specific counterexample, like $(2,-2)\in R,\ (-2,2)\in R$ while $(2,2)\notin R$ where $R$ denotes the relation defined in the exercise.
b) I think it's meant as $R=\{(x,y):x=y$ or $x=-y\}$. It's reflexive, symmetric and also transitive. To see these more directly we can restate $R=\{(x,y):|x|=|y|\}$.
c) It's not reflexive, not symmetric ($(1,2)\in R$ but $(2,1)\notin R$), and not even transitive: $(1,2),\ (2,4)\in R$ but $(1,4)\notin R$.
d) It's reflexive: for every real number $x$ we have $x\cdot x\ge 0$. It's also symmetric because multiplication is commutative, but it's not transitive: can you find three elements to witness this like above?
e) It's not reflexive: e.g. $(1,1)\notin R=\{(x,y):xy=0\}$ but symmetric, and also not transitive.
f) It's very similar to e) because $xy=0\iff x=0$ or $y=0$. Now $(1,1)$ happens to be in $R=\{(x,y):x=1$ or $y=1\}$ but no other $(x,x)$ pair, e.g. $(2,2)\notin R$.