Alright. It seem to be the best way that I try to go about this by
proving one case is false since for quantifier, it requires that all
arbitrary pair must be true else its false. I get the hang of that now
I think.
One more question. Don't be angry if I'm too slow to catch up. Here I
have the same. The set of all real numbers where (x,y) E R iff: x=3D2y
But before i make any more mistake, let me ask you to clear something
for me. Although the relation is on the set of all real number, I
cannot just use any pair of all real number right? Because this
statement : where (x,y) E R iff: x=3D2y : mean that any those pair of
real number have to be the one that satisfy x=3D2y. If it doesn't
they're no good to me. Right? So the set of all real numbers is a pool
i can choose from correct?
Based on that assumption here is what I did for finding symmetry for
(x,y) E R iff: x=3D2y
my symmetric definition say: AxAy((x,y)E R-->(y,x)E R)
So i attempt to find a counter example by finding a set that is in the
relation make true the premise but make the conclusion false. Eg. When
the set(x,y)=3D(8,4) e R. the definition say I should have (4,8). But 4!
=3D2*8. :. not symmetric.
Antisymmetric: AxAy(((x,y)E R ^ (y,x) E R)-->(x=3Dy)), since I can
never have a set (x,y) in the relation and (y,x) in the relation at
the same time as the previous seem to suggest(i know this is a generic
argument that i should avoid) . I cannot use ((x,y)E R ^ (y,x) E R)--
>(x=3Dy). The only time I can use that is when both (x,y) is 0,0. But
as you try so many times to tell me, showing one truth doesn't mean
true for all. So I won't conclude it is Antisymmetric just base on
case 0,0.
So maybe i can make use of an logical equivalent of Antisymm which say
=83pAx=83py[(x!=3Dy)-->((x,y) e R--> -(y,x) e R)]
(8!=3D4)-->((8,4) e R -->-(4,8) e R) so doesn't prove since its not a
counter example. Since I can't seem to prove that its not
antisymmetric I think it is antisymmetric. But is too generic.
For Reflexive: I am utterly confused on this one too. First glance I
say it is not reflexive. Because I can prove by counter example if
x=3D2, since 2!=3D2*2 .: not reflexive
but (x,x) from any all real numbers is not in the relation x=3D2*x, so
it doesn't do me good? This is where my real number pool question
above really hits. The only real number in the relation is 0. So this
relation I have only the set (0,0) R=3D{(0,0)} even though I'm ask from
all real numbers. And since (0,0) is the only set in the relation, and
0=3D2*0. Then it is reflexive?
Thanks for your time.
I know you're losing patient by my not getting this :)
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