>
> > To determine whether R is symmetric, you must begin
> by assuming you
> > have an ARBITRARY pair (x,y) of real numbers that
> IS in the
> > relation. That is, either x+y = 0, or x-y=0. You
> want to check
> > whether this condition guarantees that (y,x) will
> also be in the
> > relation. So you are asking:
> >
> > If (either x+y=0 or x-y=0), does it follow that
> necessarily we
> > will have (either y+x=0 or y-x=0) ?
>
> Let me get this straight. Suppose I have choose
> (-3,3) as my
> arbitrary pair, which satifisy the premise (x+y=0).
> From the symmetric definition which have ( x+y =0 v
> x-y=0)-->(y+x =0 v
> y-x=0) then the implication is true
> since for pair (y,x) it satisfy (y+x=0) or (-3+3=0)
> which one part of
> the conclusion...
>
> And that is enough to show symmetry. So unlike
> reflexive, i do not
> have to show for every value of x, in here, i just
> choose one
> arbitrary pair and show that the implication is true.
>
> Does this sound like I understand you correctly?
> Thank you.
No, it is not a correct mathematical proof. Even if the proof for general
case is analoguos with the proof for x=-3( or other particular case), you
have to prove for the general case.
for any x and y who satisfy the condition (x+y=0 or x-y=0), we also have
(y+x=0 or y-x=0).
or in a more formal way:
any x, any y ((x+y=0 or x-y=0) ---> (y+x=0 or y-x=0)).
|
