> 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.
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