In article
<40fae25b-cadb-4313-8d34-9400c24a8229@[EMAIL PROTECTED]
>,
Discrete Quest <don.scuquest@[EMAIL PROTECTED]
> wrote:
>
>> 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,
No. You are already on the wrong track.
You do NOT get to choose an "arbitrary pair". This argument must hold
for ALL pairs, so you do NOT get to "pick one".
You CANNOT prove symmetry by exhibiting an example.
The best you can do with an example is try to see why the argument
might hold ->in general<-.
> 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.
No, it is NOT enough to show symmetry. All you did was show that both
(3,-3) and (-3,3) are in the realtion, but you did NOT show that
WHENEVER a pair (x,y) is in the realtion, then the pair (y,x) will
also be in the relation.
>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.
No, you are wrong. "Arbitrary" here does not mean "you get to pick one
and check that and only that". It means it must hold for ANY pair that
satisfies the premise.
>Does this sound like I understand you correctly?
No, it sounds like you are still utterly confused about quantifiers.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
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