In article
<7fcaa3b7-d291-414a-807a-18ffefba9421@[EMAIL PROTECTED]
>,
Discrete Quest <don.scuquest@[EMAIL PROTECTED]
> wrote:
>I meant..
[...]
>> 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.
You are making the common mistake of thinking that the mathematical
phrase "for any" or "for an arbitrary" means "pick one". It does not.
In mathematics, "one arbitrary pair" does not mean "pick one and check
that." What it means is "it works for each and every pair, no matter
what the pair is." So in order to prove that something holds "for an
arbitrary pair", you need to verify that it works for ->each and every
pair<-, not just for one.
In the case of relations on the real numbers, NONE of the properties
of reflexivity, symmetry, anti-symmetry, and transitivity can be
verified simply by looking at a single specific example. All of these
properties must hold "for all" choices.
So, to show reflexivity, you have to show that ->for every<- real
number x, (x,x) is in R. Sometimes we rephrase this by saying that
"(x,x) is in R for arbitrary x"; that does not mean "(x,x) is in R for
some pick of x"; it means "(x,x) is in R, no matter what x is"; that
is, "for any x."
To show symmetry, you have to show that ->for every two real
numbers<- x and y, if (x,y) is in R then (y,x) must also be in R. To
show anti-symmetry you have to show that ->for every two real
numbers<- x and y, if both (x,y) and (y,x) are in R, then x must be
equal to y. And to show transitivity you have to show that ->for every
three real numbers x,y,z<-, if (x,y) and (y,z) are both in R, then
(x,z) will also be in R.
Of them, "reflexivity" ->is<- a little different from the rest,
because the statement of reflexivity is a simple assertion: it says
that a certain pair WILL be in R. On the other hand, the other three
properties are ->implications<-, not assertions of fact. What they say
is that ->if<- a certain condition holds, ->then<- another condition
must hold. So in order for symmetry to be true, for example, given any
two real numbers x and y, if (x,y) is not in R then you are done
(because the premise of the implication fails), or else if (x,y) is in
R then you need for (y,x) to also be in R.
Think about the a professor saying "Every student in my class has an
A". Can the professor prove this assertion to his Dean by just
bringing a particular student over and showing that this one student
has an A? No. He would have to parade each and every student and show
that each and every one has an A. On the other hand, in order for the
Chair of the Department to show that the professor is lying, the Chair
does only have to show a single student who does not have an A.
Likewise, to prove that the relation is reflexive you have to check
(x,x) holds for EVERY x; now, since there are infinitely many reals,
you won't be able to do it by staring at all the pairs one at a time,
so instead you need to give an argument to show that it does work
->always<-. That's the "proof" you are looking for.
However, in order to prove that the relation is ->not<- reflexive, you
->do<- only need to show a single, specific, number x such that (x,x)
is not in R. Similarly with the other three properties: none of them
can be shown to hold by giving a single example, but each of them can
be shown to be false by exhibiting a single specific example (and in
fact, the only real way to show the property does not hold is to show
a specific example; one should not present a generic argument saying
something "does not have to happen", or "there is no way to ensure it
happens", etc. Just show an example of it not happening).
--
======================================================================
"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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