In article
<9d79e824-5323-4ad0-93e8-4650065ddab4@[EMAIL PROTECTED]
>,
Discrete Quest <don.scuquest@[EMAIL PROTECTED]
> wrote:
>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.
This is poor phrasing. You mean, if a property is meant to hold for
ALL pairs, then it is not possible to establish it by looking at just
ONE pair.
> 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
You are not being very precise here either. The set you are
considering is not a set of real numbers, it is a set of PAIRS of real
numbers.
This got garbled in my version. It was x = 2y.
So you let R be the set of all PAIRS of real numbers (x,y) such that
x=2y. Okay.
>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?
What is it you are trying to "use" the pair for?
The set R is the collection of ALL pairs (x,y) of real numbers in
which x = 2y. You can certainly produce any number of examples of such
pairs, but the point is that R is supposed to contain ALL such pairs.
Of the four basic properties of relations, NONE OF THEM can be
established to be "true for R" by simply looking at a specific examle
of a specific pair of real numbers, because ALL OF THEM are statements
about something happening FOR ALL pairs of real numbers (except for
reflexivity, which is a statement about something happening FOR ALL
real numbers).
> 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.
This is completely garbled. That statement is a description of what
pairs are and what pairs are not in R. It says: a pair (x,y) is an
element of the set R if and only if that pair satisfies the condition
x=2y.
Think of R like a Club, and the people in the city are all the pairs
(x,y). Not everyone is allowed to belong to the Club: the Club has
standards, after all. Like all Clubs, there are conditions for a
person to satisfy if they are to be members of the club. And the rules
of the Club R are: if a person (x,y) belongs to the club, then
x=2y. And if x=2y, then the person belongs to the club. This is a rule
that applies to EVERY person that lives in the city. Some will be
members of the club, some will not.
>If it doesn't
>they're no good to me.
If the pair (x,y) does not satisfy the condition x = 2y, then that
pair does not belong to the club. That does not mean the pair is "no
good to you". It just means it does not belong to the club.
What you want to keep in mind, though, is that the conditions of
Symmetry, Anti-Symmetry, and Transitivity are ->implications<-. They
are like the rules for "legacies" in the clubs (you know, if your dad
was a member of the club then you are a member of the club?). The
symmetry condition, for example, is that if (x,y) is an element of the
club (by virtue of satisfying x = 2y), then you would also need (y,x)
to be an element of the club.
If you want to show that this club does NOT meet this legacy rule,
then you need to exhibit a specific example of a pair (x,y) such that
x=2y (so that (x,y) is in the club), but (y,x) is not in the club. In
that sense, there is a pair that will be useful to you but is not in
the set. However, you CANNOT show that this club is not symmetric by
exhibiting a pair (x,y) such that NEITHER (x,y) nor (y,x) are in the
club. As in my example above: you cannot show that the club does NOT
follow the rule "if your dad was in the club then you are in the club"
by showing a man who is not a member and whose dad was ->also<- not a
member: you need someone who is not a member but whose dad ->was<- a
member. So here, think of (x,y) as the "dad" and (y,x) as the "son."
You are trying to check if this club satisfies this legacy rule;
either you have to show that whenever the dad is in the club the son
will also be in the club, or you need to exhibit one dad who IS in the
club but whose son is not.
>Right? So the set of all real numbers is a pool
>i can choose from correct?
You are NOT "choosing real numbers". You are not choosing
->anything<-, in fact. You are trying to see whether the club
satisfies certain rules or not.
>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)
That's not "your" definition of symmetry. That's THE definition of a
relation being symmetric.
>So i attempt to find a counter example by finding a set
No, not a set. A pair.
>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.
Yes. This relation is not symmetric, and one example ot show this is
the case is the pair (4,8), which is in the club, but whose "son"
(8,4) is not in the club.
>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) .
Suppose that (x,y) and (y,x) are both in R. If this can never happen,
then the implication would be true. However, unfortunately, you are
wrong in asserting that this never happens in this instance. You
->can<- have a pair (x,y) such that both (x,y) and (y,x) are in the
set. For example, x=y=0.
Suppose both (x,y) and (y,x) are in R. Then x=2y, and
y=2x. Substituting into the first equation, yo uget
x=2y=2(2x)=4x. Thus, x=4x, or 3x=0. Thus, we must have x=0, hence
y=0. So in fact, (0,0) is the ONLY pair that has this property. Thus,
it follows that if (x,y) and (y,x) are both in R, then x=y=0, hence
x=y. So this relation IS anti-symmetric.
>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.
Make up your mind. If you "cannot use it", then you cannot use it. You
cannot both be unable to ever use it, AND be able to use it sometimes.
>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.
Okay, you are correct that you cannot conclude the relation is
anti-symmetric by saying:
Well, look at (0,0): you have (0,0) in R, and you have (0,0) in R,
and 0=0.
However, you ->CAN< prove the relation is anti-symmetric as I did
above: show that the ONLY WAY in which the premise "(x,y) and (y,x)
are both in R" can be met is if x=y=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)]
Does your "-" represent negation? If so, yes, you can do this.
>(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.
That's not "too generic". It is simply not a valid logical
argument. "I cannot do it, therefore I think it cannot be done" is
called the Fallacy of Ignorance. And, being a Fallacy, it is not a
valid argument.
Suppose x is different from y and (x,y) is in R. Then x=2y by virtue
of being in R, and x is not zero, since if x were equal to 0 then you
would have 0 = 2y, from which you would know that y=0, from which you
would conclude that x=y, contradicting our hypothesis that x is
different from y.
If (y,x) is in R, then y=2x, from which you get x = 4x as before, from
which you get 4x-x = 0, from which you get 3x = 0, from which you get
x=0, which contradicts our assumption that x and y are different. This
contradiction arises from assuming that (y,x) is in R, hence you
conclude that (y,x) is not in R. And this shows that, IF x is
different from y, and (x,y) is in R, then (y,x) is NOT in R, which is
equivalent to showing the relation is anti-symmteric.
>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
Yes. You are done. for reflexicity.
>but (x,x) from any all real numbers
You realize that "any all real numbers" is nonsense? That expression
does not even make sense!
> is not in the relation x=3D2*x, so
>it doesn't do me good?
Again: in order to prove that Reflexivity does NOT hold, it is enough
to EXHIBIT one single example of a real number x for which (x,x) is
not in R. That you have done, so you are done. I have no idea what it
is you are confused about, given that the sentence you wrote does not
even parse as English.
You do NOT need to show that (x,x) is not in R for ALL real
numbers. To show that a relation NOT reflexive you only need to show
ONE, SPECIFIC, real number a such that (a,a) is not in R. And you did
that: 2 is a real number, and (2,2) is not in R, so R cannot be
reflexive. That is all.
>This is where my real number pool question
>above really hits. The only real number in the relation is 0.
This is again nonsense as written. Numbers are not "in the relation"
by themselves. The relation is BETWEEN TWO NUMBERS.
What you mean is that 0 is the only real number that is RELATED TO
ITSELF. Yes; that is indeed correct. So what? A relation is reflexive
if and only if EVERY real number is related to itself. As you noted,
not every real number is related to itself (some are, or to be more
precise, exactly one real number ->is<- related to itself, but so
what? You need ALL of them to be, and since they are not, the relation
is not reflexive).
>So this
>relation I have only the set (0,0) R=3D{(0,0)}
No, you do not.
What you have is that the only pair of the form (x,x) that is in the
set R is (0,0). But R is NOT equal to the set {(0,0)}; R contains lots
of OTHER pairs as well.
> even though I'm ask from
>all real numbers. And since (0,0) is the only set in the relation,
First, (0,0) is an ordered pair. Second, (0,0) is the only ordered
pair FROM AMONG THE PAIRS IN WHICH THE FIRST AND SECOND COORDINATES
ARE EQUAL that is in R, but it is not the only ordered pair in R.
Going back to the "club" analogy, you are looking at a specific family
(the "Jones-Jones", say), and are wondering if EVERY member of the
family is in the club. Turns out that no, just ONE member of the
family is the club. Does that mean that the club has only one member?
No; it means the club has only one member FROM THE JONES-JONES
family.
In the club R there are lots of other members, like (2,4), and
(-3,-6), and (pi,2pi). It's just that none of those other members are
in the Jones-Jones family.
--
======================================================================
"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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