In article
<9687317a-4ec4-4d5c-b43d-c2f7e3208a3e@[EMAIL PROTECTED]
>,
Discrete Quest <don.scuquest@[EMAIL PROTECTED]
> wrote:
>I have some question i hope anyone could clarify for me.
>I am asked whether a relation R on set of all real numbers is
>reflixve, symmetric, antisymetric and/or transitive where (x,y) E R
>iff:
This statement is incomplete. Where (x,y) is in the relation if and
only if ->what<-?
>I have the formal defintions for each type I'm still not understanding
>them.
The formal definition of ->what<-? Of the terms "reflexive",
"symmetric", "antisymmetric" and "transitive"?
>one the question is
>x=+-y,
Oh, I see. Your relation is
(x,y) in R if and only if x=y or x=-y
and you want to know if this relation is reflexiv, symmetric,
antisymmteric, or transitive.
> I reason that this is reflexive because this relation is same
>as (x+y=0 or x-y=0)
True.
> and if I use x=1 as example, then 1+1!=0 but 1-1=0 so its true.
This is not good enough to show the relation is reflexive. It only
shows you that (1,1) is in R (that is, that 1 is related to itself).
The property of the relation R being reflexive is not that there is
ONE element that is related to itself. For R to be reflexive you need
EACH AND EVERY REAL NUMBER to be related to itself. That is, EVERY
real number x must be related to Itself. So checking 1 is not enough
to show that R is reflexive.
A single example of a single number a which is not related to itself
would be enough to guarantee the relation is NOT reflexive; but no
single example will suffice to show that the relation ->is<-
reflexive.
So what you ->really<- need to answer is:
Is it the case that, FOR EVERY REAL NUMBER x, either x+x=0 or else
x-x=0?
The answer, of course, is "yes", because no matter what x is, x-x=0,
hence (x,x) is in R. So R is relfexive.
>I say this is not symmetric because: if I use (1,2) by definition
>(2,1) would have to be in it, but
No, this does not work. The definition of "symmetric" is an
implication, and here your example does NOT satisfy the premise. For R
to be symmetric, you need to show that
IF (x,y) is in R, THEN (y,x) is also in R.
But (1,2) is NOT in R, so you cannot use this to argue that this
implication is false. The only way you can show that R is NOT
symmetric would be to exhibit a pair (x,y) which IS in R, but such
that (y,x) is NOT in R. Since neither (1,2) nor (2,1) are in R, this
example does you absolutely no good whatsever.
>1+2 !=0 or 1-2!=0 also. So that case is false. Remember that this on
>the set of all real numbers.
>So is my reason okay?
No. It is completely wrong (in both instances).
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) ?
If the answer is "yes", then the relation is symmetric. If the answer
is "no", then you should be able to come up with a specific example
which DOES satisfiy the premise but not the conclusion; your example,
however, fails to satisfy BOTH the premise AND the conclusion, so it
is no good.
>If you could explain also why this is antisymetric,
It is NOT antisymmetric.
In order to be antisymmetric, the relation R would have to satisfy
that:
If (x,y) and (y,x) are both in R, then we must have x=y.
Translating this into the condition you have, you are saying:
If both (x+y=0 or x-y=0) AND (y+x=0 or y-x=0), THEN we must have
x=y.
But this is false: take, for example, x=1, y=-1. Then we have (x,y) in
R, because x+y=0; and we have (y,x) in R, because y+x=0, but it is NOT
true that x=y, so the relation is not anti-symmetric.
> and transitive
To determine whether the relation is transitive or not, you have to
check whether:
Is it true that if (x,y) and (y,z) are both in R, with x,y,z
arbitrary real numbers, then (x,z) will be in R?
Note, you cannot just pick any x, y, and z and just check to see if
(x,z) is in R: they MUST also satisfy the premise of having (x,y) in R
and (y,z) in R. So what you are really asking is:
Suppose that x,y,z are real numbers such that:
(i) either x+y=0 or x-y=0; AND
(ii) either y+z=0 or y-z=0.
Does it then follow that either x+z=0 or x-z=0 ?
Well... If x+y=0 and y+z=0, then (x+y)-(y+z)=0-0=0, and since
(x+y)-(y+z)=x-z, yes, the conclusion would follow.
If x+y=0 and y-z=0, then 0=0-0 = (x+y)-(y-z)=x+z, so again, yes, the
conclusion would follow.
What if x-y=0 and y-z=0? What if x-y=0 and y+z=0? If in both these
cases you are also able to deduce that you will have x+z=0 or x-z=0,
then the answer is "yes", and the relation is transitive.
On the other hand, if there is one fo those situations in which you
CANNOT seem to conclude that eitheer x+z=0 or x-z=0, that will suggest
where to search for a specific trio of integers x, y, and z such that
(x,y) and (y,z) are in R, but (x,z) is not in R.
>also that would really put off on a good start.
Actually, you are not off to a good start, so you really need to go
back and start over.
--
======================================================================
"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
|
