Question

A relatedTo A = {(1,1), (1,2), (2,2), (3,4),
(4,1), (4,4)} This statement is AntiSymmetric; But, I don't understand why. Help?

Answer 1

Antisymmetric
 For every element x or y in the set, for every case where
(x,y) and (y,x) are both in the relation, x = y.

Answer 2

so for it to by antisymmetric we must have if R(a,b) and R(b,a) then a=b.

The relation is vacuously true since you do not have (2,1) or (4,3)...You have the elements to satisfy the hypothesis.

Answer 3

oops didn't see the rest of the set of (1,4)

Answer 4

The relation being antisymmetric is vacuously true is what I meant

Answer 5

remember the following true table:
p q if p then q
T T T
T F F
F T T
F F T

----
we have the third/fourth line going on

Answer 6

I am not seeing it. If we have (1,2) should we not also have (2,1). That would kinda of seem like x=y, then.

Answer 7

we don't have (2,1) in your set is what I'm saying

Answer 8

and, I am still trying to figure out how you drew up that truth table. I understand the truth table. But, where did it come from? What are the p's and q's?!

Answer 9

we can't fulfill the hypothesis of your statement
it makes no different if the conclusion is true or not

Answer 10

p is the hypothesis of your statement (the if part)
q is the conclusion of your statement (the then part)
p is (a,b) in R and (b,a) in R
q is a=b

Answer 11

p is false in your case since you not have (a,b) in R and (b,a) in R

Answer 12

so it makes no difference if q is false or true according to the true table
the statement is true

Answer 13

Hold on, let me read over this and see if I can't soak it in.

Answer 14

@freckles Okay, so in order for this to not be anti symmetric we need (x,y)(y,x) for every element in the set. Not just one?

Answer 15

A={(1,1),(1,2),(2,2),(3,4),(4,1),(4,4),(2,1)}
this one would not be antisymmetric
since we (1,2) and (2,1) we do have 1 does not equal 2

Answer 16

I just added a point to your original set to make it not antisymmetric

Answer 17

Wouldn't (1,2)(2,1) mean 1=2?

Answer 18

do you really think 1 is 2 though ?

Answer 19

1=2 is not true

Answer 20

1 is a different number then 2

Answer 21

Antisymmetric
For every element x or y in the set, for every case where
(x,y) and (y,x) are both in the relation, x = y.

Answer 22

x is different than y but says x=y here

Answer 23

right so it is NOT antisymmetric since we have to have 1=2 for it to be antisymmetric

Answer 24

but 1=2 is totally false

Answer 25

for 1=2 we need (1,2)(2,1) no?

Answer 26

how else can you make a=b (which i have written 1=2 because we were talking about the elements 1 and 2 (1,2)(2,1))

Answer 27

Do you see where my confusion is?

Answer 28

(1,2) and (2,1) is in R is true
but 1=2 is not true

Answer 29

why are you forcing it to be antisymmetric when it is not ?

Answer 30

if it doesn't satisfy the definition it just doesn't satisfy
you cannot make 1 be 2 ever

Answer 31

I can see it's not antisymmetric!

Answer 32

ok good

Answer 33

For every element x or y in the set, for every case where (x,y) and (y,x) are both in the relation, x = y. Okay, this is the definition from my book.

What I am trying to clarify is that for (1,2)(2,1), by this definition, it seems to me to say that 1=2. Because, in the first pair x=1 and y=2 in the second x=2 and y=1. The end of the definition says that when this happens x=y.

Answer 34

A = {(1,1), (1,2), (2,2), (3,4),(4,1), (4,4)}
A is antisymmetric since we cannot satisfy the hypothesis of antisymmetric (think true table)
B={(1,1),(1,2),(2,2),(3,4),(4,1),(4,4)}
B is not antisymmetric because we were able to satisfy the hypothesis of antisymmetric but we ere not able to satisfy the conclusion of the definition (think true table again if you need)

Answer 35

I can see from what you said it's not antisymetric. That's no problem. There is no 2,1 and that's good enough for me. But, I don't understnad the last part of that definition.

Answer 36

the then part says a has to be equal to b
but we have 1 is not 2
if 1 was 2 then yes it would by antisymmetric

Answer 37

but as I said we know 1 is not 2

Answer 38

if they were the same they would have a difference of 0
2-1=1 <--not 0

Answer 39

2 is greater than 1 actually

Answer 40

so it cannot be equal to

Answer 41

are you understanding the definition of antisymmetric is:
(x,y) in R and (y,x) in R then x=y
you understand we do not have this conclusion is true for that set B, right since 2 is not 1?

Answer 42

Okay, so I think I may have figured out where I was getting confused. (x,y) and (y,x) means x=y because both y and x have two values

Answer 43

we plug into the definition and got
True for hyp and false for conlusion which makes the statement false

Answer 44

if they had 1 value they could not be equivalent. I concede that 1>2

Answer 45

erm

Answer 46

2>1 :P

Answer 47

Hyp = (1,2) = T and (2,1) doesn't exist so False, yes? This doesn't imply a true conclusion.

Answer 48

ok I was trying to think of a set that is antisymmetric that has the hyp and conclusion satisfy
(and I know this example is going to be a dumb example but it is all I can of so far:
C={(x,1),(1,x) | x=1} is antisymmetric
why?
well it satisfies the hyp: we do have (x,1) and (1,x) is in the relation
we also have the conclusion since x=1
:p

Answer 49

right I think I get what you are saying...
B={(1,2),(2,1)} is not antisymmetric because when we try to plug into the definition which is:
if (x,y) and (y,x) are in the relation
then x=y
we get
if (2,1) and (1,2) are in the relation (which is true)
then 1=2 (which is false)

truth table tells us then this statement is false
so it is not antisymmetric

Answer 50

Yes, I think you can see what my issue was now. I was trying to plug this into the definition; but, I was only giving x and y 1 value. They, have two though. It might have made more sense if I had said "x has values 1 and 2. y has values 1 and 2. therefore, x=y."

Answer 51

so based on what you are you think E={(1,2),(2,2)} is antisymmetric or not antisymmetric?
also how about D={(x,y) in IntegerXInteger|y=x-1 or x=y-1}

Answer 52

E is antisymmetric. I'm still trying to figure out D. I'm assuming it would be antisymmetric also, because there is only 1 ordered pair.

Answer 53

That is right on E

lol I'm lost on D too... I kinda made it up...
well 5 and 6 integers which means
and 5=6-1
so (6,5) is in D (because of the y=x-1 thing)
is (5,6) also in D?
well 5=6-1 from the x=y-1 thing
so it does seem to be antisymmetric

--
thinking... a little more

Answer 54

I'm thinking, too. Am trying to, at least :P

Answer 55

ok let's look at the lines
y=x-1
and
x=y-1
solving the second equation for y gives y=x+1
f(x)=x-1 and g(x)=x+1 are inverses since
f(g(x))=f(x+1)=x+1-1=x
and
g(f(x))=g(x-1)=x-1+1=x

so since f and g are inverses that means
since (a,b) is in D then (b,a) is in D
that is for example:
since (7,8) is in D then (8,7) is in D
but 7 is not 8
err so I think is not antisymmetric

Answer 56

so I don't know why I said it was antisymmetric earlier
I was looking at (5,6) and (6,5) being in D
but 6 is not 5

Answer 57

that should have been the end of it

Answer 58

6>5!

Answer 59

:D

Answer 60

lol yes :p

Answer 61

that was interesting question
slightly harder

Answer 62

I've gota get some sleep Dr. Freckles. I'm sure I'll have more questions for you tomorrow if you're around. And, I'll leave this question up so I can re-read your brainteaser tomorrow. Believe it or not, this stuff does interest me. I'm just not very good at it-- yet...

Answer 63

i can understand the definitions being confusing at first

Answer 64

anyways goodnight

Answer 65

Good night :)