# Relations- reflexive, symmetric, anit-symmetric, transitive

• kevinsweeney
In summary, R2 is transitive because for (x,y) in R2 and (y,z) in R2, there is a (x,z) in R2. R5 is transitive because there are no terms (x,y) and (y,z) that imply (x,z) in R5, making it unproven to be false and therefore true. The remaining statements are not correct.
kevinsweeney
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Suppose that

R1={(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)},
R2={(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)},
R3={(2,4),(4,2)} ,
R4={(1,2),(2,3),(3,4)},
R5={(1,1),(2,2),(3,3),(4,4)},
R6={(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)},

Determine which of these statements are correct.
Check ALL correct answers below. A. R3 is symmetric
B. R5 is transitive
C. R6 is symmetric
D. R2 is reflexive
E. R1 is not symmetric
F. R4 is symmetric
G. R4 is transitive
H. R4 is antisymmetric
I. R3 is reflexive
J. R1 is reflexive
K. R3 is transitive
L. R5 is not reflexive
M. R2 is not transitiveA. R3 is symmetric as (x,y) implies (y,x). So (4,2),(2,4)
B. R5 is transitive as there is no (x,y) (y,z)
C. R6 isn't symmetric as there is (1,4) but no (4,1)
D. R2 ia reflexive as (1,1),(2,2),(3,3),(4,4) are elements
E. R1 isn't symmetric as (4,4) isn't an element
F. R4 isn't symmetric as there is no (2,1) for (1,2)
G. R4 is transitive as there is no (x,y) (y,z) relation
H. R4 isn't antisymmetric as its not symmetric
I. R3 isn't reflexive as there is no (2,2) for (2,4)
J. R1 is reflexive as there is as all (x,x) are satisfied
K. R3 isn't transitve as there is no (2,2) for (2,4),(4,2)
L. R5 is reflexive as all (x,x) satisfied
M. R2 is transitive as (1,1) for (1,2),(2,1)

Hi, I got a mark of 0% for this answer, and have no idea where I am going wrong. Can anybody help? Thanks Kevin

Determine which of these relations are reflexive. The variables x, y, x', y' represent integers.

A. x∼y if and only if xy is positive.
B. x∼y if and only if x+y is even.
C. x∼y if and only if x−y is a multiple of 10.
D. x∼y if and only if x−y is positive.
E. x∼y if and only if x+y is odd.

A. xy is positive, so xy>0. So if x >=1, then (x,x) is satisfied. True
B. If x+y is even, then x and y are both even or both odd. So odd(1,1) is even and even(2,2) is even. Therefore true
C. x-y is a multiple of 10, but x-x is always 0, so its false.
D. x-y is positve, but if x(1) and y(0) then (x,x) is 0, not positive, therefore false.
E. x+y is odd therefore x is odd and y is even, or x is even and y is odd. But even + even, and odd + odd is always even. Therefore false.

Again..I got a mark of 0% for this answer, and have no idea where I am going wrong. Can anybody help?

x∼y if and only if x−y is a multiple of 10 is true and so is x∼y if and only if x+y is even :)

If it makes you feel any better I'm stuck on question 2.

_VexatioN said:
If it makes you feel any better I'm stuck on question 2.

I got these sorted. What part of q2 are you stuck with?

I got Q2 eventually; I also got Q5 but Q's 3 and 4 are beyond me.
Question 3 goes right over my head,
I keep getting Question 4 wrong and no clue why.

Question 4:

(1 pt) Determine which of these relations are reflexive. The variables x, y, x', y' represent integers.

A.
x∼y if and only if x+y is positive.
B. x∼y if and only if xy≥0 \).
C. x∼y if and only if x+y is odd.
D. x∼y if and only if xy is positive.
E. x∼y if and only if x−y is a multiple of 10.I think that B and D are reflexive, not sure about E. I get it wrong whether or not I tick E though.

D is not reflexive as 0(0) for (x,x) is not positive.
E is reflexive as x -x will always be 0. 0 is a multiple of EVERY number.
yes, b is reflexive as (0)0 is always >= 0.And how about q3?

kevinsweeney said:
D is not reflexive as 0(0) for (x,x) is not positive.
E is reflexive as x -x will always be 0. 0 is a multiple of EVERY number.
yes, b is reflexive as (0)0 is always >= 0.And how about q3?

Suppose that

R1={(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)},
R2={(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)},
R3={(2,4),(4,2)} ,
R4={(1,2),(2,3),(3,4)},
R5={(1,1),(2,2),(3,3),(4,4)},
R6={(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)},

Determine which of these statements are correct.
Check ALL correct answers below.A. R2 is not transitive
B. R4 is antisymmetric
C. R4 is symmetric
D. R3 is transitive
E. R3 is reflexive
F. R4 is transitive
G. R1 is not symmetric
H. R1 is reflexive
I. R2 is reflexive
J. R6 is symmetric
K. R5 is transitive
L. R3 is symmetric
M. R5 is not reflexive

----------------------------------------

Don't even understand the notation in this one, despite reading the notes twice.

i can't see your answers, so can't tell where you are going wrong? Can you put up a screeshot/grab

Sorry, I probably should have just marked the ones I ticked.

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_VexatioN said:
Suppose that

R1={(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)},
R2={(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)},
R3={(2,4),(4,2)} ,
R4={(1,2),(2,3),(3,4)},
R5={(1,1),(2,2),(3,3),(4,4)},
R6={(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)},

Determine which of these statements are correct.
Check ALL correct answers below.A. R2 is not transitive
B. R4 is antisymmetric
C. R4 is symmetric
D. R3 is transitive
E. R3 is reflexive
F. R4 is transitive
G. R1 is not symmetric
H. R1 is reflexive
I. R2 is reflexive
J. R6 is symmetric
K. R5 is transitive
L. R3 is symmetric
M. R5 is not reflexive

----------------------------------------

Don't even understand the notation in this one, despite reading the notes twice.
_VexatioN said:
Sorry, I probably should have just marked the ones I ticked.

A) r2 is transitive as for (x,y)[1,2](y,z)[2,1] there is a (x,z)[1,1]
k) r5 is transiitve as there is no terms For (x,y)(yz) implying (x,z), therefore the theory is true because we can't prove it false(maths logic)

everything else looks right

A) Even though it doesn't apply to the whole set?
K) >Not proven to be false. > Therefore true.

You have to be joking me with the logic for those two! Thanks so much for the help, as you may have noticed I would have been screwed if you haven't have helped me with those. Dave C., Coffee on me if you ever want one.

kevinsweeney said:
A) r2 is transitive as for (x,y)[1,2](y,z)[2,1] there is a (x,z)[1,1]
k) r5 is transiitve as there is no terms For (x,y)(yz) implying (x,z), therefore the theory is true because we can't prove it false(maths logic)

everything else looks right
_VexatioN said:
A) Even though it doesn't apply to the whole set?
K) >Not proven to be false. > Therefore true.

You have to be joking me with the logic for those two! Thanks so much for the help, as you may have noticed I would have been screwed if you haven't have helped me with those. Dave C., Coffee on me if you ever want one.

A) But it does apply for the whole set. There is no other terms fitting the criteria of (x,y)(y,z) only (1,2)(2,1) as you can see y is 2 in both. Can you show me another (x,y)(y,z) in the set?
k) we came across that logic in truth tables if you remember, we had let's say x implies y. If x is false, that means y is automatically true, as we can't prove otherwise.

kevinsweeney said:
A) But it does apply for the whole set. There is no other terms fitting the criteria of (x,y)(y,z) only (1,2)(2,1) as you can see y is 2 in both. Can you show me another (x,y)(y,z) in the set?
k) we came across that logic in truth tables if you remember, we had let's say x implies y. If x is false, that means y is automatically true, as we can't prove otherwise.

A) So as long as any elements {x,y,z} in the set follows the pattern (x,y) , (y,z), (x, z) the set is transitive?
The reason I thought it wasn't transitive is that there's nothing linking (2,2), (3,3), or (4,4) to each other.

_VexatioN said:
A) So as long as any elements {x,y,z} in the set follows the pattern (x,y) , (y,z), (x, z) the set is transitive?
The reason I thought it wasn't transitive is that there's nothing linking (2,2), (3,3), or (4,4) to each other.
...No, its not {x,y,z} in the set(as that would imply we could take any 3 numbers of our choice!), its the elements linked by "y" with (x,y) (y,z)...y just has to be in the 2nd position in the first set, and 1st position in the second set.
There is actually another example in that set. (1,1)(1,2). as you can see y is 1 here, so do we have (x,z)[1,2]? Yes

kevinsweeney said:
...No, its not {x,y,z} in the set(as that would imply we could take any 3 numbers of our choice!), its the elements linked by "y" with (x,y) (y,z)...y just has to be in the 2nd position in the first set, and 1st position in the second set.
There is actually another example in that set. (1,1)(1,2). as you can see y is 1 here, so do we have (x,z)[1,2]? Yes

"y just has to be in the 2nd position in the first set, and 1st position in the second set"

and this has to be true for EVERY set in the function; or ANY set in the function?

_VexatioN said:
"y just has to be in the 2nd position in the first set, and 1st position in the second set"

and this has to be true for EVERY set in the function; or ANY set in the function?
yes, it must be true for EVERY 2 pairs to which it applies to

kevinsweeney said:
Suppose that

R1={(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)},
R2={(1,1),(1,2),(2,1),(2,2),(3,3),(4,4)},
R3={(2,4),(4,2)} ,
R4={(1,2),(2,3),(3,4)},
R5={(1,1),(2,2),(3,3),(4,4)},
R6={(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)},

Determine which of these statements are correct.
Check ALL correct answers below. A. R3 is symmetric
B. R5 is transitive
C. R6 is symmetric
D. R2 is reflexive
E. R1 is not symmetric
F. R4 is symmetric
G. R4 is transitive
H. R4 is antisymmetric
I. R3 is reflexive
J. R1 is reflexive
K. R3 is transitive
L. R5 is not reflexive
M. R2 is not transitiveA. R3 is symmetric as (x,y) implies (y,x). So (4,2),(2,4)
B. R5 is transitive as there is no (x,y) (y,z)
C. R6 isn't symmetric as there is (1,4) but no (4,1)
D. R2 ia reflexive as (1,1),(2,2),(3,3),(4,4) are elements
E. R1 isn't symmetric as (4,4) isn't an element
F. R4 isn't symmetric as there is no (2,1) for (1,2)
G. R4 is transitive as there is no (x,y) (y,z) relation
H. R4 isn't antisymmetric as its not symmetric
I. R3 isn't reflexive as there is no (2,2) for (2,4)
J. R1 is reflexive as there is as all (x,x) are satisfied
K. R3 isn't transitve as there is no (2,2) for (2,4),(4,2)
L. R5 is reflexive as all (x,x) satisfied
M. R2 is transitive as (1,1) for (1,2),(2,1)

Hi, I got a mark of 0% for this answer, and have no idea where I am going wrong. Can anybody help? Thanks Kevin
i think R4 is not Transitif because it has (1,2), (2,3) then (1,3) is not in the R4. so does (2,3), (3,4) then (2,4) and its not in R4 either. it has (x,y), (y,z) in the relation but it doesn't has (x,z)

## 1. What is a reflexive relation?

A reflexive relation is a type of binary relation in mathematics where every element is related to itself. In other words, for all elements a in a set S, the relation R is reflexive if and only if (a,a) is an element of R.

## 2. How is a symmetric relation defined?

A symmetric relation is a binary relation in mathematics where if (a,b) is in the relation, then (b,a) is also in the relation. In other words, the order of the elements doesn't matter, as long as both elements are related to each other in some way.

## 3. What is an anti-symmetric relation?

An anti-symmetric relation is a binary relation in mathematics where if (a,b) is in the relation and (b,a) is also in the relation, then a must equal b. In other words, if both elements are related to each other, they must be the same element.

## 4. Can a relation be both symmetric and anti-symmetric?

No, a relation cannot be both symmetric and anti-symmetric. This is because in a symmetric relation, the order of the elements doesn't matter, while in an anti-symmetric relation, the elements must be the same for both to be related.

## 5. How is a transitive relation defined?

A transitive relation is a binary relation in mathematics where if (a,b) is in the relation and (b,c) is also in the relation, then (a,c) is also in the relation. In other words, if two elements are related to each other and one of those elements is related to a third element, then the first element is also related to the third element.

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