# Math Help - 4 Relations Problems!

1. ## 4 Relations Problems!

1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) $E$ R if and only if

a) x + y = 0.
b) x = ±y.
c) x - y is a rational number.
d) x = 2y.
e) xy >= 0.
f) xy = 0.
g) x = 1.
h) x = 1 or y = 1.

2) Let R1 = {(1, 2), (2, 3), (3, 4)} and R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} be relations from {1, 2, 3} to {1, 2, 3, 4}. Find R1 - R2.

Would the answer of # 2 be Ø or {Ø} ?

3) Show that the closure of a relation R with respect to a property P, if it exists, is the intersection of all the relations with property P that contain R.

4) Use Algorithm 1 to find the transitive closures of these relations on {a, b, c, d, e}.

a) {(a, b), (a, c), (a, e), (b, a), (b, c), (c, a), (c, b), (d, a), (e, d)}

Thank You very much for your help!
Anu

2. Originally Posted by anu
1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) $E$ R if and only if

a) x + y = 0.
i will illustrate what you need to do with this one. the rest of the questions in this section are just applying the definitions again.

Definition: A relation $R$ defined on a set $A$ is called reflexive if $xRx$ for every $x \in A$ (that is, every element in the set is related to itself under the relation).

since $x + x = 0$ is not always true, each element in the real numbers does not relate to itself under this relation. thus this relation is not reflexive

Definition: A relation $R$ defined on a set $A$ is called symmetric if whenever $xRy$, then $yRx$ for all $x,y \in A$

we are dealing with real numbers, thus we have the property of commutativity of addition.

so if $x + y = 0$ then $y + x = 0$ thus we have $xRy \implies yRx$ as desired. this relation is symmetric

Definition: A relation $R$ defined on a set $A$ is called anti-symmetric if whenever $xRy$ and $yRx$ then $x = y$ for all $x,y \in A$

now if $x + y = 0$ and $y + x = 0$ it does not imply that $x = y$

case in point, consider x = 1 and y = -1

thus, the relation is not anti-symmetric

Definition: A relation $R$ defined on a set $A$ is called transitive if whenever $xRy$ and $yRz$ then $xRz$ for all $x,y,z \in A$

now if $x + y = 0$ and $y + z = 0$ it means that $x = z$. but we know that the relation is not reflexive, and therefore $x$ does not relate to itself. so the relation is not transitive.

case in point: $x = z = 1$ and $y = -1$

then we have $xRy$ and $yRz$ (that is, x + y = 1 - 1 = 0, and y + z = -1 + 1 = 0), but $x \not{R}z$ since $x + z = 1 + 1 = 2$. this is an implication, so if we have a true statement implying a false statement, the whole implication is false. thus our conclusion follows

try the rest. as you see, i just relied solely on the definitions, nothing fancy

3. Originally Posted by anu
2) Let R1 = {(1, 2), (2, 3), (3, 4)} and R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} be relations from {1, 2, 3} to {1, 2, 3, 4}. Find R1 - R2.

Would the answer of # 2 be Ø or {Ø} ?
i don't recall doing this in class when we were doing relations (that is, to subtract one relation from another). i suppose we treat them as sets though. so i interpret R1 - R2 to mean we take the set R1 and throw out all the elements that appear in R2 (is that correct?) thus, under this assumption, the answer would be $\emptyset$ NOT $\{ \emptyset \}$

4. Originally Posted by anu

3) Show that the closure of a relation R with respect to a property P, if it exists, is the intersection of all the relations with property P that contain R.
i have no idea what to do here

4) Use Algorithm 1 to find the transitive closures of these relations on {a, b, c, d, e}.

a) {(a, b), (a, c), (a, e), (b, a), (b, c), (c, a), (c, b), (d, a), (e, d)}
what is this "Algorithm 1" you are referring to? that sounds like it's something specific to your text or your class

5. Originally Posted by Jhevon
what is this "Algorithm 1" you are referring to? that sounds like it's something specific to your text or your class
Ignore!!! sorry

6. Originally Posted by Jhevon
i will illustrate what you need to do with this one. the rest of the questions in this section are just applying the definitions again.

Definition: A relation $R$ defined on a set $A$ is called reflexive if $xRx$ for every $x \in A$ (that is, every element in the set is related to itself under the relation).

since $x + x = 0$ is not always true, each element in the real numbers does not relate to itself under this relation. thus this relation is not reflexive

Definition: A relation $R$ defined on a set $A$ is called symmetric if whenever $xRy$, then $yRx$ for all $x,y \in A$

we are dealing with real numbers, thus we have the property of commutativity of addition.

so if $x + y = 0$ then $y + x = 0$ thus we have $xRy \implies yRx$ as desired. this relation is symmetric

Definition: A relation $R$ defined on a set $A$ is called anti-symmetric if whenever $xRy$ and $yRx$ then $x = y$ for all $x,y \in A$

now if $x + y = 0$ and $y + x = 0$ it does not imply that $x = y$

case in point, consider x = 1 and y = -1

thus, the relation is not anti-symmetric

Definition: A relation $R$ defined on a set $A$ is called transitive if whenever $xRy$ and $yRz$ then $xRz$ for all $x,y,z \in A$

now if $x + y = 0$ and $y + z = 0$ it means that $x = z$. but we know that the relation is not reflexive, and therefore $x$ does not relate to itself. so the relation is not transitive.

case in point: $x = z = 1$ and $y = -1$

then we have $xRy$ and $yRz$ (that is, x + y = 1 - 1 = 0, and y + z = -1 + 1 = 0), but $x \not{R}z$ since $x + z = 1 + 1 = 2$. this is an implication, so if we have a true statement implying a false statement, the whole implication is false. thus our conclusion follows

try the rest. as you see, i just relied solely on the definitions, nothing fancy
I kinda know all this but still confused!

7. Originally Posted by anu
I kinda know all this but still confused!
type up your solution for (b) and we'll see how you do

8. Originally Posted by Jhevon
type up your solution for (b) and we'll see how you do
b) x = +-y

not reflexive b/c x cannot equal both + y and -y, is symmetric b/c x would equal y or -y and then y or -y would equal x; is not antisymmetric and it is transitive

am i right?

9. Originally Posted by anu
b) x = +-y

not reflexive b/c x cannot equal both + y and -y, is symmetric b/c x would equal y or -y and then y or -y would equal x; is not antisymmetric and it is transitive

am i right?
check reflexive again. is $x = \pm x$?

10. Originally Posted by Jhevon
check reflexive again. is $x = \pm x$?
so it is reflexive, to check if it is reflexive we just plug in x for the y?

Originally Posted by anu
1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) $E$ R if and only if

a) x + y = 0.
b) x = ±y.
c) x - y is a rational number.
d) x = 2y.
e) xy >= 0.
f) xy = 0.
g) x = 1.
h) x = 1 or y = 1.
a) symmetric
b) reflexive, symmetric, transitive

c) wait is 0 a rational number? i think yes so it is reflexive, antisymmetric b/c x or y can be not rational numbers and still have a result of a rational number.

d) antisymmetric?

e) reflexive, antisymmetric b/c xy >= 0 but x and y individually can be less than zero, right?

f) reflexive, antisymmetric b/c x or y has to be zero but not the other?

g) none

h) symmetric?

11. Originally Posted by anu
so it is reflexive, to check if it is reflexive we just plug in x for the y?
no. to check reflexivity, you just check if any element relates to itself. any element x relates to itself in the case of x = +/- y

yes, 0 is a rational number

12. Originally Posted by anu
1) Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) $E$ R if and only if

a) x + y = 0.
b) x = ±y.
c) x - y is a rational number.
d) x = 2y.
e) xy >= 0.
f) xy = 0.
g) x = 1.
h) x = 1 or y = 1.
Originally Posted by anu
a) symmetric
b) reflexive, symmetric, transitive

c) wait is 0 a rational number? i think yes so it is reflexive, antisymmetric b/c x or y can be not rational numbers and still have a result of a rational number.

d) antisymmetric?

e) reflexive, antisymmetric b/c xy >= 0 but x and y individually can be less than zero, right?

f) reflexive, antisymmetric b/c x or y has to be zero but not the other?

g) none

h) symmetric?

So how did i do for the above?

13. Originally Posted by anu
c) x - y is a rational number.
x - x = 0 which is rational, so it is reflexive

if x - y is rational, then y - x is rational, so it is symmetric

if x - y is rational and y - x is rational then it doesn't necessarily mean x = y. case in point $x = 1, y = 2$

so it is not anti-symmetric

it is transitive

(note here that x, y and z themselves do not have to be rational. for instance $x = 1 + \sqrt{2}$, $y = z = \sqrt{2}$ fulfills the conditions)

d) x = 2y.
x is not equal to 2x in general, so it's not reflexive.

if x = 2y then y = (1/2)x so it is not symmetric

if x = 2y and y = 2x then x = -y so it would seem not to be anti-symmetric. but when you think about it, the only way this would work is if x = y = 0. so i'd say it is anti-symmetric

if x = 2y and y = 2z then x = 4z so it is not transitive.

e) xy >= 0.
xx = x^2 >= 0 so it is reflexive

xy = yx, so if xy >= 0 then yx >= 0 so it is symmetric

if xy >=0 and yx >= 0 it obviously does not mean that x = y so it is not antisymmetric (come up with your own counter-example here. it is good practice to come up with reasons for why something does not fulfill the criteria, we do that by finding a counter-example)

if xy >= 0 and yz >= 0 then xz >= 0 (why is this?) so it is transitive

f) xy = 0.
obviously this is not reflexive

if xy = 0 then yx = 0 so it is symmetric

it is obviously not anti-symmetric

it is not transitive. if xy = 0 and yz = 0 it does not mean xz = 0

consider x = z = 1, y = 0

g) x = 1.
the relation is {(1,1)} it is reflexive, symmetric, antisymmetric and transitive (recall we have implications for the definitions here, the conditions can be fulfilled vacuously)

h) x = 1 or y = 1.
again, this seems to fulfill the conditions vacuously

now i'm tired, i'm going to bed

14. Thanks a Million Jhevon

15. ## Can anyone solve this problem for me ??? Thanks.

3) Show that the closure of a relation R with respect to a property P, if it exists, is the intersection of all the relations with property P that contain R.