Question

Anyone know what Let A = {1, 2, 3,4 }, and let R be a binary relation on A × A given by ((a,b),(c,d))∈R means? The A x A confuses me Anyone know what Let A = {1, 2, 3,4 }, and let R be a binary relation on A × A given by ((a,b),(c,d))∈R means? The A x A confuses me @Mathematics

Answer 1

A x A is a cartesian product

Answer 2

http://en.wikipedia.org/wiki/Binary_relation

Answer 3

example:
\[B=\{1,2\}\]
say we want to find \[B \times B\]
so \[B \times B=\{1,2\} \times \{1,2\}=\{(1,1),(1,2),(2,1),(2,2)\}\]

Answer 4

Oh so all the possible combinations of them?

Answer 5

((a,b),(c,d))∈R? And this just shows you what form you need to write it in?

Answer 6

((a,b),(c,d))∈R? And this just shows you what form you need to write it in?

Answer 7

the entire question actually also says and if and only if a divides c and b divides d, I don't see how a can ever divide see if we assume a is 1

Answer 8

another example:
\[C=\{a,b,c\} \]
\[B=\{1,2\}\]
\[C \times B=\{(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)\}\]
\[B \times C=\{(1,a),(2,a),(1,b),(2,b),(1,c),(2,c) \}\]
do you see the difference?

Answer 9

Oh yeah I see the difference

Answer 10

well ok so how is R defined
R is anything in the form ((a,b),(c,d)) if and only if a divides c and b divides d
where (a,b) and (c,d) are elements of A cross A
?

Answer 11

so we need to look in the set A cross A such that we can find two ordered pairs in that set that satisfy a|c and b|d

Answer 12

Yeah I'm suppose to show it's an order

Answer 13

example
i will name two pairs of ordered pairs that is an element of R
you should try to find the rest
\[((2,1 ),(4,2)) \in R\]
see 2|4 and 1|2

Answer 14

but does 2/4?

Answer 15

2 divides 4 is 2|4 which means there is some integer k such that 2k=4

Answer 16

Oh I see okay so after I get all those pairs,I use them to find it the binary is reflexive, transitive and symmetric?

Answer 17

\[a|c=> ak=c , k \in \mathbb{Z}^+\]

Answer 18

Okay I see, but do you know what transitive means? sorry I keep asking so many questions you explain them really well

Answer 19

I get that it's like x,y y,z so z,x is true or something but still kind of confused b/c what if I only have

Answer 20

like (a,b) and [c,d)

Answer 21

am I going to assume (a,b) (b,c) c,d so a,d?

Answer 22

i kind of don't understand what you're asking

Answer 23

are you suppose to just find all the elemetns of A crosss A
or you trying to show R is a relation

Answer 24

I'm suppose to show that R is an orde rso show that it's symmetric, reflexive and transitive but I don't really understand transitive

Answer 25

and also doesn't everything divide with 1,1?

Answer 26

so isn't it all elements?

Answer 27

1 divides everything

Answer 28

So I would have 16 elements?

Answer 29

so would I have to go like I got (1,1) (1,2) (1,3) (1,4) (2,1) (2,2) (2,3) etc. so say 2,1 and 1,2 I'd get 2,2

Answer 30

b/c (x,y) (y,z) therefore (x,z) right so 2,2 is in my set but would I have to do this test for all the elements?

Answer 31

A = {1, 2, 3,4 }
\[A \times A=\{1,2,3,4\} \times \{1,2,3,4\}\]
\[=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),\]
\[(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\}\]

Answer 32

\[((1,1),(i,j)) \in R\]
where i=1,2,3,4 and j=1,2,3,4
this is true since 1|i and 1|j

Answer 33

\[((1,2),(i,j)) \in R\]
where i=1,2,3,4 and j=4

Answer 34

oops j=2,4

Answer 35

since 2 divides itself

Answer 36

\[((1,3),(i,j)) \in R\]
where i=1,2,3,4 and j=3

Answer 37

\[((1,4),(i,j)) \in R\]
where i=1,2,3,4 and j=4

Answer 38

\[((2,1),(i,j)) \in R\]
where i=2,4 and j=1,2,3,4

Answer 39

\[((2,2),(i,j)) \in R\]
where i=2,4 and j=2,4

Answer 40

\[((2,3),(i,j)) \in R\]
where i=2,4 and j=3

Answer 41

\[((2,4),(i,j)) \in R\]
where i =2,4 and j=4

Answer 42

\[((3,1),(i,j)) \in R\]
where i=3 and j=1,2,3,4

Answer 43

\[((3,2),(i,j)) \in R\]
where i=3 and j=2,4
--------------------------
\[((3,3),(i,j)) \in R\]
where i=3 and j=3 (i wonder can we use on element from A cross A twice)
--------------------------
\[((3,4),(i,j)) \in R\]
where i=3 and j-4 (i wonder...)
--------------------------
\[((4,1),(i,j)) \in R\]
where i=4 and j=1,2,3,4
--------------------------
\[((4,2),(i,j)) \in R\]
where i=4 and j=2,4
--------------------------
\[((4,3),(i,j)) \in R\]
where i=4 and j=3 ( i wonder ...)

Answer 44

Oh and (4,4) is everything except anything with 3 in it

Answer 45

we either have ((4,4),(4,4)) is in R
or we don't say it is in R
thats what all those wondering statements were about

Answer 46

i think we do say it is in R

Answer 47

maybe zarkon will say something

Answer 48

okay but then what do I do those i,j pairs?

Answer 49

do I now use those to find R is an order?

Answer 50

oh i thought you were looking for all the elements of R

Answer 51

Oh sorry if my question is not making sense. I feel bad for making you write all that out

Answer 52

I'm just trying to show that R is an order

Answer 53

Oh it says show so I assumed it does but it's alright I'll try to figure it out. Thanks for trying and spending time trying to explain it!! Really appreciate it

Answer 54

for all x, y and z in X it holds that if xRy and yRz then xRz is the definition for transtive
so we have that we need to show
\[(r,s)R(t,u) \& (t,u)R(x,y)=> (r,s)R(x,y)\]
lets see if this makes a difference with our example
(1,1)R(2,4) and (2,4)R(4,4)=>(1,1)R(4,4)
ok that works

Answer 55

my early definition was alittle off

Answer 56

so (1,1)R(4,4) needs to be in the set of elements?

Answer 57

Sorry what does the R stand for when you write it like 1,1)R(4,4)

Answer 58

so anyways we need to prove this so...
if (r,s)R(t,u) and (t,u)R(x,y) then we have
r|t and s|u
and
t|x and u|y
we need to show (r,s)R(x,y)=> r|x and s|y

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

so we have r|t and t|x
this means
ri=t and ta=x for some positive integers i and a
so we are trying to show rp=x
(ri)a=x
r(ia)=x
ia is an integer since i and a are integers and the set of integers in closed by multiplication
so we have shown r|x
now we nned to show s|y
so we have
s|u and u|y
s|u=> sv=u and u|y=> ue=y for some positive integers v and e
so we need to show s|y=> sq=y for some integer q
(sv)e=u
s(ve)=u
ve is an integer since the set of integers is closed by multiplication
so s|u
we have now proved the transtive property for this relation R

Answer 59

you asked what does xRy mean
The statement (x,y) ∈ R is read "x is R-related to y", and is denoted by xRy or R(x,y).

Answer 60

Oh so we don't need to show each element physically? I understand what you did I can do the same with reflexive and anti symmetric right

Answer 61

yes

Answer 62

you don't need to go through each element of R to see
you can do one example for each property if you want
thats what i did before i proved the transitive part

Answer 63

Okay thanks so much for your help!! Wish I could give you more medals, really appreciate it thanks again!

Answer 64

but it is not nessarcy

Answer 65

ok