Question

@satellite73 got a question will post in a minute...

Answer 1

Let \(R\) be an equivalence relation on \(A\) and let \(S\) be an equivalence relation on \(B\). Define a \(T\) on \( A \times B\) by \(((a_1,b_1),(a_2,b_2)) \in T \leftrightarrow (a_1,a_2) \in R\) and \((b_1,b_2) \in S\). Prove that \(T\) is an equivalence relation. \\

We need to prove that \(T\) is an equivalence relation.\\

Definition 6.2.3 states that \(R\) is an equivalence relation if \(R\) is reflexive, symmetric, and transitive. \\

R is reflexive if \((\forall x \in S)(x,x) \in R]\)\\
R is symmetric if \((\forall x,y \in S)[(x,y) \in R \rightarrow (y,x) \in R]\)\\
R is transitive if \((\forall x, y, z \in S)[((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R]\)\\

For \(T\) to be reflexive, we need \((\forall (a_1),(b_1) \in S)[((a_1,b_1),(a_1,b_1)) \in T \)\\
For all \(T\) to be symmetric, we need \((\forall (a_1,b_1),(a_2,b_2) \in S)[(a_1,b_1),(a_2,b_2) \in T \rightarrow (a_2,b_2),(a_1,b_1) \in T\)\\
For all \(T\) to be transitive, we need \((\forall (a_1,b_1),(a_2,b_2),(a_3,b_3) \in S]((a_1,b_1),(a_2,b_2) \in T \land ((a_2,b_2),(a_3,b_3)) \in T \rightarrow ((a_1,b_1), (a_3,b_3)) \in T\)\\

We have proven that \(T\) is an equivalence relation.

Answer 2

I don't know if I got this right :/

Answer 3

@wio ^^

Answer 4

I think it is wrong.

Answer 5

Start with proving that \(T\) is reflexive.

Answer 6

thought so.... so reflexive is \[(\forall x \in S)(x,x) \in R] \]

Answer 7

by def.

Answer 8

We can start out with the assumptions that \[
\forall a\in A\quad (a,a)\in R\quad \land \quad \forall b \in B\quad (b,b) \in S
\]

Answer 9

D: oh I see make it separate.. so I'll do the same thing for symmetric and transitive.

Answer 10

\[\forall a\in A\quad (a_1,a_1)\in R\quad \land \quad \forall b \in B\quad (b_1,b_1) \in S\]

Answer 11

Don't put any subscripts, we don't need them for the reflexive proof at least.

Answer 12

If you are going to use subscripts, then you should have said \(\forall a_1\in A\)

Answer 13

Oh ._. oh right because we have just x x for reflexive...

Answer 14

so that's just one .... damn I want to say element but that's not the right term

Answer 15

Now we can say straight out that \[
(a,a)\in R\land (b,b)\in S \iff ((a,b),(a,b))\in T
\]Because we defined \(T\) this way.

Answer 16

This alone completes the reflexive proof.

Answer 17

For the next proof, we need to show: \[
\forall a_1,a_2\in A\ \forall b_1,b_2\in B\quad ((a_1,b_1),(a_2,b_2))\in T \implies ((a_2,b_2),(a_1,b_1))\in T
\]

Answer 18

alright so... for symmetric

\[(\forall x,y \in S)[(x,y) \in R \rightarrow (y,x) \in R] \]
by definition ....
Also we have

\[(a_1,a_2) \in R \] and \[(b_1,b_2 ) \in S\]
so
\[(\forall a,b \in A)[ a_1,a_2 \in R \rightarrow (a_2,a_1) \in R\]

Answer 19

We can do a direct proof here.

Start with:\[
\forall a_1,a_2\in A\ \forall b_1,b_2\in B\quad ((a_1,b_1),(a_2,b_2))\in T
\]This can expanded by definition of \(T\) to: \[
(a_1,a_2)\in R\quad \land\quad (b_1,b_2)\in S
\]

Answer 20

\[(\forall (a_1,b_1),(a_2,b_2) \in S)[(a_1,b_1),(a_2,b_2) \in T \rightarrow (a_2,b_2),(a_1,b_1) \in T \] and then expand...

Answer 21

Then at this point we can use the symmetric properties of \(R\) and \(S\).

Answer 22

No, I have said this many times before... you can't prove a statement by assuming it is true.

Answer 23

\[(\forall a_1 ,a_2 \in A)[(a_1,a_2) \in R \rightarrow (a_2,a_1) \in R]\]

Answer 24

Are you listening to me?

Answer 25

yeah I'm just making attempts... ._.

Answer 26

\[(\forall b_1,b_2 \in B)[(b_1,b_2) \in S \rightarrow (b_2,b_1) \in S]\]

Answer 27

First of all, when you want to prove \[
Q\to P
\]You have to do: \[
Q\\
\to R \\
\to S \\
\to P
\]You can't do: \[
Q\to P \\
R\to S
\]

Answer 28

D: why?

Answer 29

Because, you can prove many false things that way.

Answer 30

oh ._.

Answer 31

So when I say:
\(\color{blue}{\text{Originally Posted by}}\) @wio
For the next proof, we need to show: \[
\forall a_1,a_2\in A\ \forall b_1,b_2\in B\quad ((a_1,b_1),(a_2,b_2))\in T \implies ((a_2,b_2),(a_1,b_1))\in T
\]\(\color{blue}{\text{End of Quote}}\)


Then this meand we start with: \[
\forall a_1,a_2\in A\ \forall b_1,b_2\in B\quad ((a_1,b_1),(a_2,b_2))\in T
\]And we manipulate it until we get:\[
((a_2,b_2),(a_1,b_1))\in T
\]

Answer 32

\[(\forall a_1 ,a_2 \in A)[(a_1,a_2) \in R \rightarrow (a_2,a_1) \in R] \]
\[(\forall b_1,b_2 \in B)[(b_1,b_2) \in S \rightarrow (b_2,b_1) \in S] \]

since we
have to expand the definition of \[(a_1,a_2)\in R\quad \land\quad (b_1,b_2)\in S \]
right?

Answer 33

*symmetric definition..

Answer 34

Yes.

Answer 35

Wait, can I just show you really quick?

Answer 36

sure

Answer 37

\(\color{blue}{\text{Originally Posted by}}\) @wio
Start with:\[
\forall a_1,a_2\in A\ \forall b_1,b_2\in B\quad ((a_1,b_1),(a_2,b_2))\in T
\]This can expanded by definition of \(T\) to: \[
(a_1,a_2)\in R\quad \land\quad (b_1,b_2)\in S
\]Since \(R\) and \(S\) are symmetric, we can say: \[
(a_2,a_1)\in R\quad \land\quad (b_2,b_1)\in S
\]Then using the definition of \(T\) once more \[
((a_2,b_2),(a_1,b_1))\in T
\]

Answer 38

We didn't get that second line through the property of symmetry, we got it through the definition of \(T\).

Answer 39

We got the third line by using the symmetry property.

Answer 40

definition of T one more... is it
\[(\forall (a_1,b_1),(a_2,b_2) \in S)[(a_1,b_1),(a_2,b_2) \in T \rightarrow (a_2,b_2),(a_1,b_1) \in T \]?
or

\[\forall a_1,a_2\in A\ \forall b_1,b_2\in B\quad ((a_1,b_1),(a_2,b_2))\in T \]

Answer 41

@UsukiDoll Just pop'in in don't mind me...but....*BroHoof*

Answer 42

awww Twilight! How nice of you to join us

Answer 43

I've heard Sweetie Belle was learning magic during Twilight Time.

Answer 44

@wio so for transitive we need to use \[a_1,a_2,a_3 \] since the original def. had x y z

Answer 45

\[(a_1,a_2,a_3)\in R\quad \land\quad (b_1,b_2,b_3)\in S\]

Answer 46

\[(\forall x, y, z \in S)[((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R] \]
for \[a_1,a_2,a_3 \in R \] we have
\[(\forall a_1, a_2, a_3 \in a)[((a_1,a_2) \in R \land (a_2,a_3) \in R) \rightarrow (a_1,a_3) \in R]\]
for
\[b_1,b_2,b_3 \in S\]
we have
\[(\forall b_1, b_2,b_3 \in B)[((b_1,b_2) \in S \land (b_2,b_3) \in S) \rightarrow (b_1,b_3) \in S]\]

Answer 47

@UsukiDoll Sweetie Belle has improved alot since...ya know....the "incident".

But right now i'm trying to figure out what is this math you are doing....
it looks rather, complicated...

Answer 48

Twilight, it's called Intro to Advanced Mathematics and believe me it SUCKSSSS!!!!!! D: ...but I've read the sections in advance and somehow I could do these proof problems....little did I know that I needed more that those three little lines that I posted above.

Answer 49

\[\forall a_1,a_2,a_3\in A\ \forall b_1,b_2,b_3\in B\quad ((a_1,b_1),(a_2,b_2))\in T \]......ugh the part for the transitive of T isn't fitting well with the original definition .

Answer 50

@UsukiDoll Are you in the University of Gifted unicorns, or Magic Kindergarten?
Because Keep this Between you and me but I don't understand this....and i cant find anything in my books.... :(

Answer 51

errr well ummm heheeeheheheehehhehhe shouldn't you be with Apple Bloom and her potions?

Answer 52

@wio did I get the transitive part right?

Answer 53

... i think he's afk .. :/

Answer 54

Yes very true....they need a improvement...I just got cough up on this Laptop of mine....i best be going /)

Answer 55

Start by writing the statement you are trying to prove.

Answer 56

for transitive
\[(\forall x, y, z \in S)[((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R] \]
\[a_1,a_2,a_3 \in R \] we have\[(\forall a_1, a_2, a_3 \in a)[((a_1,a_2) \in R \land (a_2,a_3) \in R) \rightarrow (a_1,a_3) \in R] \]

\[b_1,b_2,b_3 \in S \]
We have
\[(\forall b_1, b_2,b_3 \in B)[((b_1,b_2) \in S \land (b_2,b_3) \in S) \rightarrow (b_1,b_3) \in S] \]

Answer 57

@wio ^

Answer 58

\[(\forall (a_1,b_1),(a_2,b_2),(a_3,b_3) \in S]((a_1,b_1),(a_2,b_2) \in T \land ((a_2,b_2),(a_3,b_3)) \in T \rightarrow ((a_1,b_1), (a_3,b_3)) \in T \]

Answer 59

Hold on...

Answer 60

The statement to prove should not contain R and S, and you might need to use multiple lines to write it.

Answer 61

not contain R and S?!

Answer 62

Nope.

Answer 63

:O what!!!

Answer 64

The proof will contain \(R\) and \(S\), but the statement you are trying to prove should not contain them.

Answer 65

You only need \(T\).

Answer 66

how is that possible???

Answer 67

wait wait wait.... is this a one liner proof?

Answer 68

for transitive?

Answer 69

hmmm... (\[a_1,b_1) \] implies T \[(a_3,b_3)\]?

Answer 70

I'm not asking for a proof yet. I'm asking for the statement to be proven.

Answer 71

Use the definition of transitivity on \(T\).

Answer 72

\[(\forall (a_1,b_1),(a_2,b_2),(a_3,b_3) \in S]((a_1,b_1),(a_2,b_2) \in T \land ((a_2,b_2),(a_3,b_3))\]\[ \in T \rightarrow ((a_1,b_1), (a_3,b_3)) \in T \]

Answer 73

for R is transitive...
\[(\forall x, y, z \in S)[((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R]\]
for T ... previous post...

Answer 74

\[(\forall a_1, a_2, a_3 \in a)[((a_1,a_2) \in R \land (a_2,a_3) \in R) \rightarrow (a_1,a_3) \in R] \]
\[(\forall b_1, b_2,b_3 \in B)[((b_1,b_2) \in S \land (b_2,b_3) \in S) \rightarrow (b_1,b_3) \in S] \]

Answer 75

if I only need T then that's A X B

Answer 76

Okay, let's start out really simply...

Answer 77

\[
\forall c_1, c_2, c_3\in A\times B\quad (c_1, c_2)\in T\land (c_2,c_3)\in T\implies (c_1,c_2)\in T
\]

Answer 78

Now we can expand it so that \(c_i = (a_i, b_i)\).

Answer 79

So: \[
\forall a_1,a_2, a_3\in A\quad \forall b_1,b_2,b_3\in B
\]If\[
((a_1,b_1),(a_2,b_2))\in T\land ((a_2,b_2),(a_3,b_3))\in T
\]Then \[
((a_1,b_1),(a_3,b_3))\in T
\]This is what we are trying to prove..

Answer 80

do I have to use transitive def... twice for \[((a_1,b_1),(a_2,b_2))\in T\land ((a_2,b_2),(a_3,b_3))\in T \]

Answer 81

mm this looks like symmetric twice since there are two elements.. what if
\[(\forall x,y \in S)[(x,y) \in R \rightarrow (y,x) \in R] \]

\[(\forall (a_1,b_1)( a_2,b_2) \in T)[((a_1,b_1),( a_2,b_2)) \in T \rightarrow (( a_2,b_2),(a_1,b_1)) \in T] \]
and

\[(\forall (a_2,b_2),(a_3,b_3) \in T)[((a_2,b_2),(a_3,b_3)) \in T \rightarrow ((a_3,b_3)(a_2,b_2) \in T] \]

Answer 82

like that?

Answer 83

You don't EVER have to restate the transitive property for \(S\) and \(R\). You can use it outright.

Answer 84

like just use T directly?

Answer 85

........ :/

Answer 86

No

Answer 87

You expand T into S and R.

Answer 88

Then what can I use? :/

Answer 89

You apply the transitive property on S and R.

Answer 90

Then you simplify the result back to T

Answer 91

\[(\forall a_1, a_2, a_3 \in a)[((a_1,a_2) \in R \land (a_2,a_3) \in R) \rightarrow (a_1,a_3) \in R] \]
Transitive for R
\[(\forall b_1, b_2,b_3 \in B)[((b_1,b_2) \in S \land (b_2,b_3) \in S) \rightarrow (b_1,b_3) \in S] \]
Transitive for S....

Answer 92

so far so good? o r errrrrrrrr

Answer 93

Sure

Answer 94

This is one step in the proof.

Answer 95

hmm now to simplify x.x to get it back into T

Answer 96

\[((a_1,b_1),(a_2,b_2))\in T\land ((a_2,b_2),(a_3,b_3))\in T \] ?

Answer 97

First of all, what do you want to end up with?