[QUOTE=science_man_88;247948]Proposition: ~ is reflexive

Proof: we need to show that x~x. I based my statement on :

if x "is a child of" y and x "is a child of" y implies x "is a sibling of"x but I know this isn't acceptable.[/QUOTE]

You don't need to prove reflexivity. You [i]defined[/i] your relation as reflexive.

[QUOTE=science_man_88;247973]because x~x is trivially true by my definition of ~ and x~z implies x~z is also trivially trues hence the whole thing is trivially true ?[/QUOTE]

Perfect!

[QUOTE=Mr. P-1;247976]Perfect![/QUOTE]

I forgot to reply to CRG's questioning. If and when you're positive I know enough to move on feel free. Next preliminary is equivalence classes.

Oh come on this stuff is easy. It may be foundational, but drawing it
out by way of detours and excursions is silly. If you understand
something like a relationship R being a subset of S x S just means that
aRb is the same as (a,b) is in set R (R is used two ways here),
then an equivalence relation E is just a relation that is reflexive,
symmetric, and transitive, which just means that for any x,y,z in the
universal set U the following three facts hold:
x ~ x (or xEx) for any x in U
x ~ y implies y ~ x
x ~ y and y ~ z implies x ~ z
so that an equivalence relation defines and is equivalent to a partition.

No problem.

[QUOTE=davar55;248105]Oh come on this stuff is easy. It may be foundational, but drawing it
out by way of detours and excursions is silly. If you understand
something like a relationship R being a subset of S x S just means that
aRb is the same as (a,b) is in set R (R is used two ways here),
then an equivalence relation E is just a relation that is reflexive,
symmetric, and transitive, which just means that for any x,y,z in the
universal set U the following three facts hold:
x ~ x (or xEx) for any x in U
x ~ y implies y ~ x
x ~ y and y ~ z implies x ~ z
so that an equivalence relation defines and is equivalent to a partition.

No problem.[/QUOTE]

some of us aren't that smart.

[QUOTE=science_man_88;248114]some of us aren't that smart.[/QUOTE]

Please don't take my post the wrong way.
I would only suggest to you that those who lead
you down a winding path may have troubles of
their own.

one partition of a column/row of a chessboard is white squares in the column/row and black squares in the column/row

one binary relation that could be drawn is "is the same color as"

if square1 is the same color as square3 and square3 is the same color as square5 then square1 is the same color as square5, this is true and shows transitivity.

proof needed for:~ is reflexive

Proof: square1 is the same color as square1 as multicolored squares aren't on a chessboard. this is true proving reflexivity

Proof needed for:~ is symmetric

Proof : if square1 is the same color as square3 then square3 is the same color as square1. this is true proving symmetry

as the above are all true this is an equivalence relation.

[QUOTE=science_man_88;248128]one partition of a column/row of a chessboard is white squares in the column/row and black squares in the column/row

one binary relation that could be drawn is "is the same color as"

if square1 is the same color as square3 and square3 is the same color as square5 then square1 is the same color as square5, this is true and shows transitivity.

proof needed for:~ is reflexive

Proof: square1 is the same color as square1 as multicolored squares aren't on a chessboard. this is true proving reflexivity

Proof needed for:~ is symmetric

Proof : if square1 is the same color as square3 then square3 is the same color as square1. this is true proving symmetry

as the above are all true this is an equivalence relation.[/QUOTE]

You're giving instances rather than demonstrating for all instances.

To prove relation R is an equivalence relation E, must show

(1) REFLEXIVITY

xRx for all x in U which is same as (x,x) in R subset-of U x U.
So any relation R in which equality (which is the most important
equivalence relation) is a sub-set (i.e. in which all (x,x) in R for x in U)
must trivially satisfy reflexivity, i.e. xRx for all x in U.

(2) SYMMETRY

xRy implies yRx (must demonstrate for your particular R for all x,y in U)

(3) TRANSITIVITY

xRy and yRz implies xRz (must demonstrate for your R for all x,y,z in U)

When demonstrate (1),(2), and (3) you've proven R is an equivalence.


I never said it was trivial, just simple AFTER YOU'VE STUDIED HARD MATH.

In which thread did the illustrious and well respected RDS affectionately
known herein as RDS post these homework problems?

[QUOTE=science_man_88;247299]I don't see why they use the word on to describe it yet but I might soon ( maybe it's so it doesn't get confused with [TEX]\in[/TEX] (meaning in ?). My next line of questioning is on the equation part, Does this mean for all [TEX]x \in S[/TEX] there's a [TEX]y \in S[/TEX] such that the relation on S always holds ? I'll be back around 8-9 am (UTC -4:00 time).[/QUOTE]

I can't use Tex yet, so I do my math in ascii.

A relation ON a set S means OVER that set, meaning the instances of x,y
are elements OF set S.

Whereas an element x being IN set S is the first terminology of set theory.

Use the math symbols which Tex provides accordingly.

[QUOTE=CRGreathouse;247327]No. For every x you find all y such that x is related to y (x ~ y) and call the set of all such y's "[x]".

Suppose I define ~ over the real numbers as "x ~ y if and only if x - y is an integer". Then 0.2 ~ 1.2, 0.2 ~ 2.2, 0.2 ~ 3.2, etc., so that [0.2] = {..., 0.2, 1.2, 2.2, ...}.

Is this ~ an equivalence relation? (Check if it has the three properties.)[/QUOTE]

Here CRG is giving a specific relation R.

You can prove it satisfies the three properties, hence IS an equivalence.
An important one.

There are others over the integers.