And more discrete math fun!

[helios]

There's quite a few second questions. Can you be more specific?

[ResidentBiscuit]

Sorry, the VERY second one haha. This one I've been stumped on

S is the set of all real numbers and xRy means that x^2 = y^2


All I can take from this is that |x| = |y|, but what does that matter?

EDIT: Same kind of question as the first, I'm just supposed what properties it shows and why. I can't even get started on this one

[Zhuge]

So what pairs are in R? Try writing some examples and think about the properties; reflexivity, symmetry, antisymmetry, etc.

[ResidentBiscuit]

That's all that was supplied for the problem. The only relation I got was

xRy means that x² = y²

EDIT: Just found the superscript option :D Much better than using ^ for exponents

[helios]

The way it's stated would imply that the relationship covers R^2 (I mean, the cartesian product of the real set over itself).

Well, it is reflexive, since the square of any number is the same as itself.
It's also transitive symmetric, since (x^2=y^2) <=> (y^2=x^2).

Finally,
x^2=y^2 and y^2=z^2 => x^2=z^2
|x|=|y| and |y|=|z| => |x|=|z|
(x=y or x=-y) and (y=z or y=-z) => (x=z or x=-z)

x=y:
((x=z or x=-z) => (x=z or x=-z)) = true

x=-y:
((-x=z or -x=-z) => (x=z or x=-z)) = true

|x|!=|y|
(((x=y or x=-y) and (y=z or y=-z)) => (x=z or x=-z)) = (((false or false) and ?) => ?) = true

So it's also transitive.

[ResidentBiscuit]

Well that all makes sense actually. Seeing it written out like that helped a lot.

Does it matter if z is not said to be in this relation? Or is z just implied?

[helios]

What do you mean? z is just a variable I used in the reasoning. It has nothing to do with the relation.

[ResidentBiscuit]

So, x and y are just variables that could be any real number?

[helios]

Correct.

[ResidentBiscuit]

So, wouldn't this mean that the absolute value of any value in this set is equal to any other value?

[helios]

I don't know what that sentence means. Can you write it as a set comprehension? For example, this last relation is written {(x,y) in R^2 : x^2=y^2}

[ResidentBiscuit]

Sorry I don't know how to express my thought in a set notation.

Set S is a set of all real numbers:

If x^2 always = y^2, and x and y are any element of set S, this would mean that
|x| will always = |y|. This would in turn imply that all elements of set S have the same absolute value.

This is all brand new to me, so bare with my misunderstanding.

[helios]

This would in turn imply that all elements of set S have the same absolute value.

This is not possible, since you already stated that S the same as the real set.

(Several times I'll use "in" as short for "is an element of" in the following text.)

Consider R^2, which is the set of all pairs (x,y) where x,y in R. That is, R^2 contains every combination of two real numbers, such as (0,0), (0,1), (0,0.5), (e,pi), (pi,e), and so on.
The relation P is a subset of R^2, composed of all the pairs of reals whose squares are equal, such as (1,1), (1,-1), (-1,1), (-1,-1), (-4,4), and so on.
It's true that x and y are members of R, and it's also true that |x|=|y| (within the context of P), but it's obviously not true that |x|=|y| for all x,y in R. The statement |x|=|y| only makes sense when treated within the context of the definition of P, which states that: (x,y) in P iff |x|=|y|, where x,y in R. In other words, |x|=|y| for all elements (x,y) of P, since that is the definition of P.

[firedraco]

If x^2 always = y^2, and x and y are any element of set S, this would mean that
|x| will always = |y|. This would in turn imply that all elements of set S have the same absolute value.

No, x and y are not any element, helios is selecting some x and y such that (x,y) is an element of R.

[ResidentBiscuit]

Oooh! Oh my god! That just clicked for me! So it's reflexive, and symmetric, but not transitive. I was trying to say the relation WAS the whole set of real numbers, which I think is why I was sooo confused

[hamsterman]

I can't find a decent math forum so here I am again!

I've had some use of http://www.physicsforums.com/forumdisplay.php?f=4
Although you seem to be doing fine here, so maybe that doesn't matter..