Math Question

[John JD Dorian]

All--

Denoting the cardinality of a set X by |X|, we know that |X| = |Y| iff there is a bijection between X and Y.

Obviously, there is no bijection between Q and R, as |Q| = aleph_0 and |R| = c.

However, |R x R| = |R| * |R| = c * c = c,
and |R x Q| = |R| * |Q| = c * aleph_0 = c,

so |R x R| = |R x Q| and thus there is a bijection between the two sets.

Can anyone think of one?

Meaning, define f : R x R --> R x Q, such that for every (x_1, x_2) in R^2, f((x_1, x_2)) = (y_1, y_2) where y_1 is in R and y_2 in Q.

Since there is no bijection between R and Q, it's hard to see how this would look. Any takers?

[crimsonandclovr]

I can be of no help to you, but I'm here to offer my sympathies.

[tonyC4L]

All--

Denoting the cardinality of a set X by |X|, we know that |X| = |Y| iff there is a bijection between X and Y.

Obviously, there is no bijection between Q and R, as |Q| = aleph_0 and |R| = c.

However, |R x R| = |R| * |R| = c * c = c,
and |R x Q| = |R| * |Q| = c * aleph_0 = c,

so |R x R| = |R x Q| and thus there is a bijection between the two sets.

Can anyone think of one?

Meaning, define f : R x R --> R x Q, such that for every (x_1, x_2) in R^2, f((x_1, x_2)) = (y_1, y_2) where y_1 is in R and y_2 in Q.

Since there is no bijection between R and Q, it's hard to see how this would look. Any takers?
This is pretty interesting. I don't know enough about set theory to verify this is all accurate, but I'll assume that it is and that we can find a bijection. My first guess would be to define f: R x R --> R x Q by f(x,y) = (xy, x/y). Certainly xy is in R and x/y is in Q, but I'm not sure that it's one-to-one or onto. I have to go to class now but I'll think about this some more.

[John JD Dorian]

This is pretty interesting. I don't know enough about set theory to verify this is all accurate, but I'll assume that it is and that we can find a bijection. My first guess would be to define f: R x R --> R x Q by f(x,y) = (xy, x/y). Certainly xy is in R and x/y is in Q, but I'm not sure that it's one-to-one or onto. I have to go to class now but I'll think about this some more.

Certainly it isn't one-to-one, since f(0,y)=0 for all y, and we would have to also define separately f(x,0) for all x.

[John JD Dorian]

This is pretty interesting. I don't know enough about set theory to verify this is all accurate, but I'll assume that it is and that we can find a bijection. My first guess would be to define f: R x R --> R x Q by f(x,y) = (xy, x/y). Certainly xy is in R and x/y is in Q, but I'm not sure that it's one-to-one or onto. I have to go to class now but I'll think about this some more.

also, f(sqrt2, 1) is not even in R x Q.

[tonyC4L]

Certainly it isn't one-to-one, since f(0,y)=0 for all y, and we would have to also define separately f(x,0) for all x.

also, f(sqrt2, 1) is not even in R x Q.
heh... yeah, i guess I was just thinking integers or something. i'm really not good at this kinda stuff lol, i'm more of a PDEs/applied math guy.

[concernedparent]

This is discrete mathematics I'm guessing? I'm in an intro class and we've got plenty of stuff like this. Let me stare at the original post for a while and see if anything comes to me.