Equivalence relations problem (algebra)

[Pearce_09]

Z = all integers

A = Z; m is related to n, means that $$m^2 - n^2$$ is even;
B = {0,1}

I already proved that this is a equivalence relation, but i just don't know how to;

I need to find a well defined bejection
sigma : $$A_= -> B$$


I hope this makes sense.. i wrote it up as well as I can.

 

[JasonRox]

Z = all integers
A = Z; m is related to n, means that $$m^2 - n^2$$ is even;
B = {0,1}
I already proved that this is a equivalence relation, but i just don't know how to;
I need to find a well defined bejection
sigma : $$A_= -> B$$
I hope this makes sense.. i wrote it up as well as I can.

You want a bijection from A to B?

Is that was you are looking for?

Note: If that is the case, that isn't possible.

 

[Pearce_09]

I'm confused, why wouldn't that be possible??

 

[Pearce_09]

A is an equivalence class,
so yes i am looking for
$$A_= -> B$$

 

[JasonRox]

A has more than two elements, where B has two elements.

If I got the definition of bijection, which I think I do, then it's not possible.

 

[JasonRox]

I could be reading the definition of the set A incorrectly.

 

[fourier jr]

no, i think A is really Z/~ where ~ is the equivalence relation given. as with any equivalence relation, the underlying set (in this case the integers) gets partitioned into cosets, in this case the one where m^2 - n^2 is even & one where it's odd, so A:=Z/~ has two elements. make sigma(one coset) = 0 or 1 & sigma(other coset) = 0 or 1 (so sigma is bijective). i don't know what the actual bijection is but that's probably how it's done. might take some fiddling.

 

[JasonRox]

no, i think A is really Z/~ where ~ is the equivalence relation given. as with any equivalence relation, the underlying set (in this case the integers) gets partitioned into cosets, in this case the one where m^2 - n^2 is even & one where it's odd, so A:=Z/~ has two elements. make sigma(one coset) = 0 or 1 & sigma(other coset) = 0 or 1 (so sigma is bijective). i don't know what the actual bijection is but that's probably how it's done. might take some fiddling.

Oh, I thought you said all elements (m and n) that satisfy that and that are in Z, which would be written as a pair is in A.

You know what. I'm really confused.

 

[Pearce_09]

yes Fourier jr, that's the idea..
thanks guys for giving me a push in the right direction

 

[fourier jr]

another way to write it is

m~n <==> $$m^2 \equiv n^2 (mod2)$$ ie $$m^2 - n^2 \equiv 0 (mod2)$$

so you've got 2 possibilities: m~n or not, in which case $$m^2 - n^2 \equiv 1 (mod2)$$

(OOPS i just gave the solution away i think )

edit: only jasonrox is allowed to read this!

 

[Pearce_09]

im actually going to post another question similar to this..
thanks

 

[JasonRox]

another way to write it is
m~n <==> $$m^2 \equiv n^2 (mod2)$$ ie $$m^2 - n^2 \equiv 0 (mod2)$$
so you've got 2 possibilities: m~n or not, in which case $$m^2 - n^2 \equiv 1 (mod2)$$
(OOPS i just gave the solution away i think )
edit: only jasonrox is allowed to read this!

I see what's going on now. Thanks.