Math Is Fun Forum

LRS

Guest

Showing there is a unique function.

Let R be the relation:

R = {(x,y) : x = y (mod 5)} where x,y are integers.
which one can verify that R is an equivalence relation.

We define an equivalence class as [x] = { y in A : yRx }
We define A/R as the set of all equivalence classes of elements of A. In other words:
A/R = {[x] : x in A}

(a) Show that there is a unique function h: Z/R -> Z/R such that for every integer x, h([x]) = [x^2]
(b) Show that there is no function h: Z/R -> Z/R such that for every integer x, h([x]) = [2^x]

Could I please get some help with this problem? At least some idea in order to get started?

Thanks.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Showing there is a unique function.

Start by representing your elements of Z/R as [0], [1], [2], [3], and [4].  Where must f send 0?  Where must f send 1?  etc.

Once you know this, you need to prove that f would also send all of the elements in [0] to the same equivalence class that 0 goes to.  And similarly with [1], [2], [3], and [4]. Note that you don't have to do them one at a time, you can argue it in general.

To prove uniqueness, you must prove that you only have one choice for such a function.  Since you're squaring elements, this is so straight forward that it may be hard to do (or even seem silly to say).

For (b), you want to find two elements in the same equivalence class, call them x and y (remember this means that x = y (mod 5) ) such that h(x) is not in the same equivalence class as h(y).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."