# Show function equality is false

1. Sep 21, 2004

### cateater2000

Hi there I'm having trouble with this question, any tips would be great.

Show the following is false, by giving an example of a function(s) f:X->A for which the equalities fail:

f(Y&Z)=f(Y)&f(Z);

2. Sep 21, 2004

### mathwonk

try some examples and let us see what you come up with.

remember a function can be any way at all of mapping the elements of one set to the elements of another set. so start with some real simple finite sets and define the maps in various ways.

3. Sep 21, 2004

### cateater2000

I came up with something like this but it doesn't seem right.

f: (x,y)->{x+1,y-1}

X={(1,2)}
y={(0,0)}

f(X) gives us (2,1)
f(Y) gives us (1,-1)
so f(X) & f(Y) is just (1)

But f(X&Y) i'm confused because X&Y = 0
I'm not sure if I'm on the right track or not, thanks again for the help.

4. Sep 22, 2004

### matt grime

XnY is empty not zero, they are different.
You're constructing the function so you can make it so that this problem goes away.

Suppose that X and Y are sets of two elements that have one element in common, say

{1,2} and {2,3}

what about it f were the map that takes 1 and 3 to 'a', and 2 to 'b' what happens there?