# Thread: Prove that f is 1-1 or give a counterexample

1. ## Prove that f is 1-1 or give a counterexample

Z=integers

f: Z(mod 7)-> Z(mod 6) defined by f([x](mod 7))=[x](mod 6).

2. Hmm. I would look at what 0 maps to, and what 6 maps to. Does that give you an idea?

3. Sorry, my modular arithmetic is sketchy at best. Is the argument that since 0mod7 and 0mod6 both map to 0, but since 6mod7 maps to 6 and 6mod6 maps to 0, that the function is not 1-1?

4. I'm not sure that's the right way of looking at it. The function appears to change the modulus of the arithmetic you're doing from 7 to 6. A function is 1-1 if and only if, for any two distinct members of the domain, x and y, f(x) and f(y) are distinct. If you think about a function graphically, it must pass the vertical line test in order to be a function (any vertical line only intersects the function in at most one location). To be 1-1, it must pass a horizontal line test: any horizontal line must intersect the function in at most one location. So, your functions maps 0 to where, and 6 to where?

5. Ok, so since f([0]mod 7) = 0mod6 = 0, and f([6]mod7) = 6mod6 = 0, this means two distinct elements in the domain Z7 both map to the same element in the codomain Z6 which makes the fuction not 1-1. Is this argument correct?

6. Looks good to me!