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

• Jun 25th 2010, 04:06 PM
jsteel2
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).
• Jun 25th 2010, 04:49 PM
Ackbeet
Hmm. I would look at what 0 maps to, and what 6 maps to. Does that give you an idea?
• Jun 25th 2010, 05:28 PM
jsteel2
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?
• Jun 25th 2010, 05:42 PM
Ackbeet
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?
• Jun 25th 2010, 06:00 PM
jsteel2
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?
• Jun 26th 2010, 02:01 AM
Ackbeet
Looks good to me!