Mod. Arithmetic Proof: I don't see flaws in my logic, but it isn't working out.

[jdinatale]

1. Homework Statement .
1. Let a and b be constant integers with a \not = 0, and let the mapping f : Z \rightarrow Z be defined by F(x) = ax + b. Determine all values of a such that f is a bijection. Prove that the aforementioned values are the only possible values resulting in a bijection.

The logic in my proof makes sense, but my conclusion that ax \cong 0 \mod a doesn't make sense because that statement will always be true.

Homework Equations

N/A

The Attempt at a Solution

 

[HallsofIvy]

What's wrong with always being true? The only way f(x)= ax+ b will not be bijective is if there exist x_1 and x_2 such that x_1\ne x_2 but ax_1+ b= ax_2+ b. What does that tell you about a?

 

[jdinatale]

What's wrong with always being true? The only way f(x)= ax+ b will not be bijective is if there exist x_1 and x_2 such that x_1\ne x_2 but ax_1+ b= ax_2+ b. What does that tell you about a?

I understand that it will always be 1-1.
But my whole issue is that you could construct a f(x) = ax + b that is not onto. For example, consider f(x) = 4x + 3. There is no x \in \mathbf{Z} such that f(x) = 5.

That's my whole problem.

 

[fzero]

You seem to have lost the b\mod a term in your argument. Considering it should lead you to the correct condition.

 

[jdinatale]

You seem to have lost the b\mod a term in your argument. Considering it should lead you to the correct condition.

Thanks for point that out, I did mess up there. It should be ax - b \cong b \mod a. I don't quite see how this helps because that implies that a must divide ax - 2b. I don't see how that tells me anything about what a must be. I mean it tells us that a must divide 2b for this to be true. <br /> <br /> But anyways, at the end of the day, we know a must divide t - b. The only number guaranteed to divided every integer k = t - b for all integers t and b is 1. The problem is finding a mathematical argument for that. Given any other a, I could always find a number such that the function isn&#039;t a bijection.

 

[jdinatale]

I apologize for double posting, but it will no longer let me edit my last post. I believe I have found a solution, but I would greatly appreciate it if you could look through it and try to find errors.