1. ## homework help

For each ordered pair ( a, b) of integers define a mapping alpha (a, b) : Z . Z by alapha a, b( n) = an + b.

( a) For which pairs ( a, b) is alpha (a, b) onto?

( b) For which pairs ( a, b) is alpha (a, b) one- to- one?

2. Is it $\alpha(a,b): \mathbb{Z} \rightarrow \mathbb{Z}$ defined by $\alpha(a,b)(n) = an+b$ ?

For (a), consider $|a|>1$ ...can $\alpha(a,b)$ be onto? Then examine remaining cases for $a$.

In (b), again concentrate on the value of $a$ (parameter $b$ acts like a shift, it can't affect whether the function is one-to-one or not). There are not many cases in which the function is not one-to-one, can you see them?

3. for part a is there a easier way to prove this proof instead looking at the problem from both sides of iff and only if

I have this so far for part a

right side: an+b= b+1
an=1
a=plus/minus 1

4. Originally Posted by ratterlgirl10
For each ordered pair ( a, b) of integers define a mapping alpha (a, b) : Z . Z by alapha a, b( n) = an + b.

( a) For which pairs ( a, b) is alpha (a, b) onto?

( b) For which pairs ( a, b) is alpha (a, b) one- to- one?
This notation is very hard to read.
Here is a basic fact about linear functions.
$f(x) = ax + b\;\& \;a \ne 0$ is both onto and one-to-one.

But I don't understand your notation.

5. Originally Posted by ratterlgirl10
I have this so far for part a

right side: an+b= b+1
an=1
a=plus/minus 1
That's actually a good idea! So now you know that for $\alpha(a,b)$ to be onto, it is necessary that $a=1$ or $a=-1$. It is now trivial to verify that this is also sufficient.

Is the rest clear?