# homework help

• Aug 30th 2009, 03:52 PM
ratterlgirl10
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?
• Aug 30th 2009, 05:59 PM
Taluivren
Is it $\displaystyle \alpha(a,b): \mathbb{Z} \rightarrow \mathbb{Z}$ defined by $\displaystyle \alpha(a,b)(n) = an+b$ ?

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

In (b), again concentrate on the value of $\displaystyle a$ (parameter $\displaystyle b$ 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?
• Aug 30th 2009, 07:52 PM
ratterlgirl10
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
• Aug 31st 2009, 03:18 AM
Plato
Quote:

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.
$\displaystyle f(x) = ax + b\;\& \;a \ne 0$ is both onto and one-to-one.

But I don't understand your notation.
• Aug 31st 2009, 04:36 AM
Taluivren
Quote:

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 $\displaystyle \alpha(a,b)$ to be onto, it is necessary that $\displaystyle a=1$ or $\displaystyle a=-1$. It is now trivial to verify that this is also sufficient.

Is the rest clear?