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?

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- Aug 30th 2009, 03:52 PMratterlgirl10homework 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 PMTaluivren
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 PMratterlgirl10
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 AMPlato
- Aug 31st 2009, 04:36 AMTaluivren