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#### wubie

##### Guest

It will be easier to first post the question:

[Note: there are many parts to the original question. This is only part of it].Let R be the set of real numbers. Consider the function delta from R^{2}to R^{2}, defined for all points with coordinates (x,y), by the formula:

(x,y)delta = (x+1, -y)

Prove that delta has infinite order. (It is not enough to state the definition of infinite order. You must give a reason such as "the first coordinated of (x,y)delta^{n}is ...").

I am not sure how to proceed here. Firstly I know that the above function is a translation. I also know that if delta has an infinite order then delta

^{n}cannot equal the identity for any n which is an element of Z.

Now I can see that x delta

^{n}= (x+n, (-y)

^{n}) and that delta

^{n}will never return to (x,y).

But I am not sure how to prove this. If anything is unclear, please ask and I will try to clarify my question.

Do I first prove by induction on n that

x delta

^{n}= x + n for all x, n > 0

and if n > 0 then x + n is not equal to x so delta

^{n}does not equal the identity?

Any help is appreciated. Thankyou.