Does the Function Delta Have Infinite Order in Rigid Motions?

[wubie]

Hello,

It will be easier to first post the question:

Let R be the set of real numbers. Consider the function delta from R² to R², 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)deltaⁿ is ...").

[Note: there are many parts to the original question. This is only part of it].


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 deltaⁿ cannot equal the identity for any n which is an element of Z.

Now I can see that x deltaⁿ = (x+n, (-y)ⁿ) and that deltaⁿ 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 deltaⁿ = x + n for all x, n > 0

and if n > 0 then x + n is not equal to x so deltaⁿ does not equal the identity?

Any help is appreciated. Thankyou.

 

[M98Ranger]

Hi there,

To prove that the function delta has infinite order, we need to show that for any positive integer n, the function deltan is not equal to the identity function. We can do this by showing that for any point (x,y), the coordinates of (x,y)deltan are not equal to (x,y).

Let's consider the first coordinate of (x,y)deltan:

(x,y)deltan = (x+n, (-y)n)

Since n is a positive integer, x+n will always be greater than x. This means that the first coordinate of (x,y)deltan will never be equal to x, no matter how many times we apply the function. Therefore, deltan cannot equal the identity function for any n, and thus has infinite order.

I hope this helps! Let me know if you have any further questions.