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.

 

[anathema]

Hello,

Thank you for your question. To prove that delta has infinite order, we need to show that for any positive integer n, the composition of delta with itself n times, denoted as deltan, is not equal to the identity function.

Let's start by considering the first coordinate of (x,y)deltan. We can see that (x,y)deltan = (x+n, (-y)n). As n increases, the first coordinate of (x,y)deltan will also increase, meaning that deltan will never return to the original x-coordinate. This is because the function delta is a translation that adds 1 to the first coordinate, and this addition will continue for every subsequent composition of delta with itself.

Now, let's assume that deltan equals the identity function for some positive integer n. This would mean that (x,y)deltan = (x,y) for all (x,y) in R2. However, we have just shown that this is not possible since the first coordinate of (x,y)deltan will always be different from the original x-coordinate. Therefore, deltan cannot equal the identity function for any positive integer n, proving that delta has infinite order.

I hope this helps to clarify the proof for you. Let me know if you have any further questions.

 

[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.