# The Group of Rigid Motions (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

W

#### wubie

##### Guest
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.

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving