# Thread: Image/preimage transformation

1. ## Image/preimage transformation

Prove If $\displaystyle \alpha \cong \beta$ and $\displaystyle T(\alpha)=\alpha ', T(\beta)=\beta '$ where $\displaystyle T$ is an isometry, then $\displaystyle \alpha ' \cong \beta '$

I think I need to start with alpha prime and get to alpha....but im not sure how to show this...

2. Originally Posted by sfspitfire23
Prove If $\displaystyle \alpha \cong \beta$ and $\displaystyle T(\alpha)=\alpha ', T(\beta)=\beta '$ where $\displaystyle T$ is an isometry, then $\displaystyle \alpha ' \cong \beta '$

I think I need to start with alpha prime and get to alpha....but im not sure how to show this...
There is no context here. What are $\displaystyle \alpha,\beta$? Is this an isometry between metric spaces or one in just geometry?

3. sorry, just geometry

4. Originally Posted by sfspitfire23
sorry, just geometry
Ok. So what are $\displaystyle \alpha,\beta$?

5. Ah, they are point sets

6. Originally Posted by sfspitfire23
Ah, they are point sets
and lastly...what does it mean for two point sets to be congruent?

7. two sets where one is the image of the other over a transformation...

8. Hm, then it would seem i could do this proof by definition almost i believe