Image/preimage transformation

• Jan 18th 2010, 05:51 PM
sfspitfire23
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...
• Jan 18th 2010, 05:56 PM
Drexel28
Quote:

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?
• Jan 18th 2010, 06:14 PM
sfspitfire23
sorry, just geometry
• Jan 18th 2010, 06:28 PM
Drexel28
Quote:

Originally Posted by sfspitfire23
sorry, just geometry

Ok. So what are $\displaystyle \alpha,\beta$?
• Jan 18th 2010, 06:42 PM
sfspitfire23
Ah, they are point sets
• Jan 18th 2010, 06:44 PM
Drexel28
Quote:

Originally Posted by sfspitfire23
Ah, they are point sets

and lastly...what does it mean for two point sets to be congruent?
• Jan 18th 2010, 06:58 PM
sfspitfire23
two sets where one is the image of the other over a transformation...
• Jan 18th 2010, 08:33 PM
sfspitfire23
Hm, then it would seem i could do this proof by definition almost i believe