Image/preimage transformation

**sfspitfire23** · January 18th 2010, 05:51 PM

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

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

**Drexel28** · January 18th 2010, 05:56 PM

Originally Posted by sfspitfire23

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

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 $\alpha,\beta$ ? Is this an isometry between metric spaces or one in just geometry?

**sfspitfire23** · January 18th 2010, 06:14 PM

sorry, just geometry

**Drexel28** · January 18th 2010, 06:28 PM

Originally Posted by sfspitfire23

sorry, just geometry

Ok. So what are $\alpha,\beta$ ?

**sfspitfire23** · January 18th 2010, 06:42 PM

Ah, they are point sets

**Drexel28** · January 18th 2010, 06:44 PM

Originally Posted by sfspitfire23

Ah, they are point sets

and lastly...what does it mean for two point sets to be congruent?

**sfspitfire23** · January 18th 2010, 06:58 PM

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

**sfspitfire23** · January 18th 2010, 08:33 PM

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

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