Question

(x,y)-->(-x,y) is an isometry true or false..

Answer 1

isometries mean the map preserves distances... consider that \(d(x_1,y_1,x_2,y_2)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) is invariant under the map \(x\mapsto -x\):$$d(-x_1,y_1,-x_2,y_2)=\sqrt{(-x_1+x_2)^2+(y_1-y_2)^2}$$
and we know \((x_1-x_2)^2=(x_2-x_1)^2\) hence this is indeed an isometry

Answer 2

you should give the metric space along with the mapping so that we don't have to guess what metric space you are using.

Answer 3

uh, the metric space is pretty damn obvious... \(\mathbb{R}^2\) with the usual metric \(d(\mathbf{x},\mathbf{y})=\|\mathbf{x}-\mathbf{y}\|\)

Answer 4

it does not have to be that...it probably is...but it most definitely does not have to be the usual metric

Answer 5

it is on my test paper and thats all what is says...

Answer 6

nobody said it doesn't have to be but using common sense reasoning it's obvious :p

Answer 7

actually I don't think anyone even tried to suggest \(\mathbb{R}^2\) *must* be equipped with the usual metric... I did say *usual* not *mandatory*

Answer 8

you did declare that it was "pretty damn obvious". That seems to suggest that the usual metric is that one that should be used here (which we don't know if that is the case). In many instances the usual metric is not even the first metric covered in a course on metric spaces (discrete metric for example).