Rotation function

[dpa]

Hi all,

Q. A function takes (x,y) and gives (y,x). Is this function injective?

For any function to be injective, f(x,y)=f(x',y')=>(x,y)=(x',y').
But here, I get,
(y,x)=(y',x')
How can I show the function is injective? It appears to be one.

Thank You.

 

[LCKurtz]

Hi all,

Q. A function takes (x,y) and gives (y,x). Is this function injective?

For any function to be injective, f(x,y)=f(x',y')=>(x,y)=(x',y').
But here, I get,
(y,x)=(y',x')
How can I show the function is injective?

Right so far. Can you conclude (x,y) = (x',y') from that?

 

[dpa]

the ordered pairs are equal means that we can write y=y' and x=x' which in tern mean that
(x,y)=(x',y')

Is this fine.

Thank You.

 

[LCKurtz]

Yes, that's all there is to it.