How to show this function is injective?

Question

Define $h: \mathbb Z \to [0,1]$ by $h(z)= z \pi - [z\pi]$ where $[.]$ denotes the floor function.

I want to show this function is injective, I have tried both the approaches using $h(z_1)= h(z_2)$ and try to get $z_1= z_2$ and the another one $z_1\neq z_2 $ to get $h(z_1)\neq h(z_2)$ but stuck in both the methods.

Can anyone please help me out?

score 3 · Accepted Answer · answered Oct 30 '17 at 20:07

3

If $h(z_1) = h(z_2)$ then $$ z_1\pi - [z_1\pi] = z_2\pi - [z_2\pi] $$ and so $$ z_1\pi - z_2\pi= [z_1\pi] - [z_2\pi] $$ so $$ (z_1 - z_2)\pi = [z_1\pi] - [z_2\pi] $$ so \dots (I am sure you can figure out the rest.)

answered Oct 30 '17 at 20:07

Thomas

43,555

modulus of L.H.S is greater than $1$ but modulus of RHS is less than $1$. Right? – User Oct 30 '17 at 20:09
1

On the left hand side you have something that is an integer times $\pi$. On the right you have an integer, so ... – Thomas Oct 30 '17 at 20:11
Oh okay, Silly me! – User Oct 30 '17 at 20:12
1

Thanks a lot!! L) – User Oct 30 '17 at 20:12
@User: You are welcome. – Thomas Oct 30 '17 at 20:15

How to show this function is injective?

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