$1-1$ function

Let $f:\mathbb{N} \rightarrow [0, 1)$ be a function defined as $f(n)=\{ 2^{n+1/2} \}$ where $\{ \cdot\}$ denotes the fractional part. Prove that $f$ is $1-1$.

Solution

Let $m_1, \; m_2 \in \mathbb{N}$ such that $f(m_1)=f(m_2)$. Thus:

\begin{align*}
f(m)=f(n) &\Leftrightarrow  \left \{ 2^{m_1+1/2} \right \}=\left \{ 2^{m_2+1/2} \right \} \\
 &\Leftrightarrow 2^{m_1+1/2} - \left \lfloor 2^{m_1+1/2} \right \rfloor = 2^{m_2+1/2} - \left \lfloor 2^{m_2+1/2} \right \rfloor \\
 &\Leftrightarrow 2^{m_1+1/2} - 2^{m_2+1/2} = \left \lfloor 2^{m_1+1/2} \right \rfloor  - \left \lfloor 2^{m_2+1/2} \right \rfloor \\
 &\Leftrightarrow \left ( 2^{m_1}-2^{m_2} \right ) \sqrt{2} = \left \lfloor 2^{m_1+1/2} \right \rfloor  - \left \lfloor 2^{m_2+1/2} \right \rfloor
\end{align*}

Now, the RHS is definitely an integer. If $m_1 \neq m_2$ then this would imply that $\sqrt{2}$ would be rational, which is of course an obscurity. Thus $m_1=m_2$.

The exercise can also be found at mathematica.gr