I'm wondering whether the below is a valid proof for:
$f(x) = \alpha x - E(\alpha x) < 1, \;\forall x\in\mathbb Z, \;\alpha \in \{\mathbb R \setminus \mathbb Q \}$ where $E(x)$ is integer part of $x$
I've used two steps.
Step 1
Put $\mathbb I = \{\mathbb R \setminus \mathbb Q\}$ which denotes irrational numbers.
Let $\alpha = k + r, \; k \in \mathbb Z,\; r\in(0; 1),\; r\in \mathbb I$. Then the function may be rewritten as:
$$ \begin{align} f(x) & = \alpha x - E(\alpha x) \\ & = (k+r)x - E((k+r)x) \\ &= kx + rx - E(kx + rx) \\ &= rx - E(rx) \tag 1 \end{align} $$
Since $x$ may be arbitrary we need step 2.
Step 2
Let $rx = a + b, \; a\in \mathbb Z, \; b \in \mathbb I, b \in [0, 1)$. Rewrite $(1)$:
$$ $$ \begin{align} rx - E(rx) & = a + b - E(a + b) \\ & = a + b - a - E(b) \\ &= b - E(b) \\ &= b - 0 \\ &= b \tag 2 \end{align} $$ $$
But $b \in [0, 1)$ and therefore $f(x) \in [0, 1)$ which proves $f(x) < 1$