The Conjugate Roots Theorem for Irrational Roots states that for a polynomial $f(x)$ with integer coefficients, if a root of the equation $f(x) = 0$ is expressed as $a+\alpha$, where $a\in\mathbb{Q}$ and $\alpha\in\mathbb{R}-\mathbb{Q}$ are rational and is irrational, then $a-\alpha$ is also a root of the equation. However, some additional explanation should be added to this theorem so that the following contradiction, which I found, will not occur.
Let $\alpha+1=\beta\in\mathbb{R}-\mathbb{Q}$.
Then, $a+\alpha=(a-1)+\beta$
Since the root of $f(x)=0$ is expressed as $(a-1)+\beta$, by the above theorem,
$(a-1)-\beta=(a-2)+\alpha$ is also a root of the equation $f(x)=0$
Could you explain why this contradiction occurred and how to interpret this theorem to prevent the contradiction?