Let $r_1, r_2, \ldots, r_k$ be the rational roots of a polynomial $P$ of degree $n$ that has $x$ as a root. Since $x$ is irrational, it does not equal any $r_i$.  Let $c_1 > 0$ be the minimum of $\size {x - r_i}$. If there are no $r_i$, let $c_1 = 1$. Now let $\alpha = \dfrac p q$ where $\alpha \notin \set {r_1, \ldots, r_k}$.  Then: {{begin-eqn}} {{eqn | l = \map P \alpha       | o = \ne       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \size {\map P \alpha}       | o = \ge       | r = \frac 1 {q^n}       | c = as $\map P \alpha$ is a multiple of $\dfrac 1 {q^n}$ }} {{eqn | ll= \leadsto       | l = \size {\map P x - \map P \alpha}       | o = \ge       | r = \frac 1 {q^n}       | c = because $\map P x = 0$ }} {{end-eqn}} Suppose: :$\displaystyle \map P x = \sum_{k \mathop = 0}^n a_k x^k$ Then: {{begin-eqn}} {{eqn | l = \map P x - \map P \alpha       | r = \sum_{k \mathop = 0}^n a_k x^k - \sum_{k \mathop = 0}^n a_k {\alpha }^k       | c =  }} {{eqn | ll= \leadsto       | r = \sum_{k \mathop = 0}^n a_k \paren {x^k - {\alpha }^k }       | c = }} {{eqn | ll= \leadsto       | r = \sum_{k \mathop = 1}^n a_k \paren {x^k - {\alpha }^k }       | c = $x^0 - {\alpha }^0 = 0$ }} {{eqn | ll= \leadsto       | r = \sum_{k \mathop = 1}^n a_k \paren {x - \alpha} \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i       | c = Difference of Two Powers }} {{eqn | ll= \leadsto       | r = \paren {x - \alpha} \sum_{k \mathop = 1}^n a_k \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i       | c =  }} {{end-eqn}} '''Case 1''': If $\size {x - \alpha} \le 1, \alpha \notin \set {r_1, \ldots, r_k}$, then {{begin-eqn}} {{eqn | l = \size {\alpha } - \size{x }       | o = \le       | r = \size {x - \alpha}       | c = Reverse Triangle Inequality }} {{eqn | ll= \leadsto       | l = \size {\alpha } - \size{x }       | o = \le       | r = 1       | c = Transitive }} {{eqn | ll= \leadsto       | l = \size \alpha       | o = \le       | r = \size x + 1       | c = rearranging }} {{end-eqn}} Therefore {{begin-eqn}} {{eqn | l = \size {\displaystyle \map P x - \map P \alpha }       | o = \le       | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \displaystyle \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 - i} \alpha^i }       | c = Triangle Inequality }} {{eqn | ll= \leadsto       | o = \le       | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \displaystyle \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 - i} \paren {\size x + 1 }^i }       | c = substituting $\size x + 1$ for $\alpha$ }} {{eqn | ll= \leadsto       | o = \le       | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \displaystyle \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 } \paren {\frac {\size x + 1 } {x } }^i }       | c = moving $ x^{-i} $ over }} {{eqn | ll= \leadsto       | o = \le       | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k x^{k - 1 } } \displaystyle \sum_{i \mathop = 0}^{k - 1} \size {\paren {1 + \frac 1 x }^i }       | c = moving $x^{k - 1 }$ out of the sum }} {{eqn | ll= \leadsto       | o = \le       | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k x^{k - 1 } } \frac {\paren {1 + \frac 1 x }^k - 1 } {\paren {1 + \frac 1 x } - 1 }       | c = Sum of Geometric Sequence }} {{eqn | ll= \leadsto       | o = \le       | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k x^{k } } \paren {\paren {1 + \frac 1 x }^k - 1 }       | c =  }} {{eqn | ll= \leadsto       | o = \le       | r = \size {x - \alpha } \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \paren {\paren {\size {x } + 1 }^k - \size {x }^{k } }       | c = Distribute }} {{end-eqn}} To summarize: :$\size {\map P x - \map P \alpha} \le \size {x - \alpha} c_x$ where: :$\displaystyle c_x = \displaystyle \sum_{k \mathop = 1}^n \size {a_k } \paren {\paren {\size {x } + 1 }^k - \size {x }^{k } }$ So for such $\alpha$: :$\size {x - \alpha} \ge \dfrac {\size {\map P x - \map P \alpha} } {c_x} \ge \dfrac 1 {c_x q^n}$ '''Case 2''': If $\size {x - \alpha} \gt 1, \alpha \notin \set {r_1, \ldots, r_k}$ then: :$\size {x - \alpha} \gt 1 \ge \dfrac 1 {q^n}$ '''Case 3''': If $\alpha \in \set {r_1, \ldots, r_k}$, then: :$\size {x - \alpha} \ge c_1 \ge \dfrac {c_1} {q^n}$ $c = \begin{cases} \dfrac 1 {c_x} & : \size {x - \alpha} \le 1, \alpha \notin \set {r_1, \ldots, r_k} \\ 1 & : \size {x - \alpha} \gt 1, \alpha \notin \set {r_1, \ldots, r_k} \\ c_1 & : \alpha \in \set {r_1, \ldots, r_k} \end{cases}$ Then: :$\size {x - \dfrac p q} \ge \dfrac c {q^n}$ for all $\dfrac p q$. {{qed}}