Suppose that $x = p - \sqrt{q}$ where $p$ and $q$ are rational and $q \geq 0$. Can I always find rational numbers $\tilde p, \tilde q$ such that $\tilde q \geq 0$ and \begin{align*} x = \tilde p + \sqrt{ \tilde q} \end{align*}
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If $\sqrt q$ is rational. Writing $p - \tilde p = \sqrt q + \sqrt {\tilde q}$ transforms your question into a duplicate of https://math.stackexchange.com/questions/136594/. – player3236 Mar 02 '21 at 17:16
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Not unless $\sqrt{q}$ is rational.
Rewrite the equation $p - \sqrt{q} = \overline{p} + \sqrt{\overline{q}}$ as $$ x = \sqrt{q} + \sqrt{\overline{q}} \tag 1$$ where $x = p - \overline{p}$, and square both sides: $$ x^2 = q + 2 \sqrt{q \overline{q}} + \overline{q}^2$$
Since all other terms are rational, $\sqrt{q \overline{q}}$ must be rational. If $r = \sqrt{q \overline{q}}$, that says $\overline{q} = r^2/q$. Equation (1) then becomes $$x = \sqrt{q} + r/\sqrt{q} $$ so that $$ x \sqrt{q} = q + r $$ Now $\sqrt{q}\ge 0$ and $\sqrt{\overline{q}} \ge 0$. Unless they are $0$, $x > 0$ and $\sqrt{q} = (q+r)/x$ is rational.
Robert Israel
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No, except for the trivial case $\sqrt{q}\in\mathbb{Q}$.
Hint: $$p-\sqrt{q}=\bar{p}+\sqrt{\bar{q}}\\ \bar{q}=(p-\bar{p}-\sqrt{q})^2=(p-\bar{p})^2+q-2(p-\bar{p})\sqrt{q}$$
This forces $p=\bar{p}$ since $\sqrt{q}$ is irrational, but then $-\sqrt{q}=+\sqrt{\bar{q}}$, a contradiction.
Chrystomath
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