Prove or disprove that $\sqrt{4+\pi} + \sqrt{4-\pi} \in \mathbb{Q}$.


closed as off-topic by Did, mrp, Henrik, Anthony Carapetis, projectilemotion Jun 12 '17 at 12:02

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    $\begingroup$ What have you tried? What tools and facts do you have at your disposal? Are you, for instance, allowed to use the fact that $\pi$ is trancendental? $\endgroup$ – Arthur Jun 12 '17 at 10:33
  • $\begingroup$ Try using a similar approach to proving that $\sqrt2$ is irrational? $\endgroup$ – lioness99a Jun 12 '17 at 10:37
  • $\begingroup$ Also see this related question $\endgroup$ – lioness99a Jun 12 '17 at 10:39
  • $\begingroup$ If $\sqrt{4+\pi}+\sqrt{4-\pi}\in\mathbb{Q}$, $\pi$ would be algebraic. $\endgroup$ – Jack D'Aurizio Jun 12 '17 at 16:50

Suppose that:

$$\sqrt{4 + \pi} + \sqrt{4 - \pi} = r \in \mathbb{Q}$$

If you square that expression, you get:

$$4 + \pi + 2 \sqrt{16 - \pi^2} + 4 - \pi = r^2$$

If you simplify this a little bit you get:

$$2 \sqrt{16 - \pi^2} = r^2 - 8$$


$${\pi^2} = 16-{({{r^2-8}\over {2}})^2} \in\mathbb{Q} $$

...which is not true because $\pi^2$ is not a rational number. Therefore, your expression is not a rational number.

EDIT: If you have to prove that $\pi^2$ is irrational, try the opposite. Say that:

$$\pi^2={p\over q}$$

where $p,q$ are whole numbers. You get the following equation:

$$q \pi^2 - p = 0$$

But $\pi$ is a transcendent number and cannot appear as a solution of any equation of the form:

$$q x^2 - p = 0 $$

with $p,q \in \mathbb{Z}$. So our initial assumption that $\pi^2$ is a rational number is definitely false.


Let's say that $\sqrt{4 + \pi}+\sqrt{4 -\pi} = q \in \mathbb{Q}$

Square and rearrange to get: $2\sqrt{16 - \pi^2} = q^2 - 4$ [equation 1]

If you're allowed to assume that $\pi$ is transcendental, then square equation 1 again to get:

$4(16 - \pi^2) = (q^2 - 4)^2$, which can be rearranged to give a polynomial with integral coefficients satisfied by $\pi$, hence leading to a contradiction.

If you can't assume the transcendence of $\pi$, then you have to prove $\pi^2$ is irrational, which can be done using the (not exactly elementary) technique here: http://planetmath.org/piandpi2areirrational


Hint. Let $q=\sqrt{4+\pi} + \sqrt{4-\pi}\in \mathbb{Q}$. Then $$q^2=4+\pi+4-\pi+2\sqrt{4+\pi}\cdot\sqrt{4-\pi}=16+2\sqrt{16-\pi^2},$$ which implies that $\pi^2=?$. Finally recall that $\pi^2$ is irrational.

  • $\begingroup$ Just to be clear, the irrationality of $\pi$ does not imply the irrationality of $\pi^2$. You need the fact of the transcendence of $\pi$ for that inference. $\endgroup$ – Deepak Jun 12 '17 at 10:48
  • $\begingroup$ @Deepak Thanks. I forgot the square. $\endgroup$ – Robert Z Jun 12 '17 at 11:52

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