Is $\sqrt{4+\pi} + \sqrt{4-\pi} \in \mathbb{Q}$?

Question

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

Answer 1

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:

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

or:

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

...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.

Answer 2

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.

Answer 3

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