# Sum of irrational numbers

Well, in this question it is said that $$\sqrt[100]{\sqrt3 + \sqrt2} + \sqrt[100]{\sqrt3 - \sqrt2}$$, and the owner asks for "alternative proofs" which do not use rational root theorem. I wrote an answer, but I just proved $$\sqrt[100]{\sqrt3 + \sqrt2} \notin \mathbb{Q}$$ and $$\sqrt[100]{\sqrt3 - \sqrt2} \notin \mathbb{Q}$$, not the sum of them. I got (fairly) downvoted, because I didn't notice that the sum of two irrational can be either rational or irrational, and I deleted my (incorrect) answer. So, I want help in proving things like $$\sqrt5 + \sqrt7 \notin \mathbb{Q}$$, and $$(1 + \pi) - \pi \in \mathbb{Q}$$, if there is any "trick" or rule to these cases of summing two (or more) known irrational numbers (without rational root theorem).

Thanks.

• In general, it's hard to tell. We know not the irrationality of $e+\pi$, to give a famous example. For algebraic numbers, things are a bit easier. Commented Dec 15, 2011 at 18:18
• @J.M. Do transcedental numbers have anything that makes it harder to prove? Commented Dec 15, 2011 at 18:23
• Probably the fact that our knowledge of them is paltry compared to algebraics. Hence we have conjectures like Schnauel's. Commented Dec 15, 2011 at 18:25
• Aside from lacking any algebraic structure, transcendental numbers are also hard to work with because they're defined in terms of a property they don't have, rather than one they do. So it's impossible to prove anything about them "from the definition." That is, they're defined as numbers that aren't the root of any integer polynomial. With algebraic numbers, you can always start a proof off like "If $\alpha$ is algebraic then $a_n\alpha^n + \dots + a_1\alpha +a_0=0$ for some integers $a_i$." So right off the bat, you know something about algebraics. Commented Dec 15, 2011 at 18:57
• There is one little useful fact: if $x$ is algebraic and $y$ is transcendental, then $x+y$ is transcendental; so is $xy$, unless $x=0$. So, for example, $\pi+\sqrt2$ and $\pi\sqrt2$ are both transcendental. Commented Dec 15, 2011 at 23:34

To prove that $\sqrt5+\sqrt7$ is irrational:

$\sqrt 5+\sqrt 7=\frac{a}{b}$

$\frac{a^2}{b^2}=12+\sqrt{35}$

$\frac{a^2-12b^2}{b^2}=\sqrt{35}$

$35=\frac{(a^2-12b^2)^2}{(b^2)^2}$

$35|a^2-12b^2$

$35^2|(a^2-12b^2)^2$

$35^2|(b^2)^2$

Both the numerator and denominator are multiples of an even power of 2. Contradiction.

The method can be extended to many other sums of nth roots.

• I don't understand the last step, can you explain it in more detail? Also, I think there is a mistake in the first step: shouldn't it say $\frac{a^2}{b^2} = 12 + 2 \cdot \sqrt{35}$? Commented Jul 6, 2012 at 7:04
• If Ian knows how to prove that $\sqrt{35}$ is irrational, it is enough to note that $\sqrt{35}=\frac12\left((\sqrt5+\sqrt7)^2-12\right)$, which would be rational if $\sqrt 5+\sqrt 7$ were. Commented Sep 4, 2013 at 2:59

Assume $\sqrt a + \sqrt b = \dfrac pq$ where $p,q\in \mathbb{Z}$ and $a,b$ rational; $\sqrt a$ irrational

$$q[\sqrt a + \sqrt b] = p$$

$$q[a-b] =p[\sqrt a - \sqrt b]$$

$$\frac{q}{p}[a-b] = \sqrt a - \sqrt b$$

Therefore, $$\sqrt a + \sqrt b + \sqrt a - \sqrt b = \frac pq + \frac qp(a-b)$$

$$2\sqrt a = \frac{p^2 + q^2(a-b)}{qp}$$

$2\sqrt a = \dfrac{p^2 + q^2(a-b)}{qp}$ which is a contradiction because $\sqrt a$ is an irrational number and so cannot be written as a ratio of integers

Here is a useful trick, though it requires a tiny bit of field theory to understand: If $\alpha + \beta$ is a rational number, then $\mathbb{Q}(\alpha) = \mathbb{Q}(\beta)$ as fields. In particular, if $\alpha$ and $\beta$ are algebraic, then the degrees of their minimal polynomials are equal.

So, for example, we can see at a glance that $\sqrt{5} + \sqrt[3]{7}$ is irrational, because $\sqrt{5}$ and $\sqrt[3]{7}$ have algebraic degree $2$ and $3$ respectively.

Note that this trick doesn't work on your original example, because $\alpha=\sqrt[100]{\sqrt{3} + \sqrt{2}}$ and $\beta=\sqrt[100]{\sqrt{3} - \sqrt{2}}$ do have the same degree. But we can also use field theory: since $\alpha \beta = 1$, if $\alpha+\beta$ is rational then $\alpha$ and $\beta$ satisfy a rational quadratic. However, $\alpha^{100}= \sqrt{3}+\sqrt{2}$ already has degree $4$ over $\mathbb{Q}$, so $\alpha$ certainly has degree bigger than $2$.