# One more elementary question, on square roots

## Main Question or Discussion Point

Hi all,

is there a general way of proving that

sqrt(r1) + sqrt(r2) + sqrt(r3) + ... + sqrt(rn) is irrational, given that none of r1, r2, r3, ..., rn is the square of a rational number?

(or is this statement even true in general?)

for the case when n = 2, the proof is quite straight-forward; i think it can be found in most elementary textbooks.

Letting sqrt(a) + sqrt(b) = r, where r is rational, we have

sqrt(a) - sqrt(b) = (a - b) / r = q, where q is rational.

Therefore adding the two equations and halving the result gives

sqrt(a) = 1/2(r + q), which is rational, contradicting our hypothesis.

i tried to extend this proof to the case n = 3, although my proof is quite clumsy and i'm not sure whether it's correct.

however, i am interested to know whether it is true for all n, and if so how it can be proved. thanks for sharing :)

## Answers and Replies

Related Linear and Abstract Algebra News on Phys.org
CRGreathouse
Homework Helper
I don't understand your proof, so let me form a 'counterexample'. Please explain what I misunderstand that allows this.

Take 2 and $6-4\sqrt2$. The former is clearly not the square of a rational; the latter is the square of 2-sqrt(2) and as such not the square of a rational. Clearly, though,

$$\sqrt2+\sqrt{6-4\sqrt2}=\sqrt2+\left(2-\sqrt2\right)=2$$

is rational. What condition did I miss?

Last edited:
arildno
Homework Helper
Gold Member
Dearly Missed
CRGreathouse:
I believe r1....rn are meant to be RATIONAL numbers.

HallsofIvy
It is sufficient to prove that if r1 and r2 are rational numbers that are not the squares of rational numbers (so that $\sqrt{r_1}$ and $\sqrt{r_2}$ are irrational, then $\sqrt{r_1}+ \sqrt{r_2}$ is irrational). That's "non-trivial" since the sum of two irrational numbers may be rational. You should be able to show that if $\sqrt{r_1}+ \sqrt{r_2}$ is rational then so is $\sqrt{r_1}- \sqrt{r_2}$. From that the result follows easily.