One more elementary question, on square roots

  • Thread starter xalvyn
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  • #1
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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 :)
 

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  • #2
CRGreathouse
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I don't understand your proof, so let me form a 'counterexample'. Please explain what I misunderstand that allows this.

Take 2 and [itex]6-4\sqrt2[/itex]. 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,

[tex]\sqrt2+\sqrt{6-4\sqrt2}=\sqrt2+\left(2-\sqrt2\right)=2[/tex]

is rational. What condition did I miss?
 
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  • #3
arildno
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CRGreathouse:
I believe r1....rn are meant to be RATIONAL numbers.
 
  • #4
HallsofIvy
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It is sufficient to prove that if r1 and r2 are rational numbers that are not the squares of rational numbers (so that [itex]\sqrt{r_1}[/itex] and [itex]\sqrt{r_2}[/itex] are irrational, then [itex]\sqrt{r_1}+ \sqrt{r_2}[/itex] is irrational). That's "non-trivial" since the sum of two irrational numbers may be rational. You should be able to show that if [itex]\sqrt{r_1}+ \sqrt{r_2}[/itex] is rational then so is [itex]\sqrt{r_1}- \sqrt{r_2}[/itex]. From that the result follows easily.
 

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