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One more elementary question, on square roots

  1. Dec 22, 2006 #1
    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 :)
  2. jcsd
  3. Dec 22, 2006 #2


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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,


    is rational. What condition did I miss?
    Last edited: Dec 22, 2006
  4. Dec 22, 2006 #3


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    I believe r1....rn are meant to be RATIONAL numbers.
  5. Dec 23, 2006 #4


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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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