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## 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 :)

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 :)