31 Oct 07
2 edits
n is a positive integer. Here is a proof that sqrt(n) can't be a non-integer rational number:
Assume this is not the case. Then there is a minimal positive integer k such that k*sqrt(n) is an integer. But k*sqrt(n)-k*floor(sqrt(n)) is a smaller positive integer having the same property.
My question: can you write a SIMILAR proof, that will show that for all positive integers m and n, the mth root of n can't be a non-integer rational? It is easy to prove this theorem if we are allowed to use the fundamental theorem of arithmetic (=every integer has a unique prime factorization). But I want a proof similar to the proof I gave for square roots.
Thanx 🙂
01 Nov 07
1 edit
Originally posted by mtthwWhy does k*sqrt(n)-k*floor(sqrt(n)) need to be a multiple of sqrt(n)?
But that's not a contradiction, as k*sqrt(n)-k*floor(sqrt(n)) is not a multiple of sqrt(n).
You haven't used any property of sqrt(n) in your proof.
Of course it is a contradiction:
k*sqrt(n)-k*floor(sqrt(n)) is a positive integer, since k*sqrt(n) and k*floor(sqrt(n)) are integers and k*floor(sqrt(n)) < k*sqrt(n).
It is smaller than k, since sqrt(n)-floor(sqrt(n)) < 1.
When multiplied by sqrt(n), the result is an integer:
[k*sqrt(n)-k*floor(sqrt(n))]sqrt(n)=kn-(k*sqrt(n))(floor(sqrt(n)))=difference between two integers=integer.
But k was supposed to be the SMALLEST positive integer that gives an integer when multiplied by sqrt(n).
Contradiction...
Now I ask if this proof can be generalized to roots other than the square root.