Irrationality proof

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 🙂

Originally posted by David113
But k*sqrt(n)-k*floor(sqrt(n)) is a smaller positive integer having the same property.

What property do you mean?

Originally posted by TheMaster37
What property do you mean?

The property of being an integer when multiplied by sqrt(n).

Originally posted by David113
The property of being an integer when multiplied by 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.

Originally posted by mtthw
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.

Why does k*sqrt(n)-k*floor(sqrt(n)) need to be a multiple of sqrt(n)?

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.

Originally posted by David113
Of course it is a contradiction:

...(snip)...

Now that's more like it. I think the steps you assumed (and have now provided) are the key parts of the proof. Certainly makes it a lot clearer!

Originally posted by David113
Why does k*sqrt(n)-k*floor(sqrt(n)) need to be a multiple of sqrt(n)?

Incidentally, this was my misunderstanding. I thought you meant that k*sqrt(n)-k*floor(sqrt(n)) had the same property as k*sqrt(n), not the same property as k. Sorry!