Question

Prove that the number `sqrt(1+sqrt(2+...+sqrt(n)))` is not rational for any natural number `n` greater than `1`?

Answer 1

See explanation...

Explanation:

Suppose:

`sqrt(1+sqrt(2+...+sqrt(n)))` is rational

Then its square must be rational, i.e.:

`1+sqrt(2+...+sqrt(n))`

and hence so is:

`sqrt(2+sqrt(3+...+sqrt(n)))`

We can repeatedly square and subtract to find that the following must be rational:

`{ (sqrt(n-1+sqrt(n))), (sqrt(n)) :}`

Hence `n=k^2` for some positive integer `k > 1` and:

`sqrt(n-1+sqrt(n)) = sqrt(k^2+k-1)`

Note that:

`k^2 < k^2+k-1 < k^2+2k+1 = (k+1)^2`

Hence `k^2+k-1` is not the square of an integer either and `sqrt(k^2+k-1)` is irrational, contradicting our assertion that `sqrt(n-1+sqrt(n))` is rational.

Answer 2

See below.

Explanation:

Assuming

`sqrt(1+sqrt(2+cdots+sqrt(n)))=p/q` with `p/q` non reductible we have

`sqrtn = ( cdots (((p/q)^2-1)^2-2)^2 cdots -(n-1)) = P/Q`

which is an absurd, because according with this result, any square root of a positive integer is rational.