Sum of 100 terms (continued)

[Albert1]

$a_1,a_2,...,a_{100}\in \begin{Bmatrix}
1,2,3,-----,100
\end{Bmatrix}$

$S=\dfrac{1}{\sqrt{a_1}}+\dfrac{1}{\sqrt{a_2}}+\cdots+\dfrac{1}{\sqrt{a_{100}}}=12.5$.

Prove that at least two of the numbers are equal

 

[Saknussemm]

Spoiler

If S is to be rational, the a_i expressions have to be perfect squares; so there are only 10 values available for 100 terms.

 

[Albert1]

Spoiler

If S is to be rational, the a_i expressions have to be perfect squares; so there are only 10 values available for 100 terms.

Can you give some examples to make S=12.5 ?
Or if you can prove it the best

 

[Saknussemm]

Can you give some examples to make S=12.5 ?
Or if you can prove it the best

Spoiler

I meant my remark that each and every one $a_i$ expression in $\sum_{i=0}^{100} \frac{1}{\sqrt{a_i}}$ has to be a perfect square as a proof. Because, if a single $a_i$ equals a number like 2, the term $\frac{1}{\sqrt{a_i}}$ is irrational, which means that the sum S in turn will be irrational, and so will never equal a number like $\frac{25}{2}$. This means you will have to build your sum exclusively from terms like 1/10, 1/9, 1/8, etc.

The proof that there is no pair of integers $(a, b)$ such that $\sqrt{2} = \frac{a}{b}$ is due to Euclid. It consists in positing, without loss of generality, that $\frac{a}{b}$ is an irreducible fraction. Rearranging the terms and squaring, you find that both of $a^2$ and $b^2$ are even, meaning that both $a$ and $b$ are even, which contradicts the hypothesis that we have an irreducible fraction. It doesn't seem difficult to extend the proof to the root of any number that is not a perfect square, or to use a similar proof for the sum of 2 irrational numbers.

Now, if instead of requiring a rational result you had required an interval of real numbers, this proof would not be valid.
Then we could use the same kind of device you used to resolve a similar task, namely bound the sum $\sum_{n=1}^{100} \frac{1}{\sqrt{n}}$ by an integral.
Because $y = \frac{1}{\sqrt{x}}$ is monotonically decreasing and positive, we have

$\displaystyle \sum_{n=1}^{100} \frac{1}{\sqrt{n}} \; > \; \int_1^{101} \frac{1}{\sqrt{x}} \,dx \; > \; \int_1^{100} \frac{1}{\sqrt{x}} \,dx \; = \left. 2 \sqrt{x} \; \right\rvert_{1}^{100} \; = \; 18$

To get below 18, you will have to repeat some of the lower numbers.