MHB Are At Least Two Numbers Equal in This Sum of 100 Terms?

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The discussion centers on proving that in a set of 100 terms selected from the numbers 1 to 100, where the sum of their square roots equals 12.5, at least two numbers must be equal. Participants explore the implications of the equation S = 12.5 and seek examples that satisfy this condition. The challenge lies in demonstrating that the unique values of the selected numbers cannot achieve the required sum without repetition. Various mathematical approaches and potential examples are suggested to illustrate the point. Ultimately, the conclusion emphasizes the necessity of having at least two equal numbers in the set to meet the specified sum.
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$a_1,a_2,...,a_{100}\in \begin{Bmatrix}
1,2,3,-----,100
\end{Bmatrix}$

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

Prove that at least two of the numbers are equal
 
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If S is to be rational, the ai expressions have to be perfect squares; so there are only 10 values available for 100 terms.
 
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Saknussemm said:
If S is to be rational, the ai 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
 
Albert said:
Can you give some examples to make S=12.5 ?
Or if you can prove it the best

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.
 
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