SOLUTION: Hi, here is a question that I don't know how to approach: The following fractions are written in a row: 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11. Prove that one can

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Hi, here is a question that I don't know how to approach: The following fractions are written in a row: 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11. Prove that one can      Log On

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Question 1046248: Hi, here is a question that I don't know how to approach:
The following fractions are written in a row:
1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11.
Prove that one cannot insert the signs + and - between these numbers so that the result would be 0.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Hi, here is a question that I don't know how to approach:
The following fractions are written in a row:
1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11.
Prove that one cannot insert the signs + and - between these numbers so that the result would be 0.
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Let's assume for a minute that there is a combination of signs "+" and "-" such that the result is 0.

Then 1%2F11 is the linear combination of the remaining fractions.

Write this combination with the common denominator.
As a common denominator you can choose 10!

So you will get an equality A%2F10%21 = 1%2F11 with some integer A.

Cross multiply it. You will have  A*11 = 10! with some integer A.

But such equality is not possible, since the left side is divisible by 11, while the right side is not.

Contradiction.

This contradiction disproves the original assumption.

Hence, there is no such a combination of signs . . . 

The statement is proved.

Solved.


Actually, even more strong statement is true:

      There is no linear combination of the given fractions with integer coefficients which is equal to zero.

Surely, the formulation assumes that the coefficient at 1%2F11 is not zero and is not a multiple of 11.
The proof is the same.