I have decided to study maths from scratch and am currently studying from Algebra, I.M. Gelfand.

The solution to this problem is already answered here.

Howsoever I was not able to understand the solution; neither from the aforementioned source, nor from the solution provided in the book itself.

The problem statement, along with the solution provided by the author is mentioned here.

Please help me to understand the proof of: -

Why any piece (except the two pieces in the end) contains exactly 1 mark (which may be red and green)?


1 Answer 1


Previously in the book, you saw that:

$$ \frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d} $$

Put the red and green marks in the first and third terms:

$$ \frac{a}{13} < \frac{a+c}{b+d} < \frac{c}{7} $$

Add their denominators and put the result into the second term $ b + d = 20 $:

$$ \frac{a}{13} < \frac{a+c}{20} < \frac{c}{7} $$

If the red marks are at distances $ \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, ... $, the green marks at distances $ \frac{1}{13}, \frac{2}{13}, \frac{3}{13}, ..., $ and the cuts at $ \frac{1}{20}, \frac{2}{20}, \frac{3}{20}, ... $

Using the first equation you can assert that all the $ \frac{a+c}{20} $ pieces are between a red mark and a green mark.

For example, if take a look at the second piece which goes from $ \frac{1}{20} $ to $ \frac{2}{20} $, you can say:

$$ \frac{1}{13} < \frac{2}{20} < \frac{1}{7} $$

That means your second cut is after the first green mark and before the first red mark. In fact, any cut except the first and last, will be between marks:

$$ \frac{k}{13} < \frac{k+l}{20} < \frac{l}{7} $$


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