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I'm self studying mathematics for leisure and using Gelfand's Algebra. He poses problem 43 and proves it right afterwards (see image below).

I've been struggling to understand the "proof". I understand some parts, like: the ends won't have marks, and that the same color marks cannot be on the same fractional $1/20$ piece.

However, the main argument eludes me. Please help, thanks!

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The crux of the argument is that $(k+l)/20$ is between $k/7$ and $l/13$. That is $$\frac k7<\frac{k+l}{20}<\frac l{13}$$ whenever $k/7<l/13$ and $$\frac k7>\frac{k+l}{20}>\frac l{13}$$ whenever $k/7>l/13$. This boils down to proving that $$\frac k7-\frac{k+l}{20}\qquad\text{and}\qquad \frac{k+l}{20}-\frac l{13}$$ have the same sign. I would suggest simplifying these.

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  • $\begingroup$ Thanks for your help, perhaps I'm not envisioning the problem correctly. I can't figure out what "k+l/20" means. This is the way I am seeing the problem: imgur.com/a/gxwyrm1 $\endgroup$ – Asif Kazmi Dec 29 '18 at 15:59
  • $\begingroup$ @AsifKazmi $(k+l)/20$ is just a fraction with 20 as the denominator. The stick has markings at the points $k/7$ for $k=1,2,3,4,5,6$, and at $l/13$ for $l=1,2,3,4,5,6,7,8,9,10,11,12$. The stick is cut at $1/20,\,2/20,\,3/20,\,\dots\,,\,19/20$. Here, you are showing that (say) $1/7 < 3/20<2/13$. I.e. $(k+l)/20$ is between $k/7$ and $l/13$ for all the possible values of $k,l$. $\endgroup$ – John Doe Dec 29 '18 at 16:04

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