Fractions are Hard!

4.5 Coloring Fractions

Here are all the thirds and fourths that lie between \(0\) and \(1\) colored red.

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If we put them in order of size they alternate between fourths and thirds.

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Now \(3+4=7\). Here are all the sevenths between \(0\) and \(1\) colored blue.

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Now let’s arrange all eleven fractions in order of size.

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The red numbers slip neatly in place between the blue. Pretty!

 

Let’s do this again.

 

Here are all the fifths and sixths colored red

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and all the elevenths blue

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Arranging them in order also gives an alternating pattern of colors, with alternating denominators within the reds

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Whoa! Now this is getting freaky!

 

YOUR TURN!

Write out all the fourths and fifths between \(0\) and \(1\) in red, and all the ninths in blue, and arrange them in order of size. Check that the same alternating patterns appear.

 

Try some more examples like this for yourself: Write a list of \(N\)ths and \(\left(N+1\right)\)ths of your choice in red and a list of \(\left(2N+1\right)\)ths in blue. When arranged if order, do the reds again slip neatly between the blues (with the reds alternating in denominators)?

 

EXPERIMENT:

Does this phenomenon occur only for \(N\)ths and \(\left(N+1\right)\)ths?

How about thirds and fifths in red and eights in blue? (Try them!)

How about thirds and sevenths in red and tenths in blue? (Try them too!)

Do \(\dfrac{1}{a}\), \(\dfrac{2}{a}\), …, \(\dfrac{a-1}{a}\) and \(\dfrac{1}{b}\), \(\dfrac{2}{b}\), …, \(\dfrac{b-1}{b}\) in red and \(\dfrac{1}{a+b}\), \(\dfrac{2}{a+b}\), …, \(\dfrac{a+b-1}{a+b}\) in blue always yield the same alternating patterns of colors and of denominators of reds when arranged in order of size?

 

For answers, see my September 2013 COOL MATH ESSAY.

 

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