Emily’s Fractions

Posted By on Mar 8, 2018 | 0 comments

Emily invents the threeven numbers as the subset of the odd numbers that all have three as a factor.

She knows, intuitively by looking at her counting numbers, that a third of all positive integers are divisible by three with no remainder. She can also see in her numbers that half of these numbers have already been counted in the even numbers.

So, to calculate the portion of threeven numbers:

    \[ \frac{1}{2} * \frac{1}{3} = \frac{1}{6} \]

And the portion of remaining oddish numbers (threeven oddish) is:

    \[ 1 - \frac{1}{2} - \frac{1}{3} = \frac{2}{6} \]

When she creates the fiven numbers, Emily can see that they will be a fifth of the threeven oddish numbers:

    \[ \frac{2}{6} * \frac{1}{5} = \frac{2}{30} \]

This then leaves the fiven oddish numbers:

    \[ 1 - \frac{1}{2} - \frac{1}{6} - \frac{2}{30} = \frac{8}{30} \]

Emily then makes the seven’n numbers and they will be a seventh of the fiven oddish numbers:

    \[ \frac{8}{30} * \frac{1}{7} = \frac{8}{210} \]

Leaving the seven’n oddish numbers:

    \[ 1 - \frac{1}{2} - \frac{1}{6} - \frac{2}{30} - \frac{8}{210} = \frac{48}{210} \]

The portion of the eleven’n numbers then follow:

    \[ \frac{48}{30} * \frac{1}{11} = \frac{48}{2310} \]

And the remaining oddish numbers (eleven’n oddish):

    \[ 1 - \frac{1}{2} - \frac{1}{6} - \frac{2}{30} - \frac{8}{210} - \frac{48}{2310}= \frac{580}{2310} \]

Emily calculated her fractions by instinct and the assertion that they represent portions is correct for a suitably large, finite set. When we consider all the positive integers (an infinite set) it is more correct to think of them as probabilities.

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