Early Timekeeping

Why are there sixty seconds in a minute, sixty minutes in an hour, yet twenty-four hours in a day? The answer is because modern timekeeping derives from the base-60 number system.

It is believed that Sumerians of Mesopotamia used their phalanges to count. They counted increments of 12 with one hand's four fingers, each of which has three bones, and tracked them with the other hand's five fingers: 12, 24, 36, 48, 60.

Early civilization calendars were often lunisolar, based on the phases of the moon—roughly aware of the sun's yearly 365-day orbit. Though they were somewhat imprecise, many resembled the 12-month calendar we know today. Ancient calendars, however, would often have extra days or months periodically added for alignment purposes.

The Sumerians had no tradition for referring to the length of time we call a "week," nor did they identify months. They simply observed months and years. Later, the Babylonians would put forth the notion of the "week," as well as move to use a solar, rather than lunisolar, calendar.

A history of calendars: https://en.wikipedia.org/wiki/History_of_calendars

James Garfield's Pythagorean Proof

Today I learned James Garfield, who once worked as a lawyer, Civil War General, and served as the 20th President of the United States, was math savvy and published a novel Pythagorean theorem proof.^[1]

\[ \text{Area}_{\text{trapezoid } ACED} = \frac{1}{2} \cdot (AC + DE) \cdot CE = \frac{1}{2} \cdot (a + b) \cdot (a + b) = \frac{(a + b)^2}{2} \] \[ \begin{aligned} \text{Area}_{\text{trapezoid } ACED} &= \text{Area}_{\Delta ACB} + \text{Area}_{\Delta ABD} + \text{Area}_{\Delta BDE} \\ &= \frac{1}{2}(a \times b) + \frac{1}{2}(c \times c) + \frac{1}{2}(a \times b) \end{aligned} \] \[ (a + b) \times \frac{1}{2}(a + b) = \frac{1}{2}(a \times b) + \frac{1}{2}(c \times c) + \frac{1}{2}(a \times b) \] \[ a^{2} + b^{2} = c^{2} \]

Small Pieces

We can take this in smaller pieces. First, we can find the area of the right-angled trapezoid with the following equation:

\[ \text{Area}_{\text{trapezoid}} = \frac{1}{2} \cdot (a + b) \cdot (a + b) = \frac{(a + b)^2}{2} \]

We can find the area of each of the two outer triangles with the following:

\[ \text{Area}_{\text{triangle}} = \frac{ab}{2} \]

And the area of the inner triangle with:

\[ \text{Area}_{\text{inner triangle}} = \frac{c^2}{2} \]

Proof

Reducing, we can go to the end, beginning with our substituted and now simplified area equation demonstrated above:

\[ \frac{(a + b)^2}{2} = 2 \cdot \frac{ab}{2} + \frac{c^2}{2} \]

Then we expand \( (a + b)^2 \) on the left hand side. And our equation on the right can also be simplified since we're both multiplying and dividing \( ab \) by 2:

\[ \frac{a^2 + 2ab + b^2}{2} = ab + \frac{c^2}{2} \]

Multiply both sides by 2 to eliminate denominators:

\[ a^2 + 2ab + b^2 = 2ab + c^2 \]

Lastly, subtract \( 2ab \) from both sides:

\[ a^2 + b^2 = c^2 \]

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Footnotes

1. Mathematical Treasure: James A. Garfield’s Proof of the Pythagorean Theorem ↩︎