<?xml version="1.0" encoding="UTF-8" standalone="yes"?><oembed><version><![CDATA[1.0]]></version><provider_name><![CDATA[evolutionistx]]></provider_name><provider_url><![CDATA[https://evolutionistx.wordpress.com]]></provider_url><author_name><![CDATA[evolutiontheorist]]></author_name><author_url><![CDATA[https://evolutionistx.wordpress.com/author/evolutiontheorist/]]></author_url><title><![CDATA[The Big 6 Civilizations (pt 2:&nbsp;Egypt)]]></title><type><![CDATA[link]]></type><html><![CDATA[<p><a href="https://evolutionistx.files.wordpress.com/2016/06/1024px-egypt-giza-sphinx-02.jpg"><img class="alignnone size-medium wp-image-5191" src="https://evolutionistx.files.wordpress.com/2016/06/1024px-egypt-giza-sphinx-02.jpg?w=300&#038;h=225" alt="1024px-Egypt.Giza.Sphinx.02" width="300" height="225" /></a></p>
<p><strong>2. Egypt</strong></p>
<p>I know I don&#8217;t have to tell you about Egyptian civilization, but did you know the <a href="https://en.wikipedia.org/wiki/Ancient_Egyptian_mathematics">Egyptians had maths</a>?</p>
<p>Problem number 56 from the <a href="https://en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus">Rhind Mathematics Papyrus</a> (dated to around 1650 BC):</p>
<blockquote>
<div data-shortcode="caption" id="attachment_5195" style="width: 160px" class="wp-caption alignright"><a href="https://evolutionistx.files.wordpress.com/2016/06/300px-geometrical_diagram_showing_egyptian_seked_system.jpg"><img class="size-thumbnail wp-image-5195" src="https://evolutionistx.files.wordpress.com/2016/06/300px-geometrical_diagram_showing_egyptian_seked_system.jpg?w=150&#038;h=115" alt="Egyptian seked" width="150" height="115" /></a><p class="wp-caption-text"><a href="https://en.wikipedia.org/wiki/Seked">Seked of the Great Pyramid</a></p></div>
<p>If you construct a pyramid with base side 12 [cubits] and with a seked of 5 palms 1 finger; what is its altitude?<sup id="cite_ref-Clagett_1-7" class="reference"><a href="https://en.wikipedia.org/wiki/Egyptian_geometry#cite_note-Clagett-1">[1]</a></sup></p></blockquote>
<p>Most <a href="https://en.wikipedia.org/wiki/Egyptian_geometry">Egyptian geometry </a>questions appear to deal with more mundane matters, like the dimensions of rectangular fields and round granaries, rather than pyramids. (The Egyptians had not yet worked out an exact formula for the area of a circle, but used octagons to approximate it.)</p>
<p>&nbsp;</p>
<p><a href="https://evolutionistx.files.wordpress.com/2016/06/picture-4.png"><img class="alignleft size-medium wp-image-5192" src="https://evolutionistx.files.wordpress.com/2016/06/picture-4.png?w=300&#038;h=92" alt="Picture 4" width="300" height="92" /></a>A &#8220;pefsu&#8221; problem involves a measure of the strength of the <a title="Beer" href="https://en.wikipedia.org/wiki/Beer">beer</a> made from a <a title="Hekat (unit)" href="https://en.wikipedia.org/wiki/Hekat_%28unit%29">heqat</a> of grain, called a pefsu.</p>
<p><span class="mwe-math-mathml-inline mwe-math-mathml-a11y">pefsu = (the number of  loaves of bread [or jugs of beer]) / (number of heqats of grain used to make them.) </span></p>
<p>For example, problem number 8 from the <a href="https://en.wikipedia.org/wiki/Moscow_Mathematical_Papyrus">Moscow Mathematical Papyrus</a> (most likely written between 1803 BC and 1649 BC, but based on an earlier manuscript thought to have been written around 1850 BC):</p>
<dl>
<dd>Example of calculating 100 loaves of bread of pefsu 20:</dd>
<dd>If someone says to you: “You have 100 loaves of bread of pefsu 20 to be exchanged for beer of pefsu 4, like 1/2 1/4 malt-date beer,&#8221;</dd>
<dd>First calculate the grain required for the 100 loaves of the bread of pefsu 20. The result is 5 heqat. Then reckon what you need for a des-jug of beer like the beer called 1/2 1/4 malt-date beer. The result is 1/2 of the heqat measure needed for des-jug of beer made from Upper-Egyptian grain.</dd>
<dd>Calculate 1/2 of 5 heqat, the result will be 2<span class="frac nowrap"><sup>1</sup>⁄<sub>2. </sub></span>Take this 2<span class="frac nowrap"><sup>1</sup>⁄<sub>2</sub></span> four times.</dd>
<dd>The result is 10. Then you say to him:</dd>
<dd>Behold! The beer quantity is found to be correct.<sup id="cite_ref-Clagett_1-4" class="reference"><a href="https://en.wikipedia.org/wiki/Egyptian_algebra#cite_note-Clagett-1">[1]</a></sup></dd>
</dl>
<p>&#8220;Behold! The beer quantity is found to be correct,&#8221; is one of the most amusing answers to a math problem I&#8217;ve seen.</p>
<p><a href="https://evolutionistx.files.wordpress.com/2016/06/picture-5.png"><img class="alignleft size-full wp-image-5196" src="https://evolutionistx.files.wordpress.com/2016/06/picture-5.png?w=294&#038;h=137" alt="Picture 5" width="294" height="137" /></a>The Egyptians also used <a href="https://en.wikipedia.org/wiki/Egyptian_Mathematical_Leather_Roll">fractions</a> and solved <a href="https://en.wikipedia.org/wiki/Egyptian_algebra">algebraic equations</a> that we would write as linear equations, eg, 3/2 * x + 4 = 10.</p>
<p>But their <a href="https://en.wikipedia.org/wiki/Ancient_Egyptian_multiplication">multiplication and division</a> was really weird, probably as a side effect of not yet having invented a place value system.</p>
<p>A. Let&#8217;s suppose you wished to multiply 9 * 19.</p>
<p>B. First we want to turn 9 into powers of 2.</p>
<p>C. The powers of 2 = 1, 2, 4, 8, 16, 32, 64, etc.</p>
<p>D. The closest of these to 9 is 8, and 9-8=1, so we turn 9 into 8 and 1.</p>
<p>E. Now we&#8217;re going to make a table using 1, 8, and 19 (from line A), like so:</p>
<p>1        19<br />
2        ?<br />
4         ?<br />
8         ?</p>
<p>F. We fill in our table by doubling 19 each time:</p>
<p><strong>1        19</strong><br />
2        38<br />
4         76<br />
<strong>8         152</strong></p>
<p>E. Since we turned 9 into 1 and 8 (step D), we add together the numbers in our table that correspond to 1 and 8: 19 + 152 = 171.</p>
<p>Or to put it more simply, using more familiar methods:</p>
<p>9 * 19 = (1 +8) * 19 = (19 * 1) +(19 * 8) = (19 * 1) + (19 * 2 * 2 * 2) = 171</p>
<div data-shortcode="caption" id="attachment_5193" style="width: 310px" class="wp-caption alignnone"><a href="https://evolutionistx.files.wordpress.com/2016/06/800px-neferetiabet.jpg"><img class="size-medium wp-image-5193" src="https://evolutionistx.files.wordpress.com/2016/06/800px-neferetiabet.jpg?w=300&#038;h=216" alt="Slab stela of Old Kingdom princess Neferetiabet (dated 2590–2565 BC), with number hieroglyphs" width="300" height="216" /></a><p class="wp-caption-text">Slab stela of Old Kingdom princess <a href="https://en.wikipedia.org/wiki/Ancient_Egyptian_mathematics#/media/File:Neferetiabet.jpg">Neferetiabet</a> (dated 2590–2565 BC), with number hieroglyphs</p></div>
<p>Now let&#8217;s do 247 * 250:</p>
<p>The closest power of 2 (without going over) is 128. 247 -128 = 119. 119 &#8211; 64 = 55. 55 &#8211; 32 = 23. 23 &#8211; 16 = 7. 7 &#8211; 4 = 3. 3 &#8211; 2 = 1. Whew! So we&#8217;re going to need 128, 64, 32, 16, 4, 2, and 1, and 250.</p>
<p>Let&#8217;s arrange our table, with the important numbers in bold (in this case, it&#8217;s :</p>
<p><strong>1       250</strong><br />
<strong>2        ?</strong><br />
<strong>4         ?</strong><br />
8         ?<br />
<strong>16       ?</strong><br />
<strong>32       ?</strong><br />
<strong>64       ?</strong><br />
<strong>128      ?</strong></p>
<p>So, doubling 250 each time, we get:</p>
<p><strong>1</strong>  <strong>     250</strong><br />
<strong>2       500</strong><br />
<strong>4       1000</strong><br />
8        2000<br />
<strong>16     4000</strong><br />
<strong>32     8000</strong><br />
<strong>64     16,000</strong><br />
<strong>128    32,000</strong></p>
<p>Adding together the bold numbers in the second column gets us <strong>61,750</strong>&#8211;and I probably don&#8217;t need to tell you that plugging 247 * 250 into your calculator (or doing it longhand) also gives you 61,750.</p>
<p>The advantage of this system is that the Egyptians only had to memorize their 2s table. The disadvantages are pretty obvious.</p>
<div data-shortcode="caption" id="attachment_5194" style="width: 217px" class="wp-caption alignleft"><a href="https://evolutionistx.files.wordpress.com/2016/06/papyrus_berlin_6619.jpg"><img class="size-medium wp-image-5194" src="https://evolutionistx.files.wordpress.com/2016/06/papyrus_berlin_6619.jpg?w=207&#038;h=300" alt="Berlin Papyrus" width="207" height="300" /></a><p class="wp-caption-text"><a href="https://en.wikipedia.org/wiki/Berlin_Papyrus_6619">Berlin Papyrus</a></p></div>
<p>See also the <a href="https://en.wikipedia.org/wiki/Lahun_Mathematical_Papyri">Lahun Mathematical Papyri</a>, the <a href="https://en.wikipedia.org/wiki/Egyptian_Mathematical_Leather_Roll">Egyptian Mathematical Leather Roll</a>, the <a href="https://en.wikipedia.org/wiki/Akhmim_wooden_tablets">Akhmim wooden tablets</a>, the <a href="https://en.wikipedia.org/wiki/Reisner_Papyrus">Reisner Papyrus</a>, and finally the <a href="https://en.wikipedia.org/wiki/Papyrus_Anastasi_I">Papyrus Anastasi I</a>, which is believed to be a fictional, satirical tale for teaching scribes&#8211;basically, a funny textbook, and the <a href="https://en.wikipedia.org/wiki/Berlin_Papyrus_6619">Berlin Papyrus 6619</a>:</p>
<blockquote><p>The Berlin Papyrus contains two problems, the first stated as &#8220;the area of a square of 100 is equal to that of two smaller squares. The side of one is ½ + ¼ the side of the other.&#8221;<sup id="cite_ref-6" class="reference"><a href="https://en.wikipedia.org/wiki/Berlin_Papyrus_6619#cite_note-6">[6]</a></sup> The interest in the question may suggest some knowledge of the <a title="Pythagorean theorem" href="https://en.wikipedia.org/wiki/Pythagorean_theorem">Pythagorean theorem</a>, though the papyrus only shows a straightforward solution to a single second degree equation in one unknown. In modern terms, the simultaneous equations <span class="nowrap"><i>x</i><sup>2</sup> + <i>y</i><sup>2</sup> = 100</span> and <span class="nowrap"><i>x</i> = (3/4)<i>y</i></span> reduce to the single equation in <i>y</i>: <span class="nowrap">((3/4)<i>y</i>)<sup>2</sup> + <i>y</i><sup>2</sup> = 100</span>, giving the solution <i>y</i> = 8 and <i>x</i> = 6.</p></blockquote>
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