μαθηματικοι
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Why the name "μαθηματικοι" (or "mathematikoi," if the other was Greek to you)? Well, let's see....

Pythagoras — he of the eponymous theorem — was the leader of a group that we might say (with modern eyes) looked suspiciously like a cult. It was based in Kroton, modern day Crotone in Calabria (the ball of the foot in the boot that is the Italian peninsula). The Pythagoreans consisted of two parts, the akousmatikoi (ακουσματικοι), who were the more ritualized and religious practices and the mathematikoi, who concentrated on the more scientific and mathematical aspects of their leader's thought. Note that in ancient Greet akousmatikoi meant listeners while mathematikoi meant learners.

One particular mathematikos is a special hero of mine: Hippasus of Metapontum. Apparently he leaked to the general public the mathematikoi's secret that $\sqrt{2}$ is irrational, meaning that there do not exist any whole numbers $p$ and $q$ with the property that $\sqrt{2}=p/q$. This was something of a threat to the Pythagoreans, because of the whole everything is number obsession they had, as well as the whole beauty [particularly in music — see the whole business with the monochord here] is fractions with small numerators and denominators thing of their leader's.

For this public revelation, Hippasus was tracked down and killed (or so the story goes, which some believe is quite apocryphal). Those were the days that mathematical facts made for high emotions!

In any case, I like the idea of sharing mathematics with the public, no matter the personal cost. I like the idea of the more scientific and rational branch of the Pythagoreans, not the ritualistic and religious part. And I like the idea that the word "mathematikoi", which looks so much like it should be the origin of the modern word "mathematicians," meant "learners."

Hence the name.

By the way, I'm sure your curiousity has been peaked, so here is a proof of the irrationality of $\sqrt{2}$:

Assume $\sqrt{2}=\frac{p}{q}$, where $p$ and $q$ are whole numbers with no common factors (so the fraction is written "in lowest terms").

Squaring both sides, $2=\frac{p^2}{q^2}$.

Multiplying both sides by $q^2$ gives $2q^2=p^2$.

Therefore $2\mid p^2$, which means $2\mid p$.

But if $2\mid p$ then $4\mid p^2$.

Since $4\mid p^2$ and $p^2=2q^2$, $2\mid q^2$.

Since $2\mid q^2$, $2\mid q$.

So $p$ and $q$ have a common factor: $\Rightarrow\Leftarrow$!

Therefore, $\nexists$ such $p$ and $q$ and $\sqrt{2}\notin\QQ$.

Notes:

 
 
 

Jonathan Poritz (jonathan@poritz.net) Last modified:
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