by Jacqueline Hernandez and Herman Yu CSU Long Beach, Fall of 2014 History of Mathematics (MATH 310)

Comeasurable: Two line segments are comeasurable if there exists another line segment that fits into both perfectly AB uu uuu A B = 2u 3u = 2 3

We will be using a proof by Contradiction:

We will be using a proof by Contradiction Suppose, by way of contradiction, that √2 is rational. Then √2 can be written as follows: √2 = 1 = = A B Where A and B are integers and the fraction is irreducible.

In particular, that means the side and diagonal of this square is comeasurable. √ So there exists a line segment u that measures both the side and diagonal. u

A B CD Let ABCD be the vertices of our square. Consider the diagonal AC. Then AC is of length √2 and the side AB is of length 1.

Take the point E on the diagonal AC such that AE = AB. That is, take a compass over AB and rotate the compass until it hits the diagonal AC. A B CD

A B CD

A B CD

A B CD

A B CD

A B CD E

A B CD E Now, connect the points E and B with a line to form a triangle. Since AE = AB, we have an isosceles triangle!

A B CD E F Draw the perpendicular at E until it meets the side BC at F.

A B CD E F Now, angle ACB is 45 degrees, and angle CEF is 90 degrees, so it must be the case that angle EFC is 45 degrees.

A B CD E F G Hence CE and EF are two sides of a square with diagonal CF.

A B CD E F G Recall that triangle ABE was an isosceles triangle

A B CD E F G with AE = AB

A B CD E F G So then: Angle AEB = Angle EBA

A B CD E F G So then: angle FEB = angle EBF

A B CD E F G And so: EF = FB

A B CD E F G Now, this is where things get really WILD!

A B CD E F G Recall that we started off by assuming √2 is rational. This implied that the side and diagonal of our original square were comeasurable. u

A B CD E F G So then AB and AC are both measurable by some unit u. Also, recall that AE = AB. u

A B CD E F G u So their difference EC is also measurable by u.

A B CD E F G Now EC and EF are both sides of the same square. So if EC is measurable, then so is EF. u

A B CD E F G Now EC and EF are both sides of the same square. So if EC is measurable by u, then so is EF u But EF and BF are two congruent sides of an isosceles triangle, so if EF is measurable by u, then so is BF.

A B CD E F G Now, AB and BC are two sides of the same square. Since AB is measurable by u, so is BC. u

A B CD E F G u Since, BC and BF are both measurable by u, their difference FC is also measurable by u. (Note that all highlighted line segments are measurable by u.)

A B CD E F G But now CE and CF form the side and diagonal of a square. Since both are measurable by u, we may repeat the process of constructing an even smaller square, which is still measurable by u. u

A B CD E F G H u

A B CD E F G H u

A B CD E F G H u

A B CD E F G H All of these squares have sides that are measurable by u. u

A B CD E F G H But we can keep repeating this process until we get a square with sides smaller than u. u

A B CD E F G H u

u So then u must also measure a square, with sides smaller than u. But how can something bigger fit perfectly into something smaller?

u So then u must also measure a square, with sides smaller than u. But how can something bigger fit perfectly into something smaller? The answer: it cannot! That specific square is not measurable by u. But by all our previous work (which was a lot!) we just showed that it is measurable by u...

u So then u must also measure a square, with sides smaller than u. But how can something bigger fit perfectly into something smaller? CONTRADICTION!!! The answer: it cannot! That specific square is not measurable by u. But by all our previous work (which was a lot!) we just showed that it is measurable by u...

u So then u must also measure a square, with sides smaller than u. But how can something bigger fit perfectly into something smaller? The answer: it cannot! That specific square is not measurable by u. But by all our previous work (which was a lot!) we just showed that it is measurable by u... This means u cannot actually exist, or else the universe would tear apart!

So the side and diagonal of our square is not Comeasurable. Therefore...

And so:

THE SQUARE ROOT OF 2 IS IRRATIONAL.

Q.E.D (The End )