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

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

Comeasurable: Two line segments are comeasurable if there exists another line segment that fits into both perfectly

A B

u u u u u++ +

A

B=

2u3u

=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 =√21

= =AB

Where A and B are integers and the fraction is irreducible.

In particular, that means the side and diagonal of this square is comeasurable.

√2

1

1

1

1 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

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

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

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

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

E

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

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

with AE = AB

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

So then AB and AC are both measurable by some unit u. Also, recall that AE = AB.

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

Now, AB and BC are two sides of the same square. Since AB is measurable by u, so is BC.

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

Hu

A B

CD

E

F

G

Hu

A B

CD

E

F

G

Hu

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

But we can keep repeating this process until we get a square with sides smaller than u.

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:

And so:

THE SQUARE ROOT OF 2

IS IRRATIONAL.

Q.E.D

(The End )