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 )