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