### A GEOMETRIC PROOF THAT THE SQUARE ROOT OF 2 IS IRRATIONAL

There are a number of algebraic proofs that the square root of 2 is irrational.
None of these is likely to be what the Pythagoreans used in their original
proof. Their thought processes were very geometric, and their proof was
probably geometrically based. Also, they did not actually define "rational" as
we do today. They thought of two line segments as "commensurable" if both were
an integer multiple of some smaller length.
Thus a line segment of length 3/8 and a line segment of length 5/6 were
commensurable since they were both integer multiples of a line segment of length
1/24. They assumed all pairs of line segments were commensurable (which is the
same as assuming all numbers are rational). Their proof may have gone something
like this (note that all required constructions can be done with a straightedge
and compass as preferred by Greek mathematicians):
Construct a square ABCD as shown. Draw the diagonal AC. Now if AC and AB are
commensurable,
then there is a line segment of some length, say r, such that AC=m_{1}r
and AB=m_{2}r where m_{1} and m_{2} are integers. Now
draw a circle centered at A with radius AB. Note that CB and CD are tangent to
this circle [why?]. This circle intersects AC at some point; call it P. Now
obviously AP=AB, so PC=AC-AB= m_{1}r - m_{2}r= (m_{1} -
m_{2})r so PC is also an integer multiple of r and is therefore
commensurable with AB and AC. Construct the tangent to the circle at P. Note
that this tangent is perpendicular to AC. This tangent intersects BC at some
point; call it Q. Now Angle(PCQ)=45^{o}=Angle(PQC) so Triangle(PQC) is an isosceles triangle and
therefore PC=PQ. We can construct a square with vertices C, P, and Q; call the
fourth vertex R (see the figure above).
Now since QP and QB are tangents to a circle from an external point, they are
equal, so QB=QP=PC=(m_{1} - m_{2})r=m_{4}r where we have
set m_{4}= m_{1} - m_{2}. Then QC=BC-QB=m_{2}r
- m_{4}r=(m_{2} - m_{4})r=m_{3}r (with
m_{3}=m_{2} - m_{4}) so QC is also commensurable with AB
and AC. We now have a new square whose side and diagonal are commensurable with
AB and AC. We note that
m_{1}>m_{2}>m_{3}>m_{4}. There's no
reason we can't do the same thing on square PCRQ. We can draw a circle centered
at C with radius CP, call the intersection of this circle with CQ point X,
construct the tangent there, and construct square XYZQ as shown. Then the side
and diagonal of this square are commensurable with AB and AC as well. Then we
can continue the process indefinitely constructing smaller and smaller squares
with sides and diagonals commensurable with AB and AC. However, there are only
a finite number of integer multiples of r less than m_{1}r, so we have a
contradiction.