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=ACAB= 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=BCQB=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.
Had we looked a little more carefully at QB in the above figure, we would have discovered that its length is equal to (2m_{2}m_{1}). This leads to a "TwoForOne" proof that the square root of 2 is irrational. The difference between the two proofs is a slight change in assumptions at the beginning which leads to a different ending, but the algebra in the middle is identical.
Proof 1 
Common Statements 
Proof 2 
Assume that the square root of 2 is rational. Then there exist positive integers a and b such that  
Assume that a/b is in reduced form. This is equivalent to saying that a/b is the representation of this rational number with the smallest positive denominator.  


Now a little algebra:
a^{2}=2b^{2} a^{2}ab=2b^{2}ab a(ab)=b(2ba) 

We now have a fraction equal to a/b with a smaller positive denominator than b.  We now have a fraction equal to a/b with a smaller positive denominator than b which is a contradiction.  
Let a'=2ba and b'=ab. We can use the same procedure to find a'' and b'' so that a/b=a'/b'=a''/b'' with b>b'>b''>0. There's no algebraic reason we can't continue this process indefinitely. However, there are only a finite number of positive integers less than b, so we have a contradiction. 
In the book The Golden Ratio by Mario Livio, the author suggests that due to their interest in the pentagram and pentagon, the Pythagoreans may have discovered the incommensurability of the side and diagonal of the pentagon before they discovered the incommensurability of the side and diagonal of a square. If this theory is true, then the irrationality of the golden mean (PHI) was discovered before the irrationality of the square root of 2. See Chapter 2 and Appendix 2 of the book for details, and see this page for a similar, but somewhat simpler proof.
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Date last modified: 12/23/02
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