Approximation by Rational Numbers

That the center of attention here is irrational can be seen by being reminded of this argument: If \(\sqrt{3}\) were rational, it could be written as the reduced fraction \[\sqrt{3}=\frac{a}{b}\] for integers \(a\) and \(b.\) Then by squaring both sides, the result is the equation \[a^{2}=3b^{2}.\] Now if \(a\) and \(b\) are each written as their unique product of primes, then the prime \(3\) occurs on the left side of this equation an even number of times, while on the right side, \(3\) occurs an odd number of times. Such a situation violates the Unique Factorization Theorem, so the equation just above is impossible, whence \(\sqrt{3}\) must be irrational.

Note, however, that, to six decimal places, \[\sqrt{3}\approx 1.732051=\frac{1732051}{1000000}.\] This is a reduced fraction since \(1732051\) is a prime. We will now look at three other ways to arrive at fractional approximations that are "equivalent" in that they also have six-place decimal accuracy: first, by use of the classical Greek ladder for \(\sqrt{3};\) second, by examining the convergents of a continued fraction for \(\sqrt{3};\) and, third, by using iterates of Newton's Method.