A Disquisition on the Square Root of Three - 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.