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Finally, two days ago, I succeeded - not on account of my hard efforts, but by the grace of the Lord. Like a sudden flash of lightning, the riddle was solved. I am unable to say what was the conducting thread that connected what I previously knew with what made my success possible.

In H. Eves Mathematical Circles Squared, Boston: Prindle, Weber and Schmidt, 1972.

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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.

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Wisner, Robert J., "A Disquisition on the Square Root of Three," Loci (June 2010), DOI: 10.4169/loci003514