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One of the key ideas in transcedental number theory is proving that a number is transcedental (i.e. not the root of any polynomial with integer coefficients) by showing a sequence of rational numbers that approximate it too well.

This was the case for liouville numbers- it relies on his theorem:

If $\alpha$ is an irrational algebraic number of degree $d$, then there exists a constant $C>0$ such that for all rational numbers $p/q$ (with $q>0$): $$\left |\alpha-\frac{p}{q} \right |>\frac{C}{q^d} \quad (i)$$

This theorem shows that algebraic numbers can't be approximated by rational numbers too well. Since liouville numbers are irrational, and by definition they have excellent approximations, they must be transcendental.

Now, I was wondering about the opposite case- by that I mean, proving that a number is rational by showing that rationals converge to it badly.

By Dirichlet's approximation theorem, if $\alpha$ is an irrational number then there are infinitely many fractions $p/q$ such that

$$\left |\alpha-\frac{p}{q} \right |<\frac{1}{q^2} \quad (ii)$$

This isn't the case for rational numbers, though.

In fact, assume that we have some number $\alpha$, and we prove that all rational numbers with an arbitarily large denominator $p/q$ do not satisfy $(ii)$. Then we've proved non-constructively that $\alpha$ is rational: that is, we haven't constructed integers $a,b$ such that $\alpha=a/b$, but we've shown they exist.

Has this technique has ever been applied in reality?

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  • $\begingroup$ In the statement of the Liouville result, should the exponent $n$ on the right really be the $d$ of the degree of $\alpha$? $\endgroup$ – coffeemath Jun 21 '16 at 18:20
  • $\begingroup$ You're right, it's now fixed. $\endgroup$ – Ohad Jun 21 '16 at 18:21

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