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This isn't really a GCD question, because GCD is only defined for integers. I'm interested in the the existence of a common divisor of any two non-zero real numbers. In other words can you prove or disprove the following:

Given $x,y \neq 0\in \mathbb{R}, \exists \space g \space s.t. \space x/g \in \mathbb{Z}$ and $y/g \in \mathbb{Z}$.

(I hope my math is understandable, haven't done this in awhile). It's clearly possible for many numbers, including irrational ones (e.g. for multiples of $\pi$, $g = \pi$). Is it possible for all real numbers?

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    In a way, yours is close to the Classical version. In Euclid, two line segments $AB$ and $CD$ are called *commensurable* if there is a line segment $PQ$ that "measures" both of them, meaning that $AB$ is obtainable by applying $PQ$ an integer number of times, and the same for $CD$. What we call the irrationality of $\sqrt{2}$ is for Euclid the statement that the side and diagonal of a square are incommensurable.2012-03-20

2 Answers 2

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The following conditions are equivalent for nonzero reals $x,y$

  1. There is a real $g$ such that $x/g$ and $y/g$ are integers
  2. The quotient $x/y$ is rational

Proof:

$1 \implies 2$: Since quotient of integers is rational, your condition implies that

$(x/g) / (y/g) \in \mathbb{Q}$

after clearing $g$ in denominators

$x/y \in \mathbb{Q}$.

$2 \implies 1$:

If $x/y$ is rational: $x/y=p/q$ then define $g = y/q$ (or $g = x/p$), then $x/g = xq/y = p$ and $y/g = q$ are integers. QED

So any pair of reals with irrational quotient is a counterexample, for example $x=1$ and $y=\sqrt{2}$.

Real numbers $x,y$ with rational quotient are known as commensurable. This is how irrationality was formulated in the ancient times. It has been said that diagonal of a square is not commensurable with its side.

The Euclidean algorithm for finding GCD was originally formulated on segments (reals) - it found a common measure ($g$) given segments of length $x$ and $y$.

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    Great answer, just a small addition: There is an "algorithm" for finding the [GCD of two commensurate real numbers](https://en.wikipedia.org/wiki/Greatest_common_divisor#Other_methods), $gcd(a,b) = a ~ f(b / a)$, where $f$ is [Thomae's function](https://en.wikipedia.org/wiki/Thomae%27s_function).2014-08-04
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The statement is not true. Consider $x=1,y=\pi$. If $y/g=n\in\mathbb Z$, then $g=\pi/n$ so $x/g=n/\pi\notin \mathbb Z$.