commensurable numbers

Two positive real numbers $a$ and $b$ are commensurable, iff there exists a positive real number $u$ such that

with some positive integers $m$ and $n$ .

If the positive numbers $a$ and $b$ are not commensurable, they are incommensurable.

Theorem. The positive numbers $a$ and $b$ are commensurable if and only if their ratio is a rational number $\displaystyle\frac{m}{n}$ ($m,\,n \in \mathbb{Z}$ ).

Proof. The equations (1) imply the proportion

Conversely, if (2) is valid with $m,\,n \in \mathbb{Z}$ , then we can write $$a = m\!\cdot\!\frac{b}{n}, \quad b = n\!\cdot\!\frac{b}{n},$$ which means that $a$ and $b$ are multiples of $\displaystyle\frac{b}{n}$ and thus commensurable. Q.E.D.

Example. The lengths of the side and the diagonal of square are always incommensurable.

Commensurability as relation

• The commensurability is an equivalence relation in the set $\mathbb{R}_+$ of the positive reals: the reflexivity and the symmetry are trivial; if $a\!:\!b = r$ and $b\!:\!c = s$ , then $a\!:\!c = (a\!:\!b)(b\!:\!c) = rs$ , whence one obtains the transitivity.
  
• The equivalence classes of the commensurability are of the form $$[\varrho] \,:=\, \{r\varrho\,\vdots\;\; r \in \mathbb{Q}_+\}.$$
• One of the equivalence classes is the set $[1] = \mathbb{Q}_+$ of the positive rationals, all others consist of positive irrational numbers.
• If one sets $[\varrho]\!\cdot\![\sigma] := [\varrho\sigma]$ , the equivalence classes form with respect to this binary operation an Abelian group.