Let the given ratios in least numbers be: :$a : b$, $c : d$, $e : f$ From {{EuclidPropLink|book = VII|prop = 34|title = Existence of Lowest Common Multiple}}, we can find: :$g = \lcm set {b, c}$ Let: :$g = r b$ :$g = s c$ Then let: :$h = r a$ :$k = s d$ Now either $e \divides k$ or $e \nmid k$. First, suppose $e \divides k$. Let $k = t e$. Then let $l = t f$. We have: :$h = r a$ :$g = r b$ thus from {{EuclidDefLink|VII|20|Proportional}}: :$a : h = b : g$ So from {{EuclidPropLink|book = VII|prop = 13|title = Proportional Numbers are Proportional Alternately}}: :$a : b = h : g$ Similarly: :$c : d = g : k$ and further: :$e : f = k : l$ Thus $h, g, k, l$ are continuously proportional: :in the ratio of $a : b$ :in the ratio of $c : d$ :in the ratio of $e : f$. It is necessary to prove that these are the least natural numbers with this property. Suppose $h, g, k, l$ are not the least natural numbers in the ratios of $a : b$, $c : d$, $e : f$. Let these numbers be $n, o, m, p$. Then: :$a : b = n : o$ while $a$ and $b$ are least. By {{EuclidPropLink|book=VII|prop=20|title=Ratios of Fractions in Lowest Terms}}: :$b \divides o$ :$c \divides o$ and so $o$ is a common multiple of $b$ and $c$. Therefore by {{EuclidPropLink|book = VII|prop = 35|title = LCM Divides Common Multiple}}: :$\lcm \set {b, c} \divides o$ But $g = \lcm \set {b, c}$ from above. Therefore $g \divides o$, the greater the less, which is impossible. Therefore there are no numbers $n, o, m, p$ less than $h, g, k, l$ which have the properties in question. {{qed|lemma}} Now suppose $e \nmid k$. Let $m = \lcm \set {e, k}$. Thus let: :$m = u k$ :$n = u h$ :$o = u g$ and: :$m = v e$ :$p = v f$ Thus from {{EuclidDefLink|VII|20|Proportional}}: :$h : n = g : o$ From {{EuclidPropLink|book = VII|prop = 13|title = Proportional Numbers are Proportional Alternately}}: :$h : g = n : o$ Similarly, let: But: :$h : g = a : b$ and so: :$a : b = n : o$ For the same reason: :$c : d = o : m$ Since: :$m = v e$ :$p = v f$ from {{EuclidDefLink|VII|20|Proportional}}: :$e : m = f : p$ and so from {{EuclidPropLink|book=VII|prop=13|title=Proportional Numbers are Proportional Alternately}}: :$e : f = m : p$ Therefore $n, o, m, p$ are continuously proportional: :in the ratio of $a : b$ :in the ratio of $c : d$ :in the ratio of $e : f$. It is necessary to prove that these are the least natural numbers with this property. Suppose $n, o, m, p$ are not the least natural numbers in the ratios of $a : b$, $c : d$, $e : f$. Let these numbers be $n', o', m', p'$. Then: :$a : b = n' : o'$ while $a$ and $b$ are least. By {{EuclidPropLink|book=VII|prop=20|title=Ratios of Fractions in Lowest Terms}}: :$b \divides o'$ :$c \divides o'$ and so $o'$ is a common multiple of $b$ and $c$. Therefore by {{EuclidPropLink|book = VII|prop = 35|title = LCM Divides Common Multiple}}: :$\lcm \set {b, c} \divides o'$ But $g = \lcm \set {b, c}$ from above. Therefore $g \divides o'$, the greater the less, which is impossible. Therefore there are no numbers $n', o', m', p'$ less than $n, o, m, p$ which have the properties in question. {{qed}} {{Euclid Note|4|VIII|It needs to be pointed out that {{AuthorRef|Euclid}} uses the term continued proportion here to mean something different from geometric sequence; the ratios $a : b$, $c : d$ and $e : f$ are not necessarily equal.}}