:300px Let $AB$ be a binomial. Let $CD$ be commensurable in length with $AB$. It is to be shown that $CD$ is a binomial, and that the order of $CD$ is the same as the order of $AB$. Let $AB$ be divided into its terms by $E$. Let $AE$ be the greater term. By definition, $AE$ and $EB$ are rational straight lines which are commensurable in square only. Using {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$AB : CD = AE : CF$ Therefore by {{EuclidPropLink|book = V|prop = 19|title = Proportional Magnitudes have Proportional Remainders}}: :$EB : FD = AB : CD$ But $AB$ is commensurable in length with $CD$. Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AE$ is commensurable in length with $CF$ and: :$EB$ is commensurable in length with $FD$. But by hypothesis $AE$ and $EB$ are rational. Therefore $CF$ and $FD$ are rational. From {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = EB : FD$ Therefore by {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ But by hypothesis $AE$ and $EB$ are commensurable in square only. Therefore by {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$CF$ and $FD$ are commensurable in square only. But $CF$ and $FD$ are rational. Therefore, by definition, $CD$ is a binomial. It remains to be demonstrated that $CD$ is of the same order as $AB$. We have that $AE > EB$. Then $AE^2$ is greater than $EB^2$ by either: :the square on a straight line which is commensurable with $AE$ or: :the square on a straight line which is incommensurable with $AE$. Suppose $AE^2 > EB^2$ by the square on a straight line which is commensurable with $AE$. Then by {{EuclidPropLink|book = X|prop = 14|title = Commensurability of Squares on Proportional Straight Lines}}: :$CF^2$ will greater than $FD^2$ by the square on a straight line which is commensurable with $CF$. Let $AE$ be commensurable in length with a rational straight line which has been set out. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$CF$ will likewise be commensurable in length with that same rational straight line. Thus both $AB$ and $CD$ are by definition first binomial, and therefore of the same [Definition:Order of Binomial|order]]. Let $EB$ be commensurable in length with a rational straight line which has been set out. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$FD$ will likewise be commensurable in length with that same rational straight line. Thus both $AB$ and $CD$ are by definition second binomial, and therefore of the same [Definition:Order of Binomial|order]]. Suppose that neither $AE$ nor $EB$ is commensurable in length with a rational straight line which has been set out.  Then from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :Neither $CF$ nor $FD$ will be commensurable in length with that rational straight line. It follows that both $AB$ and $CD$ are by definition third binomial, and therefore of the same [Definition:Order of Binomial|order]]. Suppose $AE^2 > EB^2$ by the square on a straight line which is incommensurable in length with $AE$. Then by {{EuclidPropLink|book = X|prop = 14|title = Commensurability of Squares on Proportional Straight Lines}}: :$CF^2$ will greater than $FD^2$ by the square on a straight line which is incommensurable in length with $CF$. Let $AE$ be commensurable in length with a rational straight line which has been set out. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$CF$ will likewise be commensurable in length with that same rational straight line. It follows that both $AB$ and $CD$ are by definition fourth binomial, and therefore of the same [Definition:Order of Binomial|order]]. Let $EB$ be commensurable in length with a rational straight line which has been set out. Then by {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$FD$ will likewise be commensurable in length with that same rational straight line. Thus both $AB$ and $CD$ are by definition fifth binomial, and therefore of the same [Definition:Order of Binomial|order]]. Suppose that neither $AE$ nor $EB$ is commensurable in length with a rational straight line which has been set out.  Then from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :Neither $CF$ nor $FD$ will be commensurable in length with that rational straight line. It follows that both $AB$ and $CD$ are by definition sixth binomial, and therefore of the same [Definition:Order of Binomial|order]]. {{qed}} {{Euclid Note|66|X}}