:400px Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line. Using {{EuclidPropLink|book = X|prop = 28|qual = Lemma 1|title = Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater}}, let: : $AB : BC = m^2 : n^2$ where $m$ and $n$ are numbers such that $m^2 - n^2$ is not square. Let $D$ be a number which is not square. Let $D$ have to neither $AB$ nor $AC$ the ratio that a square number has to another square number. Let $E$ be a rational straight line. Using {{EuclidPropLink|book = X|prop = 6|qual = Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let: :$D : AB = E^2 : FG^2$ where $FG$ is a straight line. From {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$E^2$ is commensurable with $FG^2$. We have that $E$ is rational. Therefore $FG$ is also rational. Since: : $D$ does not have to $AB$ the ratio that a square number has to another square number then: : $E^2$ does not have to $FG^2$ the ratio that a square number has to another square number. From {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$E$ is incommensurable in length with $FG$. Using {{EuclidPropLink|book = X|prop = 6|qual = Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let: :$BA : AC = FG^2 : GH^2$ where $GH$ is a straight line constructed such that $FH = FG + GH$ is itself a straight line. From {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$FG^2$ is commensurable with $GH^2$. We have that $FG$ is rational. Therefore $GH$ is also rational. Since: : $BA$ does not have to $AC$ the ratio that a square number has to another square number then: : $FG^2$ does not have to $GH^2$ the ratio that a square number has to another square number. From {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$FG$ is incommensurable in length with $GH$. Therefore $FG$ and $GH$ are rational straight lines which are commensurable in square only. Therefore by definition $FH$ is a binomial. We have that: :$D : AB = E^2 : FG^2$ and :$BA : AC = FG^2 : GH^2$ From {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$D : AC = E^2 : GH^2$ But $D$ does not have to $AC$ the ratio that a square number has to another square number. Therefore, neither does $E$ have to $GH$ the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$E$ is incommensurable in length with $GH$. We have that: :$BA : AC = FG^2 : GH^2$ Therefore: : $FG^2 > GH^2$ Let: : $FG^2 = GH^2 + K^2$ for some $K$. From {{EuclidPropLink|book = V|prop = 19|qual = Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$AB : BC = FG : K$ But:  : $AB : BC = m^2 : n^2$ where $m$ and $n$ are numbers. Therefore: :$FG : K = m^2 : n^2$ Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$FG$ is commensurable in length with $K$. Therefore $FG^2$ is greater than $GH^2$ by the square on a straight line which is commensurable in length with $FG$. But $FG$ and $GH$ are rational straight lines which are commensurable in square only. Also, neither $FG$ nor $GH$ is commensurable in length with $E$. Therefore $FH$ is a third binomial straight line. {{qed}} {{Euclid Note|50|X}}