◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-10-incommensurable-magnitudes / Proof: By Euclid

Proof: By Euclid

(related to Proposition: Prop. 10.050: Construction of Third Binomial Straight Line)

• Let the two numbers $AC$ and $CB$ be laid down such that their sum $AB$ has to $BC$ the ratio which (some) square number (has) to (some) square number, and does not have to $AC$ the ratio which (some) square number (has) to (some) square number.
• And let some other non-square number $D$ also be laid down, and let it not have to each of $BA$ and $AC$ the ratio which (some) square number (has) to (some) square number.
• And let some rational straight line $E$ be laid down, and let it have been contrived that as $D$ (is) to $AB$, so the (square) on $E$ (is) to the (square) on $FG$ [Prop. 10.6 corr.] .
• Thus, the (square) on $E$ is commensurable with the (square) on on $FG$ [Prop. 10.6].
• And $E$ is a rational (straight line).
• Thus, $FG$ is also a rational (straight line).
• And since $D$ does not have to $AB$ the ratio which (some) square number has to (some) square number, the (square) on $E$ does not have to the (square) on $FG$ the ratio which (some) square number (has) to (some) square number either.
• $E$ is thus incommensurable in length with $FG$ [Prop. 10.9].
• So, again, let it have been contrived that as the number $BA$ (is) to $AC$, so the (square) on $FG$ (is) to the (square) on $GH$ [Prop. 10.6 corr.] .
• Thus, the (square) on $FG$ is commensurable with the (square) on on $GH$ [Prop. 10.6].
• And $FG$ (is) a rational (straight line).
• Thus, $GH$ (is) also a rational (straight line).
• And since $BA$ does not have to $AC$ the ratio which (some) square number (has) to (some) square number, the (square) on $FG$ does not have to the (square) on $HG$ the ratio which (some) square number (has) to (some) square number either.
• Thus, $FG$ is incommensurable in length with $GH$ [Prop. 10.9].
• $FG$ and $GH$ are thus rational (straight lines which are) commensurable in square only.
• Thus, $FH$ is a binomial (straight line) [Prop. 10.36].
• So, I say that (it is) also a third (binomial straight line).

• For since as $D$ is to $AB$, so the (square) on $E$ (is) to the (square) on $FG$, and as $BA$ (is) to $AC$, so the (square) on $FG$ (is) to the (square) on $GH$, thus, via equality, as $D$ (is) to $AC$, so the (square) on $E$ (is) to the (square) on $GH$ [Prop. 5.22].
• And $D$ does not have to $AC$ the ratio which (some) square number (has) to (some) square number.
• Thus, the (square) on $E$ does not have to the (square) on $GH$ the ratio which (some) square number (has) to (some) square number either.
• Thus, $E$ is incommensurable in length with $GH$ [Prop. 10.9].
• And since as $BA$ is to $AC$, so the (square) on $FG$ (is) to the (square) on $GH$, the (square) on $FG$ (is) thus greater than the (square) on $GH$ [Prop. 5.14].
• Therefore, let (the sum of) the (squares) on $GH$ and $K$ be equal to the (square) on $FG$.
• Thus, via convertion, as $AB$ [is] to $BC$, so the (square) on $FG$ (is) to the (square) on $K$ [Prop. 5.19 corr.] 2.
• And $AB$ has to $BC$ the ratio which (some) square number (has) to (some) square number.
• Thus, the (square) on $FG$ also has to the (square) on $K$ the ratio which (some) square number (has) to (some) square number.
• Thus, $FG$ [is] [commensurable in length]bookofproofs$1095 with $K$ [Prop. 10.9].
• Thus, the square on $FG$ is greater than (the square on) $GH$ by the (square) on (some straight line) commensurable (in length) with ($FG$).
• And $FG$ and $GH$ are rational (straight lines which are) commensurable in square only, and neither of them is commensurable in length with $E$.
• Thus, $FH$ is a third binomial (straight line) [Def. 10.7] .
• (Which is) the very thing it was required to show.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"