Proof: By Euclid
(related to Proposition: Prop. 10.072: Sum of two Incommensurable Medial Areas give rise to two Irrational Straight Lines)
- For $AB$ is either greater than or less than $CD$.
- By chance, let $AB$, first of all, be greater than $CD$.
- And let the rational (straight line) $EF$ be laid down.
- And let $EG$, equal to $AB$, have been applied to $EF$, producing $EH$ as breadth, and $HI$, equal to $CD$, producing $HK$ as breadth.
- And since $AB$ and $CD$ are each medial, $EG$ and $HI$ (are) thus also each medial.
- And they are applied to the rational straight line $FE$, producing $EH$ and $HK$ (respectively) as breadth.
- Thus, $EH$ and $HK$ are each rational (straight lines which are) incommensurable in length with $EF$ [Prop. 10.22].
- And since $AB$ is incommensurable with $CD$, and $AB$ is equal to $EG$, and $CD$ to $HI$, $EG$ is thus also incommensurable with $HI$.
- And as $EG$ (is) to $HI$, so $EH$ is to $HK$ [Prop. 6.1].
- $EH$ is thus incommensurable in length with $HK$ [Prop. 10.11].
- Thus, $EH$ and $HK$ are rational (straight lines which are) commensurable in square only.
- $EK$ is thus a binomial (straight line) [Prop. 10.36].
- And the square on $EH$ is greater than (the square on) $HK$ either by the (square) on (some straight line) commensurable (in length) with ($EH$), or by the (square) on (some straight line) incommensurable (in length with $EH$).
- Let it, first of all, be greater by the square on (some straight line) commensurable in length with ($EH$).
- And neither of $EH$ or $HK$ is commensurable in length with the (previously) laid down rational (straight line) $EF$.
- Thus, $EK$ is a third binomial (straight line) [Def. 10.7] .
- And $EF$ (is) rational.
- And if an area is contained by a rational (straight line) and a third binomial (straight line) then the square root of the area is a second bimedial (straight line) [Prop. 10.56].
- Thus, the square root of $EI$ - that is to say, of $AD$ - is a second bimedial.
- And so, let the square on $EH$ be greater than (the square) on $HK$ by the (square) on (some straight line) incommensurable in length with ($EH$).
- And $EH$ and $HK$ are each incommensurable in length with $EF$.
- Thus, $EK$ is a sixth binomial (straight line) [Def. 10.10] .
- And if an area is contained by a rational (straight line) and a sixth binomial (straight line) then the square root of the area is the square root of (the sum of) two medial (areas) [Prop. 10.59].
- Hence, the square root of area $AD$ is also the square root of (the sum of) two medial (areas).
- So, similarly, we can show that, even if $AB$ is less than $CD$, the square root of area $AD$ is either a second bimedial or the square root of (the sum of) two medial (areas).
- Thus, when two medial (areas which are) incommensurable with one another are added together, the remaining two irrational (straight lines) arise (as the square roots of the total area) - either a second bimedial, or the square root of (the sum of) two medial (areas).
- A binomial (straight line), and the (other) irrational (straight lines) after it, are neither the same as a medial (straight line) nor (the same) as one another.
- For the (square) on a medial (straight line), applied to a rational (straight line), produces as breadth a rational (straight line which is) also incommensurable in length with (the straight line) to which it is applied [Prop. 10.22].
- And the (square) on a binomial (straight line), applied to a rational (straight line), produces as breadth a first binomial [Prop. 10.60].
- And the (square) on a first bimedial (straight line), applied to a rational (straight line), produces as breadth a second binomial [Prop. 10.61].
- And the (square) on a second bimedial (straight line), applied to a rational (straight line), produces as breadth a third binomial [Prop. 10.62].
- And the (square) on a major (straight line), applied to a rational (straight line), produces as breadth a fourth binomial [Prop. 10.63].
- And the (square) on the square root of a rational plus a medial (area) , applied to a rational (straight line), produces as breadth a fifth binomial [Prop. 10.64].
- And the (square) on the square root of (the sum of) two medial (areas), applied to a rational (straight line), produces as breadth a sixth binomial [Prop. 10.65].
- And the aforementioned breadths differ from the first (breadth), and from one another - from the first, because it is rational - and from one another, because they are not the same in order.
- Hence, the (previously mentioned) irrational (straight lines) themselves also differ from one another.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes
