Construction of First Binomial Straight Line

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Theorem

In the words of Euclid:

To find the first binomial straight line.

(The Elements: Book $\text{X}$: Proposition $48$)


Proof

Euclid-X-48.png

Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line.

Using Lemma 1 to Proposition $28$ of Book $\text{X} $: 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 rational straight line.

Let $EF$ be constructed commensurable in length with $D$.

Then $EF$ is also a rational straight line.

Using Porism to Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable, let:

$BA : AC = EF^2 : FG^2$

where $FG$ is a straight line constructed such that $EG = EF + FG$ is itself a straight line.

But:

$AB : AC = m^2 : \left({m^2 - n^2}\right)$

where $\left({m^2 - n^2}\right)$ is not square.

So from Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:

$EF^2 : FG^2 = m^2 : \left({m^2 - n^2}\right)$

$EF$ and $FG$ are commensurable in square.

But from Proposition $9$ of Book $\text{X} $: Commensurability of Squares: $EF$ and $FG$ are incommensurable in length.

Therefore $EF$ and $FG$ are rational straight lines which are commensurable in square only.

Therefore by definition $EG$ is a binomial.


We have that:

$BA : AC = EF^2 : FG^2$

while:

$BA > AC$

Therefore:

$EF^2 > FG^2$

Let:

$FG^2 + H^2 = EF^2$

for some $H$.

From Porism to Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:

$AB : BC = EF^2 : H$

But $AB : BC$ has the ratio that a square number has to a square number.

Therefore by Proposition $9$ of Book $\text{X} $: Commensurability of Squares:

$EF$ is commensurable in length with $H$.

Therefore $EF^2$ is greater than $FG^2$ by the square on a straight line commensurable with $EF$.

We have that:

$EF$ and $FG$ are rational straight lines

and:

$EF$ is commensurable in length with $D$.

Therefore $EF$ is a first binomial straight line.

$\blacksquare$


Historical Note

This proof is Proposition $48$ of Book $\text{X}$ of Euclid's The Elements.


Sources

in which a mistake appears.