Commentary on Proposition II.5¶
Back to Proposition II.5.
It will be observed that the gnomon is indicated in the figure by three separate letters and a dotted curve. This is no doubt a clearer way of showing what exactly the gnomon is than the method usual in our text-books. In this particular case the figure of the MSS. has two M’s in it, the gnomon being ΜΝΞ. I have corrected the lettering to avoid confusion.
It is easily seen that this proposition and the next give exactly the theorem already alluded to under the last propositions, namely that the difference of the squares on two straight lines is equal to the rectangle contained by their sum and difference. The two given lines are, in II. 5, the lines CB and CD, and their sum and difference are respectively equal to AD and DB. To show that II. 6 gives the same theorem we have only to make CD the greater line and CB the less, i.e. to draw C'D' equal to CB, measure C'B' along it equal to CD, and then produce B'C' to A', making A'C' equal to B'C', whence it is immediately clear that A'D' on the second line is equal to AD on the first, while D'B' is also equal to DB, so that the rectangles AD, DB and A'D', D'B' are equal, while the difference of the squares on CB, CD is equal to the difference of the squares on C'D', C'B'.
Perhaps the most important fact about II. 5, 6 is however their bearing on the
Geometrical solution of a quadratic equation.¶
Suppose, in the figure of II. 5, that AB $= a$, DB $= x$; then
Thus, if the area of the gnomon is given ($= b^2$, say), and if $a$ is given ($=$ AB), the problem of solving the equation [ ax - x^2 = b^2 ] is, in the language of geometry, To a given straight line ($a$) to apply a rectangle which shall be equal to a given square ($b^2$) and shall fall short by a square figure, i.e. to construct the rectangle AH or the gnomon NOP.
Now we are told by Proclus (on I. 44) that “these propositions are ancient and the discoveries of the Muse of the Pythagoreans, the application of areas, their exceeding and their falling-short.” We can therefore hardly avoid crediting the Pythagoreans with the geometrical solution, based upon II. 5, 6, of the problems corresponding to the quadratic equations which are directly obtainable from them. It is certain that the Pythagoreans solved the problem in II. 11, which corresponds to the quadratic equation
$$a(a - x) = x^2,$$
and Simson has suggested the following easy solution of the equation now in question,
$$ax - x^2 = b^2,$$
on exactly similar lines.
Draw CO perpendicular to AB and equal to $b$; produce OC to N so that ON $=$ CB (or $\frac{1}{2}a$); and with O as centre and radius ON describe a circle cutting CB in D.
Then DB (or $x$) is found, and therefore the required rectangle AH.
For the rectangle AD, DB together with the square on CD is equal to the square on CB, [II. 5] i.e. to the square on OD, i.e. to the squares on OC, CD; [I. 47] whence the rectangle AD, DB is equal to the square on OC, or
$$ax - x^2 = b^2.$$
It is of course a necessary condition of the possibility of a real solution that $b^2$ must not be greater than $(\frac{1}{2}a)^2$. This condition itself can easily be obtained from Euclid’s proposition; for, since the sum of the rectangle AD, DB and the square on CD is equal to the square on CB, which is constant, it follows that, as CD diminishes, i.e. as D moves nearer to C, the rectangle AD, DB increases and, when D actually coincides with C, so that CD vanishes, the rectangle AD, DB becomes the rectangle AC, CB, i.e. the square on CB, and is a maximum. It will be seen also that the geometrical solution of the quadratic equation derived from Euclid does not differ from our practice of solving a quadratic by completing the square on the side containing the terms in $x^2$ and $x$.
But, while in this case there are two geometrically real solutions (because the circle described with ON as radius will not only cut CB in D but will also cut AC in another point E), Euclid’s figure corresponds to one only of the two solutions. Not that there is any doubt that Euclid was aware that the method of solving the quadratic gives two solutions; he could not fail to see that $x =$ BE satisfies the equation as well as $x =$ BD. If however he had actually given us the solution of the equation, he would probably have omitted to specify the solution $x =$ BE because the rectangle found by means of it, which would be a rectangle on the base AE (equal to BD) and with altitude EB (equal to AD), is really an equal rectangle to that corresponding to the other solution $x =$ BD; there is therefore no real object in distinguishing two solutions. This is easily understood when we regard the equation as a statement of the problem of finding two magnitudes when their sum ($a$) and product ($b^2$) are given, i.e. as equivalent to the simultaneous equations
$$x + y = a,$$
$$xy = b^2.$$
These symmetrical equations have really only one solution, as the two apparent solutions are simply the result of interchanging the values of $x$ and $y$. This form of the problem was known to Euclid, as appears from the Data, Prop. 85, which states that, If two straight lines contain a parallelogram given in magnitude in a given angle, and if the sum of them be given, then shall each of them be given.
This proposition then enables us to solve the problem of finding a rectangle the area and perimeter of which are both given; and it also enables us to infer that, of all rectangles of given perimeter, the square has the greatest area, while, the more unequal the sides are, the less is the area.
If in the figure of II. 5 we suppose that $AD = a$, $BD = b$, we find that $CB = (a + b)/2$ and $CD = (a - b)/2$, and we may state the result of the proposition in the following algebraical form [ left(frac{a + b}{2}right)^2 - left(frac{a - b}{2}right)^2 = ab. ]
This way of stating it (which could hardly have escaped the Pythagoreans) gives a ready means of obtaining the two rules, respectively attributed to the Pythagoreans and Plato, for finding integral square numbers which are the sum of two other integral square numbers. We have only to make $ab$ a perfect square in the above formula. The simplest way in which this can be done is to put $a = n^2$, $b = 1$, whence we have [ left(frac{n^2 + 1}{2}right)^2 - left(frac{n^2 - 1}{2}right)^2 = n^2, ] and in order that the first two squares may be integral $n^2$, and therefore $n$, must be odd. Hence the Pythagorean rule.
Suppose next that $a = 2n^2$, $b = 2$, and we have [ (n^2 + 1)^2 - (n^2 - 1)^2 = 4n^2, ] whence Plato’s rule starting from an even number $2n$.