PROPOSITION 28.
To find medial straight lines commensurable in square only which contain a medial rectangle.
Let the rational straight lines
A,
B,
C commensurable in square only be set out; let
D be taken a mean proportional between
A,
B, [
VI. 13] and let it be contrived that,
as B is to C, so is D to E. [VI. 12]
Since
A,
B are rational straight lines commensurable in square only, therefore the rectangle
A,
B, that is, the square on
D [
VI. 17], is medial. [
X. 21]
Therefore
D is medial. [
X. 21]
And since
B,
C are commensurable in square only, and, as
B is to
C, so is
D to
E, therefore
D,
E are also commensurable in square only. [
X. 11]
But
D is medial; therefore
E is also medial. [
X. 23, addition]
Therefore
D,
E are medial straight lines commensurable in square only.
I say next that they also contain a medial rectangle.
For since, as
B is to
C, so is
D to
E, therefore, alternately, as
B is to
D, so is
C to
E. [
V. 16]
But, as
B is to
D, so is
D to
A; therefore also, as
D is to
A, so is
C to
E; therefore the rectangle
A,
C is equal to the rectangle
D,
E. [
VI. 16]
But the rectangle
A,
C is medial; [
X. 21] therefore the rectangle
D,
E is also medial.
Therefore medial straight lines commensurable in square only have been found which contain a medial rectangle. Q. E. D.
LEMMA I.
To find two square numbers such that their sum is also square.
Let two numbers
AB,
BC be set out, and let them be either both even or both odd.
Then since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [
IX. 24, 26] therefore the remainder
AC is even.
Let
AC be bisected at
D.
Let
AB,
BC also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers.
Now the product of
AB,
BC together with the square on
CD is equal to the square on
BD. [
II. 6]
And the product of
AB,
BC is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number the product is square. [
IX. 1]
Therefore two square numbers, the product of
AB,
BC, and the square on
CD, have been found which, when added together, make the square on
BD.
And it is manifest that two square numbers, the square on
BD and the square on
CD, have again been found such that their difference, the product of
AB,
BC, is a square, whenever
AB,
BC are similar plane numbers.
But when they are not similar plane numbers, two square numbers, the square on
BD and the square on
DC, have been found such that their difference, the product of
AB,
BC, is not square. Q. E. D.
LEMMA 2.
To find two square numbers such that their sum is not square.
For let the product of
AB,
BC, as we said, be square, and
CA even, and let
CA be bisected by
D.
It is then manifest that the square product of
AB,
BC together with the square on
CD is equal to the square on
BD. [See
Lemma 1]
Let the unit
DE be subtracted; therefore the product of
AB,
BC together with the square on
CE is less than the square on
BD.
I say then that the square product of
AB,
BC together with the square on
CE will not be square.
For, if it is square, it is either equal to the square on
BE, or less than the square on
BE, but cannot any more be greater, lest the unit be divided.
First, if possible, let the product of
AB,
BC together with the square on
CE be equal to the square on
BE, and let
GA be double of the unit
DE.
Since then the whole
AC is double of the whole
CD, and in them
AG is double of
DE, therefore the remainder
GC is also double of the remainder
EC; therefore
GC is bisected by
E.
Therefore the product of
GB,
BC together with the square on
CE is equal to the square on
BE. [
II. 6]
But the product of
AB,
BC together with the square on
CE is also, by hypothesis, equal to the square on
BE; therefore the product of
GB,
BC together with the square on
CE is equal to the product of
AB,
BC together with the square on
CE.
And, if the common square on
CE be subtracted, it follows that
AB is equal to
GB: which is absurd.
Therefore the product of
AB,
BC together with the square on
CE is not equal to the square on
BE.
I say next that neither is it less than the square on
BE.
For, if possible, let it be equal to the square on
BF, and let
HA be double of
DF.
Now it will again follow that
HC is double of
CF; so that
CH has also been bisected at
F, and for this reason the product of
HB,
BC together with the square on
FC is equal to the square on
BF. [
II. 6]
But, by hypothesis, the product of
AB,
BC together with the square on
CE is also equal to the square on
BF.
Thus the product of
HB,
BC together with the square on
CF will also be equal to the product of
AB,
BC together with the square on
CE: which is absurd.
Therefore the product of
AB,
BC together with the square on
CE is not less than the square on
BE.
And it was proved that neither is it equal to the square on
BE.
Therefore the product of
AB,
BC together with the square on
CE is not square. Q. E. D.