Euclid Book 5 Proposition 8
Statement
If a first magnitude is greater than a second (AB>CD), then the first has a greater ratio to a third than the second does (>), and the ratio of the third to the second is greater than the ratio of the third to the first (>).
AB
EF
CD
EF
EF
CD
EF
AB
Computational Explanation
GeometricScene{A.,B.,C.,D.,E.,F.},{x.,y.,z.},Style[Line[{A.,B.}],Red],Style[Line[{C.,D.}],Pink],Style[Line[{E.,F.}],Blue],EuclideanDistance[A.,B.]x.,EuclideanDistance[C.,D.]y.,x.>y.,EuclideanDistance[E.,F.]z.,>,>,>,>
EuclideanDistance[A.,B.]
EuclideanDistance[E.,F.]
EuclideanDistance[C.,D.]
EuclideanDistance[E.,F.]
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y.
z.
EuclideanDistance[E.,F.]
EuclideanDistance[C.,D.]
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Explanations
Let , be unequal magnitudes, and let be greater; let be another, chance, magnitude; I say that has to a greater ratio than has to , and has to a greater ratio than it has to .
For, since is greater than , let be made equal to ; then the less of the magnitudes , , if multiplied, will sometime be greater than .[Case 1. ]
First, let be less than ; let be multiplied, and let be a multiple of it which is greater than ; then, whatever multiple is of , let be made the same multiple of and of ; and let be taken double of , triple of it, and successive multiples increasing by one, until what is taken is a multiple of and the first that is greater than .
Let it be taken, and let it be which is quadruple of and the first multiple of it that is greater than .
Then, since is less than first, therefore is not less than .
And, since is the same multiple of that is of , therefore is the same multiple of that is of .
But is the same multiple of that is of ; therefore is the same multiple of that is of ; therefore , are equimultiples of , C.
Again, since is the same multiple of that is of , and is equal to , therefore is equal to .
But is not less than ; therefore neither is less than .
And is greater than ; therefore the whole is greater than , together.
But, together are equal to , inasmuch as is triple of , and , together are quadruple of , while is also quadruple of ; whence , together are equal to .
But is greater than , ; therefore is in excess of , while is not in excess of .
And, are equimultiples of , , while is another, chance, multiple of ; therefore has to a greater ratio than has to .
I say next, that also has to a greater ratio than has to .
For, with the same construction, we can prove similarly that is in excess of , while is not in excess of .
And is a multiple of , while , are other, chance, equimultiples of ,; therefore has to a greater ratio than has to .[Case 2. ]
Again, let be greater than .
Then the less,, if multiplied, will sometime be greater than .
Let it be multiplied, and let be a multiple of and greater than ; and, whatever multiple is of , let be made the same multiple of , and of .
Then we can prove similarly that, are equimultiples of , ; and, similarly, let be taken a multiple of but the first that is greater than , so that is again not less than .
But is greater than ; therefore the whole is in excess of , , that is, of .
Now is not in excess of , inasmuch as also, which is greater than , that is, than , is not in excess of . And in the same manner, by following the above argument, we complete the demonstration.
AB
C
AB
D
AB
D
C
D
D
C
AB
For, since
AB
C
BE
C
AE
EB
D
First, let
AE
EB
AE
FG
D
FG
AE
GH
EB
K
C
L
D
M
D
K
Let it be taken, and let it be
N
D
K
Then, since
K
N
K
M
And, since
FG
AE
GH
EB
FG
AE
FH
AB
But
FG
AE
K
C
FH
AB
K
C
FH
K
AB
Again, since
GH
EB
K
C
EB
C
GH
K
But
K
M
GH
M
And
FG
D
FH
D
M
But
D
M
N
M
D
M
D
D
N
D
M
D
N
But
FH
M
D
FH
N
K
N
And
FH
K
AB
C
N
D
AB
D
C
D
I say next, that
D
C
D
AB
For, with the same construction, we can prove similarly that
N
K
N
FH
And
N
D
FH
K
AB
C
D
C
D
AB
Again, let
AE
EB
Then the less,
EB
D
Let it be multiplied, and let
GH
EB
D
GH
EB
FG
AE
K
C
Then we can prove similarly that
FH
K
AB
C
N
D
FG
FG
M
But
GH
D
FH
D
M
N
Now
K
N
FG
GH
K
N