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Euclid Book 5 Proposition 5
Statement
Computational Explanation
Explanations
For let the magnitude
A
B
be the same multiple of the magnitude
C
D
that the part
A
E
subtracted is of the part
C
F
subtracted; I say that the remainder
E
B
is also the same multiple of the remainder
F
D
that the whole
A
B
is of the whole
C
D
.
For, whatever multiple
A
E
is of
C
F
, let
E
B
be made that multiple of
C
G
. Then, since
A
E
is the same multiple of
C
F
that
E
B
is of
G
C
, therefore
A
E
is the same multiple of
C
F
that
A
B
is of
G
F
.
[
V
.
1
]
But, by the assumption,
A
E
is the same multiple of
C
F
that
A
B
is of
C
D
.
Therefore
A
B
is the same multiple of each of the magnitudes
G
F
,
C
D
; therefore
G
F
is equal to
C
D
.
Let
C
F
be subtracted from each; therefore the remainder
G
C
is equal to the remainder
F
D
.
And, since
A
E
is the same multiple of
C
F
that
E
B
is of
G
C
, and
G
C
is equal to
D
F
, therefore
A
E
is the same multiple of
C
F
that
E
B
is of
F
D
.
But, by hypothesis,
A
E
is the same multiple of
C
F
that
A
B
is of
C
D
; therefore
E
B
is the same multiple of
F
D
that
A
B
is of
C
D
.
That is, the remainder
E
B
will be the same multiple of the remainder
F
D
that the whole
A
B
is of the whole
C
D
.
Classes
Euclid's Elements
Theorems
Geometric Algebra
EuclidBook5
