I have read about geometrical proofs of irrational numbers based on incommensurability of lengths elsewhere. But, am stuck by the line:
For, if any number of odd numbers are added to one another so that
the number of numbers added is an odd number the result is
also an odd number.
given in the book : Julian Havil, The irrationals; at pg. #22, 23. The other pages of the book: pg. 21, pg. 24, 25, pg. #26,27.
The text given in the pg.$21, 22 of the complete proof are given below, with the line in bold:
Let $ABCD$ be a square and $AC$ its diameter. I say that $AC$ is
incommensurable with $AB$ in length. For let us assume that it
is commensurable. I say that it will follow that the same num-
ber is at the same time even and odd. It is clear that the square
on $AC$ is double the square on $AB$. Since then (according to
our assumption) $AC$ is commensurable with $AB, AC$ will be to
AB in the ratio of an integer to an integer. Let them have the
ratio $DE:DF$ and let $DE$ and $DF$ be the smallest numbers which
are in this proportion to one another. $DE$ cannot then be the
unit. For if DE was the unit and is to $DF$ in the same proportion
as $AC$ to $AB, AC$ being greater than $AB, DE$, the unit, will be
greater than the integer $DF$, which is impossible. Hence $DE$ is
not the unit, but an integer (greater than the unit). Now since
$AC:AB = DE:DF$, it follows that also $AC^2 :AB^2 = DE^2 :DF^2$. But
$AC^2 = 2AB^2$ and hence $DE^2 = 2DF^2$. Hence $DE^2$ is an even num-
ber and therefore $DE$ must also be an even number. For, if it was
an odd number, its square would also be an odd number. For, if
any number of odd numbers are added to one another so that
the number of numbers added is an odd number the result is
also an odd number. Hence $DE$ will be an even number. Let then
$DE$ be divided into two equal numbers at the point $G$. Since $DE$
and $DF$ are the smallest numbers which are in the same pro-
portion they will be prime to one another. Therefore, since $DE$
is an even number, $DF$ will be an odd number. For, if it was an
even number, the number $2$ would measure both $DE$ and $DF$,
although they are prime to one another, which is impossible.
Hence $DF$ is not even but odd. Now since $DE = 2EG$ it follows
that $DE^2 = 4EG^2$ . But $DE^2 = 2DF^2$ and hence $DF^2 = 2EG^2$. There-
fore $DF^2$ must be an even number, and in consequence $DF$ is
also an even number. But it has also been demonstrated that
$DF$ must be an odd number, which is impossible. It follows,
therefore, that $AC$ cannot be commensurable with $AB$, which
was to be demonstrated.
Doubt: I feel that the line is irrelevant as the sum of odd quantities in odd number is no where to be seen.
If not, then have not understood the proof correctly.
The proof depends on the fact that the square of an odd number is odd. That follows from the statement in bold since you square an odd number by adding it to itself an odd number of times.