[70r]

       Ockham
Tuesday. 22^nd Dec^r

 Dear M^r De Morgan  I now see
exactly my mistake.  I had
overlooked that the Series in
question is not one in successive
Powers of \(x\) ['like that in page 185' inserted], but only in
successive even powers of \(x\) .

I used once to regret these
sort of errors, & to speak of
time lost over them.  But I
have materially altered my
mind on this subject.  I often
gain more from the discovery
of a mistake of this sort, than
from 10 acquisitions made at
[70v] once & without any kind of
difficulty.     

 There is still one little thing in
your Demonstration not perfectly
clear to me.  At the end
you remark that ''our result
''gave the 503\) ^\textup{d}\) Term instead of
''the 502\) ^\textup{nd}\), which arose from
''taking the whole number next
''above \(\sqrt{x+\frac{1}{4}}\) instead of an
''intermediate fraction.''       
In examining the equation
\) n=\) next whole number above

  \(\frac{5}{4}+\frac{1}{2}\sqrt{1000,000+\frac{1}{4}}\) 

 I see clearly that \(\frac{5}{4}+\frac{1}{2}(1001)\) is
greater than \(\frac{5}{4}+\frac{1}{2}\sqrt{1000,000+\frac{1}{4}}\);
that the true answer would be
\( \frac{5}{4}+\frac{1}{2}\left(1000+\frac{1}{a}\right)\), \(\frac{1}{a}\) being some
[71r] fraction.  We should then have
had,
\( n=\) nearest whole number above

 \(\frac{5}{4}+500+\frac{1}{2a}\), instead of \(=\) 
\( =\frac{5}{4}+500+\frac{1}{2}\) 
But, since \(\frac{5}{4}\) is greater than \(1\),
the result must exceed \(501\) even
if we neglected the \(\frac{1}{4}\) altogether;
and therefore at any rate \) n\)
(the next whole number above
\( \frac{5}{4}+\frac{1}{2}\sqrt{Z+\frac{1}{4}}\) ), must be \(502\), &
\( n+1\) consequently \(=503\) .
I do not therefore see that the
fact of taking \(1001\) instead of
the real square part of \(Z+\frac{1}{4}\) 
does account for the discrepancy
in question .         

[71v] I have now some question to
put respecting certain operations
with Incommensurables.  Thanks to
your Treatise I think I understand
['this subject' inserted] pretty tolerably now.  But there
are still one or two points of
Practical Application which I
am [something crossed out] busy in working up
previous to leaving the subject
altogether as a direct study, &
which I find not quite plain
sailing.    

 I have been writing out in
the Mathematical Scrap-Book,
a full explanation of the
operations with Incommensurables
analogous to those of Multiplication,
Division, Raising of Powers &c,
and a day or two ago I was
[72r] about completing it with that
analogous to the extraction of
Roots, when I found I did
not fully understand the
process, that is beyond the
consideration of one Mean
Proportional. I have written
out & enclose my explanation
for one Mean Proportional,
& my difficulty in the case
of two or more Mean
Proportionals.         

 Also, I wished now to return
to the passage, page 29, lines 8
and 9 from the top, (Trigonometry)
which first suggested to me the
necessity of studying the subject
of Incommensurables; in order
that I might see if I could
[72v] now demonstrate the Proposition
of (46), for \(\theta\) and \(\sin\theta\) Incom=
=mensurable quantities. But I do
not find that I can.  I believe
I understand the example referred
to in (4), the long & short
of which I understand to be
that if in the Right-Angled
Triangle [diagram in original] \(A\), \(B\), \(C\) are
Incommensurables, and \(V\) be any
given linear unit, then the
Ratio compounded of \(A:V\) and \(A:V\) 
added to the Ratio compounded of
\( B:V\) & \(B:V\), is equal to the
Ratio compounded of \(C:V\) and
\( C:V\) .   
With respect to the Ratio of an
Angle with it's [\textit{sic}] Sine, I began to
[73r] write it out as follows, after the
manner of pages 68 & 69 of the
Number & Magnitude:
\( \theta\) or \(\theta:1\) is the Ratio of \(\frac{AB}{AO}\) 
\( \sin\theta\) or \(\sin\theta:1\) is the Ratio of \(\frac{BM}{AO}\),
\( AB:AO\), \(BM:AO\) being
Incommensurable Ratios, what
then does \(\frac{\theta}{\sin\theta}\) really mean?    
In the first place we may
consider it to mean

 \(\theta\,\frac{1}{\sin\theta}:1\), or a Ratio
compounded of the Ratio \(\theta:1\) and
\( \frac{1}{\sin\theta}\), or compounded of the
Ratios \(AB:AO\) and \(AO:BM\) .    

But further than this I cannot
get, nor see my way at all.

I conclude that in Incommensurable
language, a Ratio equal to \(1\) or
a Ratio approximately to \(1\) can
[73v] only mean a Ratio in which
the Magnitude constituting the
Antecedent is equal to the
Magnitude constituting the
Consequent, or is constantly
approaching an equality to it,
and therefore that if we take
the above Ratio compounded
of \(AB:AO\) and \(AO:BM\), or
the Ratio \(AB:BM\), & prove
that \(AB\) constantly approaches
in equality to \(BM\), that is the
desired Demonstration.      

 I can only end by repeating
what I have often said before,
that I am very troublesome,
& only wish I could do you
any such service as you are
doing me.  Yours most truly

 A. A. L