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\centerline{MONT 107Q -- Thinking About Mathematics}
\centerline{Discussion on Diophantus}
\centerline{March 1 (and 3?), 2017}
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\noindent
{\it Background}
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The following are roughly literal translations of the next four Propositions in
Book I of the {\it Arithmetica} of Diophantus after the ones in the reading for
today.  The goal in this exercise is to ``decode'' what Diophantus is saying,
translate it into modern algebra and think about how you would go about solving
similar problems even if the data involved different given numbers.  Does what
Diophantus writes give enough of an idea of how to solve the problem in general?

Each group will be responsible for presenting their work {\it orally to the class}
later in the period on Wednesday, or possibly in class on Friday if more time is required.
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\noindent
{\bf Proposition 7.}  From the same [unknown] number, subtract two given numbers leaving
the remainders in a given proportion.
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\noindent
Let it be proposed to subtract $100$ and $20$ from some number leaving numbers
with the larger in $3^{pl}$ ratio to the smaller.  Let the number sought be 
laid out as $\s 1$.  The remainder on subtracting $100$ is $\s 1 \Lambda M^o 100$ and
the remainder on subtracting $20$ is $\s 1\Lambda M^o 20$.  We must have the greater
to the smaller in $3^{pl}$ ratio, so the larger is three times the smaller, so 
$\s 3 \Lambda M^o 300$ and $\s 1 \Lambda M^0 20$ are the same.  Let the lack be
added to both, so $\s 3$ and $\s 1 M^o 280$ are the same.  Now let equals be subtracted
from equals, and $\s 2$ and $M^o 280$ are the same.  Therefore $\s$ is $140$.

To what was set down ... If $100$ is subtracted from $140$, the remainder is $40$.
If $20$ is subtracted from $140$, the remainder is $120$ which is three times $40$.  
So the larger is three times the smaller.
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\noindent
{\bf Proposition 8.} Add to two given numbers the same [unknown] number to make
the sums have a given proportion.
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\noindent
It is necessary for the given proportion to be smaller than the proportion between
the larger and the smaller given numbers.  
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\noindent
Let it be proposed to add $M^o 100$ and $M^o 20$ to a number yielding sums in $3^{pl}$ ratio.
Let the unknown number be $\s 1$.  Then the sums are $\s 1 M^o 100$ and $\s 1 M^o 20$.
Since the larger is to be three times the smaller, $\s1 M^o 100$ and $\s3 M^o 60$
are the same.  Take away equals from equals.  The remainders are
$M^o 40$ and $\s 2$ and so $\s$ becomes $M^o 20$.  

To what was set down.  If $100$ is added to $20$, the sum is $120$.  If $20$ is added
to $20$, the sum is $40$.  And $120$ is three times $40$.  So the larger is three times the smaller.
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\noindent
{\bf Proposition 9.}  From two given numbers subtract the same [unknown] number to make the 
remainders have a given proportion.

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\noindent
It is necessary for the given proportion to be larger than the proportion between
the larger of two given numbers and the smaller one.

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\noindent
Let it be proposed to subtract the same number from $20$ and $100$
leaving remainders that are in $6^{pl}$ ratio, the greater to the smaller.  
Let the number to be subtracted from the two given numbers be $\s 1$.
The remainders are then $M^o 20 \Lambda \s 1$ and $M^o 100 \Lambda \s 1$.  We must
have that the larger is $6$ times the smaller, so $M^o 100\Lambda \s 1$ and
$M^o 120 \Lambda \s 6$ are equal.  Let the lack be added to both and let equals
be subtracted from equals.  Then $\s 5$ and $M^o 20$ are equal, so $\s$ is $M^o 4$.

To what was set down.  Let the number to be subtracted be $4$.  Then 
taking $4$ from $20$ leaves $16$.  Taking $4$ from $100$ leaves $96$.  
And $96$ is $6$ times $16$.   So the larger is $6$ times the smaller.    
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\noindent
{\bf Proposition 10.}  From the larger of two given numbers subtract a[n unknown] number
and add the same number to the smaller of the two given numbers to make the sum have
a given ratio with the difference.  
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\noindent
Let it be proposed to add a number to $20$ and subtract the same number from $100$
to yield numbers in $4^{pl}$ ratio, the larger to the smaller.  Let the number
to be added and subtracted be laid out as $\s 1$.  If we add $\s 1$, the sum is 
$\s 1 M^o 20$; if we subtract the difference is $M^o 100 \Lambda \s 1$. 
Then $M^o 400 \Lambda \s 4$ and $\s 1 M^o 20$ are equal.  Let the 
lack be added to both and subtract equals from equals.  Then $\s 5$ is equal to 
$M^o 380$ and $\s 1$ is $M^o 76$.  

To what was set down.   If $76$ is added to $20$, the sum is $96$.  If $76$ is
subtracted from $100$, the difference is $24$.  And $96$ is $4$ times $24$.
So the larger is four times the smaller.
\bye 




   
