Page:A History of Mathematics (1893).djvu/142

That it should be necessary to make such conditions seems strange to us; but it must be remembered that the monks of the Middle Ages did not attend school during childhood and learn the multiplication table while the memory was fresh. Gerbert's rules for division are the oldest extant. They are so brief as to be very obscure to the uninitiated. They were probably intended simply to aid the memory by calling to mind the successive steps in the work. In later manuscripts they are stated more fully. In dividing any number by another of one digit, say 668 by 6, the divisor was first increased to 10 by adding 4. The process is exhibited in the adjoining figure.^[3] As it continues, we must imagine the digits ${\overset {\widehat {\vert {\widehat {C~\vert ~D~}}\vert {\widehat {~S}}\vert }}{\begin{array}{|c|c|c|}&&\scriptstyle {6}\\&&\scriptstyle {4}\\\hline \scriptstyle {\not 6}&\scriptstyle {\not 6}&\scriptstyle {\not 8}\\\hline \scriptstyle {\not 6}&\scriptstyle {\not 6}&\scriptstyle {\not 8}\\\scriptstyle {\not 2}&\scriptstyle {\not 4}&\scriptstyle {\not 6}\\\scriptstyle {\not 1}&\scriptstyle {\not 8}&\scriptstyle {\not 8}\\\scriptstyle {\not 1}&\scriptstyle {\not 4}&\scriptstyle {\not 8}\\&\scriptstyle {\not 2}&\scriptstyle {2}\\&\scriptstyle {\not 4}&\\&\scriptstyle {\not 6}&\\&\scriptstyle {\not 2}&\\&\scriptstyle {\not 2}&\\\hline &\scriptstyle {\not 6}&\scriptstyle {\not 6}\\&\scriptstyle {\not 2}&\scriptstyle {\not 8}\\&\scriptstyle {\not 1}&\scriptstyle {\not 2}\\\scriptstyle {1}&\scriptstyle {1}&\scriptstyle {1}\\\end{array}}}$which are crossed out, to be erased and then replaced by the ones beneath. It is as follows: $\scriptstyle {600\div 10=60}$, but, to rectify the error, $\scriptstyle {4\times 60}$, or 240, must be added; $\scriptstyle {200\div 10=20}$, but $\scriptstyle {4\times 20}$, or 60, must be added. We now writs for $\scriptstyle {60+40+60}$, its sum 180, and continue thus: $\scriptstyle {100\div 10=10}$; the correction necessary is $\scriptstyle {4\times 10}$, or 40, which, added to 80, gives 120. Now $\scriptstyle {100\div 10=10}$, and the correction $\scriptstyle {4\times 10}$, together with the 20, gives 60. Proceeding as before, $\scriptstyle {60\div 10=6}$; the correction is $\scriptstyle {4\times 6=24}$. Now $\scriptstyle {20\div 10=2}$, the correction being $\scriptstyle {4\times 2=8}$. In the column of units we have now $\scriptstyle {8+4+8}$, or 20. As before, $\scriptstyle {20\div 10=2}$; the correction is $\scriptstyle {2\times 4=8}$, which is not divisible by 10, bet only by 6, giving the quotient 1 and the remainder 2. All the partial quotients taken together give $\scriptstyle {60+20+10+10+6+2+2+1=111}$, and the remainder 2.

Similar but more complicated, is the process when the divisor contains two or more digits. Were the divisor 27,