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Alien Encounters - Act 2: Introducing the Strangers - Scene 1

Author(s): 
Gavin Hitchcock

A lecture room in the University of Cambridge, circa 1730; ENTER NICHOLAS SAUNDERSON, Lucasion Professor of Mathematics, and famed for his didactic prowess. He is blind, having become so through contracting smallpox at the age of one. He is led by an ASSISTANT, who then writes on the blackboard when necessary, finding it difficult to keep pace and so occasioning some irritation on the part of Saunderson.



SAUNDERSON18

Good day, gentlemen!

Before we can proceed to the multiplication of algebraic quantities, we are to take notice that if the signs of the multiplicator and multiplicand be both alike, that is, both affirmative, or both negative, the product will be affirmative, otherwise it will be negative...

Thus the rule resolves itself into four cases:

first, that affirmative 4 multiplied into affirmative 3 produces affirmative 12; second, that negative 4 multiplied into affirmative 3 produces negative 12; third, that affirmative 4 multiplied into negative 3 produces affirmative 12; and lastly, that negative 4 multiplied into negative 3 produces affirmative 12.


(+4)(+3)
=
12
(-4)(+3)
=
-12
(+4)(-3)
=
-12
(-4)(-3)
=
+12

If you expect a demonstration of this rule, you must first be advertised of two things:

First: That numbers are said to be in arithmetical progression, when they increase or decrease with equal differences, as nothing, two, four, six; or six, four, two, nothing; also as three, nothing, negative three; four, nothing, negative four; twelve, nothing, negative twelve, or negative twelve, nothing, affirmative twelve.

     0, 2, 4, 6                        6, 4, 2, 0

     3, 0, -3         4, 0, -4        12, 0, -12        -12, 0, +12

Secondly: If a set of numbers in arithmetical progression, as three, two, and one, be successively multiplied into one common multiplicator, as four, or if a single number, as four, be successively multiplied into a set of numbers in arithmetical progression, as three, two and one, then the products, twelve, eight, and four, in either case, will be in arithmetical progression.

               3    2    1                                                            4    4    4

               4    4    4                                                            3    2    1

              12   8    4                                                           12   8    4

 

This being allowed (which is in a manner self-evident), we proceed to the demonstration of the rule.

First Case: That affirmative 4 multiplied into affirmative 3 produces affirmative 12 is self-evident, and needs no demonstration; or if it wanted one, it might receive it as follows: to multiply affirmative 4 by affirmative 3 is the same thing as to add 4 plus 4 plus 4 into one sum; but 4 plus 4 plus 4 added into one sum gives affirmative 12; therefore affirmative four multiplied into affirmative three, gives affirmative twelve.

Second Case: That negative four multiplied into affirmative three produces negative twelve, I shall demonstrate thus: multiply the terms of this arithmetical progression: four, nothing, negative four, into affirmative three, and the products will be in arithmetical progression, as above; but the two first products are 12 and nothing; therefore the third will be negative twelve; therefore negative four multiplied into affirmative three, produces negative twelve.

                                                              +4      0    -4

                                                              +3    +3   +3

                                                             +12     0   -12

 

The Third Case is similar to the second, except that I multiply affirmative four into affirmative three, nothing, negative three, successively...

                                                             +4     +4     +4

                                                             +3       0      -3

                                                            +12      0     -12

Fourth Case :  Lastly, to demonstrate that negative four multiplied into negative three produces affirmative twelve, multiply negative four into three, nothing, negative three, successively and the products will be in arithmetical progression; but the two first products are negative twelve and nothing, by the second case; therefore the third product will be affirmative twelve...

                                                             -4      -4      -4

                                                             +3      0      -3

                                                            -12      0     +12

These four cases may be also more briefly demonstrated thus:  affirmative four mulitplied into affirmative three produces affirmative twelve; therefore negative four into affirmative three, or affirmative four into negative three, ought to produce something contrary to affirmative twelve, that is, negative twelve: therefore negative four multiplied into negative three ought to produce something contrary to negative twelve, that is, affirmative twelve.

So that this last case, so very formidable to young beginners, appears at last to amount to no more than a common principle in Grammar, to wit, that two negatives make an affirmative; which is undoubtedly true in Grammar, though perhaps it may not always be observed in Languages.

Are there, perchance, any questions? There being no questions I beg to wish you all: Good day, gentlemen!

[EXIT led by his assistant ]
[CURTAIN]

Dummy View - NOT TO BE DELETED