THE ALTERNATE VOTE. 36/ 



does, Nanson excludes the lowest (22). 



22. Nanson then proves : that if one candidate A is the best 

 man, he cannot be rejected at the first scrutiny, and is thus sure 

 of being returned. 



The whole number of electors is supposed to be 2n. All vote 

 for one of the three candidates, and express their second pre- 

 ference. Considering- that all the 2n voted in regard to C and A, 

 if d of them prefer C to A, then 2n-d prefer A to C, since all 

 express their preferences as to A and C. 



I have ventured on some very elementary arithmetic. Perhaps 

 some of us have forgotten our elementary algebra, so please bear 

 with me once more. 



Suppose a poor man has nothing in the world but one shilling 

 in his pocket, which we shall call b, and the clothes he stands 

 in, and that they are of the value of n. That one shilling is a 

 positive quantity, and added to n makes n greater. Using the 

 algebraical signs + for addition, and — for subtraction, b being 

 a positive quantity, that is being more than nothing, the expres- 

 sion n + b is more than n. 



In place of having one shilling this man may be owing a 

 shilling, be in debt to the extent of a shilling ?>. 6 is now no longer 

 more than nothing, it is now less than nothing, it is a minus 

 quantity, a negative quantity ; and in estimating this poor man's 

 estate b so far from making n more, makes n less; and n+b 

 becomes less than n. Remembering then that any quantity is 

 positive or negative according to whether it is more or less than 

 nothing, we may express the number of voters who prefer C to 

 A by the algebraical expression 71+b. That number will either 

 be more than »?, which remember is half the number of voters, 

 when b is positive, or less than n when b is negative. In the 

 case where it is more than «, b will be more than nothing, it 

 will be positive. In the case where it is less than «, b will be 

 less than nothing, it will be a minus quantity, and this being so 

 v + b will now be less than n. So now in place of d which we 

 supposed to be the number of voters who preferred C to A, we 

 may substitute n + b ; remembering that when b is a negative 

 quantity, or as we say when b is negative, that n + b is less than 

 u- and of course v — b more than n. So much for elementary 

 algebra. 



We may say then those that prefer C to A number n + b ; those 

 who prefer A to C are therefore 2n— {n-\-b) or n — b. Let those 

 who prefer A to B be represented by n + c, and of course then 

 those who prefer B to A will be n — c;c like b being either a 

 positive or a negative quantity. So also those who prefer B to 

 C may be expressed by n + a, and those who prefer C to B by 

 n — a. 

 Those who prefer A to C are n — b. Those who prefer C to A are 



n + b. 

 Those who prefer A to C are n + a. Those who prefer C to B are 



n — a. 

 Those who prefer A to B are n + c. Those who prefer B to A are 

 n — c. 

 Using the expression AB for the number of electors or committee- 



