Pasha Zusmanovich
[writings] [soft] [teaching] [for students] [talks] [links, files, etc.] 
Problem 47, Proposed by
L.E. Dickson,
Fellow in Mathematics, University of Chicago:
Prove that (1)(1) = +1. Solutions: I. Assuming the distributive law to hold, (1)((+1) + (1)), or 0, = (1)(+1) + (1)(1). Assuming the commutative law, (1)(+1) = (+1)(1 = 1. (1) + (1)(1) = 0, or (1)(1) = +1. [L.E. Dickson] II. (1)(1) means that 1 is to be taken subtractively one time. 0  (1) = +1. (1)(1) = +1. [G.B.M. Zerr] III. 1 × a = a, 1 × (a1) = (a1) = a + 1. 1 × ((a1)a) = a + 1  (a) = a + 1 + a = 1. [M. Philbrick] ... VII. According to Wood's Elementary Algebra, 17th edition, we have (5)(3) = +15. Here 3 is to be subtracted 5 times; that is, 15 is to be subtracted. Now, subtracting 15 is the same as adding +15. Therefore, we have to add +15. Similarly, (1)(1) = +1. [W.I. Taylor, F.P. Matz] VIII. The case (a)(b) = +ab is purely conventional and consequently an assumption, which, however, does not deprive the result of its great importance to algebraic operations. [J.F.W. Scheffer] 

Illustration by Gwynedd M. Hudson, courtesy of British Library. 