I haven't written a lot of proofs so I need the opinion of the experts on my proof of a simple proposition. Here's the various properties I used: (P10) (Trichotomy law) For every number a, one and only one of the following holds: (i) a = 0, (ii) a is in the collection P, (iii) —a is in the collection P. (P7) For every number a not equal to 0, there is a number a^-1 such that a • a^-1 = a^-1 • a = 1. (P6) If a is any number, then a • 1 = 1 • a = a. If ax=a for some number "a" different from 0, then x=1.(Spivak's calculus.) I consider two cases: a>0 or a<0. By definition (Given in Spivak's calculus) : a>b if a-b is in the collection P (P being the collection of all positive numbers.) a>0 because a-o is in collection P by trichotomy law(P10) So a*x=a by P7 a*a^-1*x= a*a^-1 1*x=1 by P6 x=1 Second case : By definition : a< to b if b>a a<0 because 0>a Now, we do the same thing as the previous case. Proposition proven ! Here's Spivak's answer 1=a^-1*a=a^-1*(a*x)=(a^-1*a)*x=1*x=x Here's my interpretation line by line (just to be sure im understanding it) a^-1*a=1 a^-1*(a*x)=1 (Is my interpretation correct in saying that the number "a" can be factorized in a way that makes a=a*x, "x" being a variable which the value is to be determined) (a^-1*a)*x=1 1*x=1 x=1 So, basically, Spivak is constructing his proof. Is it correct ? Any opinions ? Thank you!