1. The problem statement, all variables and given/known data I've been steadily working through Apostol's Calculus, Volume 1 and trying to prove all of the theorems and solve all of the exercises, but I skipped a few theorems and exercises in the Field Axiom section (I 3.2 & I 3.3). The lack of continuity irks me, so I sometimes flip back to that section and reattempt the problems, only to find myself in the same predicament. I'll post a few of the problems in hope that some one can shed some light on the situation. :) [itex](a/b) + (c/d) = (ad + bc)/(bd)[/itex] if [itex]b \neq 0[/itex] and [itex]d \neq 0[/itex] [itex]-(a + b) = - a - b[/itex] 2. Relevant equations Apostol says that we can only use Axioms 1-6 and Theorems I.1 through I.4 for the first problem. This includes commutativity, associativity, distributivity, existence of identity elements, existence of negatives, existence of reciprocals as axioms, the cancellation law, and the possibility of subtraction which states that there is a unique real number [itex]x[/itex] such that [itex]a + x = b[/itex] for real numbers [itex]a[/itex] and [itex]b[/itex]. The second problem can use the Axioms and all of the theorems. 3. The attempt at a solution I proved the first equality using Theorem I.7 (the possibility of division), which is not on the included list, and some shoddy applications of associativity. But I cannot seem to solve it only using the given axioms and theorems. There exists a unique [itex]x[/itex] such that [itex](bd)x = ad + bc[/itex] and it is denoted as [itex](ad + bc)/(bd)[/itex]. Then I replace [itex]x[/itex] with [itex](a/b) + (c/d)[/itex] and it comes out to [itex](ad + bc)[/itex], so it must be equal to [itex](ad + bc)/(bd)[/itex], as asserted. To prove the second equality I noticed that [itex]-(a + b)[/itex] is the negative of [itex](a + b)[/itex]. I also assumed that the negative of a real number is unique because of the possibility of subtraction, but this might be a formally weak argument. Since we know that there is a unique real number [itex]x[/itex] such that [itex]a + x = b[/itex] for real numbers [itex]a[/itex] and [itex]b[/itex], then if we are given a real number [itex]a[/itex] and [itex]0[/itex], there is a unique number [itex]x[/itex] (denoted as the negative of [itex]a[/itex]) such that [itex]a + x = 0[/itex]. If this is not sufficient or ideal, please let me know. So then we replace [itex]x[/itex] with [itex](-a -b)[/itex] and show that [itex](a + b) + (-a -b) = 0[/itex]. It ends up working out, but I am slightly concerned that my applications of the associativity axiom were not concrete. The associativity axiom states that [itex]x + (y + z) = (x + y) + z[/itex], so we can take [itex](a + b) = x[/itex], [itex]-a = y[/itex] and [itex]-b = z[/itex] and then everything works out. Is that logically sound? I thought it was until I read the next exercise which reads [itex](a - b) + (b - c) = a -c[/itex], which seems to follow from the same argument. So perhaps I cannot assume the above application of associativity for that problem. Any help is greatly appreciated. Thanks!