..i failed to prove. [real analysis class, we've just begun and we're defining the natural numbers rigoriously as to aid us in proof writing for more difficult concepts later on] Let a and b be natural numbers. 3. If a>= b and b >=a then a = b 5. a < b if and only if a++ <= b For 3, I have: Assume a >= b and b >=a. Then by definition of order a = b + m for some natural number m, and b = a + n for some natural number n. Since a >= b, then a = (a+n) + m, which, by associativity, implies a = a + (n+m). By Lemma 1, a = a + 0, thus a + 0 = a + (n+m), so that 0 = (n+m) by the cancellation law. --no idea what to do next, or if i even did things correctly here. I had another idea here but it was circularly reasoned-- For 5, I have: a<b then a++ <= b Assume a < b. Then b > a. By definition, b = a + m for some natural number m, provided that b != a. Now we have to show that a++ <= b, or b >= a++. Then, by definition, it would suffice to show that b = (a++) + n for some natural number n. By definition of addition, b = (a+n)++. --stuck here-- Help?