I have to prove that a < b if and only if a++ ≤ b. I am using the book analysis 1 by Terence Tao, which unfortunately has no section for solutions of exercises.
both a and b are natural numbers, in this case, natural numbers start at 0. the notation a++ represent the successor of a number, for instance 0++ = 1 also, a <= b is defined as b = a + d, for some natural number d, and a < b is defined as a <= b and a != b I have tried to use the definition.
If a < b, then a != b and b = a + d, since a! = b, d > 0, then I tried to add 1 to both side and I got a++ < b++, which is obviously not what I have to prove, I have also tried with proof by contradiction but I didn't find any contradiction. Since I couldn't prove it, I tried with the converse theorem, if a++ <= b, then a < b, I used the fact that a < a++, and a++ <= b, a != a++, therefore a != b and by transitivity, a < b, I'm not sure if this is completely correct because I haven't proved yet that a < a++, and so far, the book hasn't mentioned that a < a++, though it is obvious. the axiom 2(peano's axioms) says that if a is a natural number, then a++ is also a natural number.