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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.

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  • $\begingroup$ What is the definition of order? That is, what is the definition of “$a\lt b$”? $\endgroup$ – Arturo Magidin May 17 at 23:30
  • $\begingroup$ a < b is defined as a <= b and a!= b and a <= b is defined as b = a + d for some natural number d $\endgroup$ – Donlans Donlans May 17 at 23:46
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Depending on the precise definition of the order, addition, and the properties you know, you can argue as follows:

You have $b=a+d$ for some $d$, with $d\gt 0$. Since $d\gt 0$, it is a successor, so $d=c\mathrm{++}$ for some $c\geq 0$. Thus, $$b = a+d = a+(c\mathrm{++}) = (a+c)\mathrm{++} \geq a\mathrm{++}$$ since $a+c\geq a$.

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  • $\begingroup$ thank you so much, I've been trying to solve this one for about an hour ago, I dont know why I didn't try with d = c++. $\endgroup$ – Donlans Donlans May 18 at 0:15

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