My first time on Mathematics, thank you for your attention and patience.

Eventually, I decided to have a taste of mathematical analysis, useful for my University. My weapon of choice is Terence Tao's Analysis 1. I love its style. It is making me clear a lot of things that before were cause of headaches. Unfortunately, it does not provide exercise solutions allowing me to test the product of my efforts.

I am trying to solve the Exercise 2.2.2:

Let $a$ be a positive number. Then there exists exactly one natural number $b$ such that $b{++}=a$.

The author suggests to use induction. Frankly, I have no idea how to induct on such a problem. My approach used another way:

  1. $a \neq 0$ by definition of positive natural number;

  2. Peano's Axiom 4 states that different natural numbers must have different successors;

  3. Hence, if $b{++} = a$ and $c{++} = a$ then $b = c$, contradicting the lemma stating that $\exists! (b{++} = a)$.

    The same problem is faced in this question, but I found it quite confusing. Is my approach acceptable? How should induction work on such a problem?


You have uniqueness as a consequence of axiom 4.

In order to show existence, I'd still suggest to use induction (though it is a very trivial induction proof that does not even use the induction hypothesis)

  • 1
    $\begingroup$ I am a newbie of the field. I am afraid that I am still not capable to properly apply induction on this kind of problem. The base case is $b = 0$. If $b = 0$ and $b++ = a$, by Axiom 4 no other natural number $n$ can truly provide $n++ = a$, because otherwise $n = b$. Now let's suppose that we inductively proved that $n++ = a$. I could repeat what I have done before with the base case: if $m++ = a$ them $m = n$, and this is not possible by Ax. 4, but I am not convinced by the approach I tried. $\endgroup$
    – Worice
    Sep 18 '16 at 16:07

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