# Prove that for a positive number only exists one natural number such that $b{++} = a$

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?

• 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. Sep 18 '16 at 16:07