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

Question

This is one of the first exercises from Terence Tao's Real Analysis. Guys, I am a total newbie in university level mathematics. Could you write step-by-step proof to this lemma.

I have tried to look at other answers to this question, but I don't understand the logic behind the process. E.g this answer

All we need to show is that there is a unique a such that a++=k++. Well, clearly this is true for a=k, and so there is at least one, and for any b for which b++=k++, we have by axiom 2.4 that b=a, and so this a(=k) is unique.

Axiom 2.4. Different natural numbers must have different successors; i.e., if n,m are natural numbers and n≠m, then n++≠m++. Equivalently, if n++=m++, then we must have n=m.

Answer 1

So the lemma you want to prove is:

For every natural $k$ there is exactly one natural $a$ such that $a\text{++} = k\text{++}$.

This requires two axioms. One is axiom 2.4:

If $n\text{++} = m\text{++}$, then $n = m$.

The other axiom may not be directly stated. It is a consequence of the axiom of equality. The general form of it says that if $x = y$ and you apply the same process $P$ to both $x$ and $y$, then the result will also be equal: $P[x] = P[y]$. In this case the process is taking the successor:

if $n = m$, then $n\text{++} = m\text{++}$.

The proof of the lemma has two parts:

• First, it proves that there is at least one $a$ such that $a\text{++} = k\text{++}$.
• Second, it proves that there is at most one $a$ such that $a\text{++} = k\text{++}$.

The first part is proven by noting that $k$ itself is a natural number such that $k\text{++} = k\text{++}$. I.e., if we set $a = k$, then by the axiom of equality $a\text{++} = k\text{++}$. Hence there is at least one value for which this is true.

The second part is proven by assuming that $a$ and $b$ are both natural numbers such that $a\text{++} = k\text{++}$ and $b\text{++} = k\text{++}$. But by transitivity of equality, that means $a\text{++} = b\text{++}$, and by axiom 2.4, that means $a = b$.

So if you are given two solutions, in fact, they have to be the same solution hiding behind two nom-de-plumes.