# Fifth Peano axiom — Properties of the natural numbers

This question is kind of a follow-up question to this. I am also using Terence Tao's book and I still struggle to understand why the fifth Peano axiom is valid.

Tao defines the fifth axiom in the following way.

Axiom 2.5 (Principle of mathematical induction). Let P(n) be any property pertaining to a natural number n. Suppose that P(0) is true, and suppose that whenever P(n) is true, P(n++) is also true. Then P(n) is true for every natural number n.

He then continues to explain the axiom intuitively and prove informally that now we can create the natural numbers.

The informal intuition behind this axiom is the following. Suppose P(n) is such that P(0) is true, and such that whenever P(n) is true, then P(n++) is true. Then since P(0) is true, P(0++) = P(1) is true. Since P(1) is true, P(1++) = P(2) is true. Repeating this indefinitely, we see that P(0), P(1), P(2), P(3), etc. are all true - however this line of reasoning will never let us conclude that P(0.5), for instance, is true. Thus Axiom 2.5 should not hold for number systems which contain “unnecessary” elements such as 0.5. (Indeed, one can give a “proof” of this fact. Apply Axiom 2.5 to the property P(n) = n “is not a half-integer”, i.e., an integer plus 0.5. Then P(0) is true, and if P(n) is true, then P(n++) is true. Thus Axiom 2.5 asserts that P(n) is true for all natural numbers n, i.e., no natural number can be a half-integer. In particular, 0.5 cannot be a natural number. This “proof” is not quite genuine, because we have not defined such notions as “integer”, “half-integer”, and “0.5” yet, but it should give you some idea as to how the principle of induction is supposed to prohibit any numbers other than the “true” natural numbers from appearing in N.)

This is where he loses me.

To me, it looks like Tao just took the property we need to create the natural numbers "n 'is not a half-integer'" and therefore, of course this set must be the natural numbers.

However, as far as I can tell, the property is not defined, so with that train of thought, I could also start my induction at P(1) and state that every natural number is uneven. Since 1 is uneven, 1++ = 2 is uneven as well and so forth, which of course is false.

What am I missing? Why does the fifth axiom only allow for the natural numbers as we know them? And specifically: What does "property" really mean in this context so that I cannot prove anything? There must be a certain property that only non-decimal postive numbers have, but as far as I can tell, we cannot define such a property with the five peano axioms.

• Nope. If 1 is uneven, that doesn't mean 1++ is uneven. Commented Nov 18, 2022 at 4:51

Your example that all integers are odd fails because you can prove $$P(1)$$ but you cannot prove the induction step. In particular $$P(1)$$ does not imply $$P(2)$$
When the induction axiom is stated this way it does not really say that everything we are talking about is a natural number, where we informally define a natural as one that you get by applying successor to $$0$$ a finite number of times. It is stated this way because it is the best you can do in first order logic, which has very nice proof properties. Unfortunately, it does not let you rule out that there are more numbers than the naturals we know and love. It is possible there are what are called nonstandard numbers. I could define a statement with one free variable $$Q(n)$$ that means $$n$$ is either even or odd. That statement is provable by induction. I am allowed to conclude that all numbers are either even or odd, even the nonstandard ones (if they exist). It is a large and complex subject.
• The official notation is usually given in a description of Godel's proof because it is important in the numbering. I liked Nagel and Newman for an accessible intro but my edition was 50 years ago. It is otherwise not so important except to know that the operations you have available are $S,+,\times$. The induction axiom is really an axiom scheme. For any one place predicate $P$ you have $(P(0) and \forall n P(n) \implies P(n+1))\implies \forall n P(n)$ Yes, you can start with a higher base case $k$ by just saying $n \ge k \implies P(n)$ Commented Nov 18, 2022 at 5:23