Peano axioms for natural numbers - prove 0.5 ∉ N

[ato]

i am studying real analysis from terence tao lecture notes for analysis I. http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/

from what i understand , property is just like any other statement. for example P(0.5) is P(0) with the 0s replaced with 0.5 . so the notes says (assumes ?),

$P(0.5)\textrm{ is unprovable}\Rightarrow0.5\notin N$

i mean its alright to assume something like but i just want to make sure that what i understood is correct. if it is why not just assume something like this,
$0.5\in N\textrm{ is unprovable}\Rightarrow0.5\notin N$

but i might be wrong, so in that case could you prove 0.5 ∉ N.

thank you

 

[nomadreid]

Since you did not give a page number, I cannot tell what property P you are referring to, but I would presume that P(x) is something like x =0 $\vee$$\exists$y (y$\in$N $\wedge$ x=y+1).

Before I answer the question further, please tell me what P(.) is.

 

[ato]

ok, its page 15 on Week 1: Introduction to analysis; the natural number system; induction; the integers; the rationals .

the notes has mentioned four axioms to construct the set ,
1. 0 is natural number
2. n is natural number ⇒ n++ is natural number
3. n is natural number ⇒ n++ ≠ 0
4. (n,m are natural numbers and n ≠ m) ⇒ n++ ≠ m++

as far as i understand P(.) should be ..00 if P(0) is 0000 and P(1) is 1100 .

 

[nomadreid]

OK, all clear: P is simply the variable in the axiom of induction, which can be seen formally here: http://en.wikipedia.org/wiki/Mathematical_induction#Axiom_of_induction. (Note that this does not require the domain of P to be N; it can only prove P for non-negative integers.) He omits a statement (or leaves it implicit) that "this is all the natural numbers": that is, if a number does not satisfy all the axioms, then it is not a natural number. (Here I follow the convention that 0 is included in the set of natural numbers. Some places don't, and just call N the set of non-negative integers.) So the author's reasoning is basically that 0.5 does not satisfy the axioms, hence is not a natural number. Therefore, the statements you listed are equivalent if you take P(x) being "x$\in$N".

 

[ato]

Therefore, the statements you listed are equivalent if you take P(x) being "x∈N".

i was following alright until this. do you mean this is correct,
$P(x)\textrm{ is unprovable for }x\in N\Rightarrow x\notin N$
but then P(x) would never be unprovable (hence redundant) because $P(x)$ is true for $x\in N$.
why would x∈N assumed as condition ? would not this require N to be known.

please clarify.

 

[nomadreid]

Remember my parenthetical remark that x does not have to be in N. You have introduced "$\in$"N where it wasn't before. That is, you had two statements
(1)

P(0.5) is unprovable⇒0.5∉N

and
(2)

0.5∈N is unprovable⇒0.5∉N

This latter quote is an instance of the addition that I mentioned was implicit,
(3)

"this is all the natural numbers".

and hence not surprising.
I suggested
(4)

take P(x) being "x∈N".

Applying (4) to (2), (2) morphs into (1). (And, of course, the contrary.) That's what I meant by the equivalence.
Applying (4) to your new statement

P(x) is unprovable for x∈N⇒x∉N

would give
(x∈N is unprovable for x∈N)⇒x∉N
which is a completely different statement, and would take us into that interesting area about true but unprovable statements... to quote from The Never-Ending Story (a good title for mathematics), "But that is another story and shall be told another time."

 

[ato]

got it, thanks