REEEEEE

That is why you have a base case.

>>8077604
First you show that its true for n=1
And then show that if its true for n (like n=1) its true for n+1.
Putting them together shows that n=2 is true, which implies n=3, which implies n=4, etc

>>8077612
So, using an unproven statement (the hypothesis), you construct another unproven statement during the induction step, which suddenly becomes valid due to the base case?

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Just read about the axiom of induction. I don't believe it. It's unintuitive black magic. What happens if I negate or omit it?

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>>8077604
It is like climbing a ladder.
One must be on the ladder. (Establish base case.)

If one knew that one were on a ladder, one could proceed to the next step, provided it were explained how to take that step, given the aforementioned assumption that one be on the ladder in the first place.

>>8077619
Thats why you prove the two statements.
They lead to a more general idea
>>8077629
You can't prove things for countably infinite sets
:^)

>>8077629
Does this relate to mathematical constructivism?

>>8077629

Well ordering principle is perfectly reasonable but it's surprising it cannot be proved from the other axioms.

You can prove it if you take a complete ordered field as a given, ironically.

>>8077642
I would assert that the result obtained from the inductive step is not a proof, because the hypothesis, the fact that it is true for all integers, was not proved.

>>8077649
you're not getting it.
you prove 2 things:
P(1)
P(n) => P(n+1)
having those two, you can prove P(5) by using
P(1) & P(1) => P(2) & P(2) => P(3) & ....
and same for any integer

>>8077654
Wouldn't it be necessary to prove that P(n) is true for all n?

> inb4 "the P(1) base case is true, thus making the precedent true"
then it's obviously not proven for all other n, thus preventing the "domino effect"

>>8077662

Assume it's not true, then there exists a smaller integer n for which it fails(by well ordering). But then its true for n-1 and by induction proof it's true for n, a contradiction, therefore it's true for all n.

QED,

>>8077666

I was wrong, I meant to say natural number.

>>8077662
no.
you have proved that P(1) and that P(n) implies P(n+1).
thus P(1) => P(2) and P(2) is true
so then P(2) => P(3) and P(3) is true

and so on
come on man

>>8077662
You fucking retard.
If you prove P=>Q, you've proven nothing about P or Q on their own. So when you show P(n)=>P(n+1), you're only showing that a certain relationship must hold in the situation that P(n) happens to be true. You are not saying that P(n) itself is true. However, the base case comes in handy because (P^(P=>Q))=>Q necessarily, so P(base)^(P(n)=>P(n+1)) necessarily implies P(base + 1)

>>8077662
There's a few things that is going on. First of all, the inductive hypothesis states "Let n be an integer. If theorem is true for n, then theorem is true for n+1." This is different from saying that the theorem is true for all n. It is only stating that the theorem is true for a case in which the preceding case is true. Proving that the inductive hypothesis is true is a completely valid step, as it alone does not prove the theorem. In order to use the inductive hypothesis, you need a base case, which allows for your "domino effect".

Ok. But ladders end and the snake will eventually fall of because of his belief in induction.

>>8077673
>>8077676
thanks a lot

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>>8077648
So does induction rely on the axiom of induction or is the latter only the formalisation of a true statement that can be proved otherwise (but why is it called an axiom then)?

>>8077699

The two are equivalent

>>8077604
calm down, breathe and read again.
Let us assume this hypothesis is true for AN integer N.
Which you MUST HAVE PROVEN already.

>>8077765
This is now understood. The only uncertainty I have is about the axiom (why it is required). It seems it is "only" a shortcut, to apply induction to all integers in one step.

r1 = a1
Representation = Abstract
Abstract = Representation
Induction by rule of observed constancy.
It is possible it's circular logic?
Yes, but circular logic isn't inherently wrong, it's only "wrong" when baseless presumptions are used.

>>8077793
It is not a shortcut. Although, given a valid base case and induction hypothesis, we are able to prove that our theorem holds for some finite subset of the natural numbers, by tedious construction. However, if we want to claim it holds for the (countably) infinite set of natural numbers, we require an axiom.

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>>8077642
>They lead to a more general idea
false. it is a choice to have the inference rule ''induction schema''

It is possible to believe the Peano axioms to be inconsistent, but only if you believe all models of Robinson arithmetic to be nonstandard.

In other words, if you believe every model of arithmetic has infinite descending chains.

Of course, even if you believe the latter, you can still believe PA to be consistent. There exist nonstandard models of PA.

Also, even if all models of arithmetic are nonstandard, if you believe in the consistency of ZF you believe in the consistency of PA.

So basically you have to be really daft to reject PA. You have to both reject set theory and visualize the natural numbers as having an infinite descending chain.

>>8077648
>>
>Well ordering principle is perfectly reasonable
>2016

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>>8077619
assume case n is correct
1+2+...+n = n(n+1)/2
how about n+1
[n]([n]+1)/2 + (n+1) = ....(lots of boring manipulation)
...= [n+1]([n+1]+1)/2
so that proves you can get from [n] to [n+1]

so now we have a car that has its spinning wheels lifted in the air, all we need to do is drop the thing to a base case spot and the contraption will hop +1 leaps from there on