He proved, that every system rich enough to contain “infinite arithmetics” is EITHER inconsistent (have paradoxes) EITHER have some non provable sentences.
Almost everybody thought—“Okay, okay, so we’ll always have nonprovables. It’s a shame, but what can we do?”
But this was not the only one explanation. The “Okay, okay so we can’t make the complete “infinite arithmetic” without a paradox. We must cease to even try that.”—flies at least as well.
He proved, that there is ALWAYS either at least one “A & ~A” (and therefore many) - either an unprovable theorem exists. Inside all those systems, which contain the “standard calculus”!
He didn’t prove an actual “A & ~A”, but that one always exists, if there are no unprovable theorems in those “standard calculus systems”.
He proved, that there is ALWAYS either at least one “A & ~A” (and therefore many) - either an unprovable theorem exists. Inside all those systems, which contain the “standard calculus”!
That isn’t actually grammatical English, and unfortunately some plausible guesses at reconstructing it produce things that are completely false. So here’s what Goedel actually he did: he proved that for any powerful enough system either there is at least one (hence many) A & ~A, or there are A for which neither A nor ~A can be proved by the system.
It’s not “almost correct”, it’s actually correct. I wasn’t trying to make it complete :-). Yes, the “powerful enough” criterion is about being able to represent arithmetic. But if you want an actually-complete statement then “with arithmetics” and “contain the ‘standard calculus’” really aren’t any better than “powerful enough”. (I can never remember the exact requirements and am too lazy to look them up for this sort of discussion, and it’s not like the details matter very much here.)
Finite sets of numbers have obvious problems, well explored in computer science (if you integer has be be expressed in 4 bytes, you can express only a finite set of integers. Or real numbers, for that matter).
Trivially, if your finite set ends at n, what is the sum of (n-1) + (n-2)? A plausible answer is “you can’t do that”, but that’s also an answer to any paradox whatsoever.
How do you feel about imaginary numbers, by the way?
A plausible answer is “you can’t do that”, but that’s also an answer to any paradox whatsoever.
No, this is not true. The only answer we have to paradoxes is—change you axiom system!
How do you feel about imaginary numbers, by the way?
There are no paradoxes there. At least no paradoxes which wouldn’t be present in the real numbers already.
SQRT(-1)=i is no paradox whatsoever. People might often say—“oh, well that’s paradoxical”.
No, it’s not. No A & ~ A here. It might be weird to somebody, that you can calculate square roots from negative numbers. I really don’t see why, but it might.
Weirdness and paradoxicality are two different things.
You can eliminate paradoxes by declaring operations which lead to them “illegal”
Precisely! You narrow your axiomatic system, hoping that one of contradictors will fall out. They’ve eliminated self containing sets from the Naive Set Theory and Russell’s paradox vanished. Hopefully.
Is i a number? If not, what is it?
It is not “what is it”, it is “how does it behave”. Whatever behaves as a number, is a number.
Does it have a sign?
If it behaves as it had a sign … And it does. So yes, it has a sign!
So, Gödel?
He proved, that every system rich enough to contain “infinite arithmetics” is EITHER inconsistent (have paradoxes) EITHER have some non provable sentences.
Almost everybody thought—“Okay, okay, so we’ll always have nonprovables. It’s a shame, but what can we do?”
But this was not the only one explanation. The “Okay, okay so we can’t make the complete “infinite arithmetic” without a paradox. We must cease to even try that.”—flies at least as well.
I don’t know about that. They key word is “useful”. I’m not quite ready to discard and forget Peano arithmetic.
Yes, it is useful. Like a buggy program may be useful. But there comes the time of refactoring the whole thing.
Perhaps on an entirely new architecture.
What “A & ~A” did he prove?
He proved, that there is ALWAYS either at least one “A & ~A” (and therefore many) - either an unprovable theorem exists. Inside all those systems, which contain the “standard calculus”!
He didn’t prove an actual “A & ~A”, but that one always exists, if there are no unprovable theorems in those “standard calculus systems”.
That isn’t actually grammatical English, and unfortunately some plausible guesses at reconstructing it produce things that are completely false. So here’s what Goedel actually he did: he proved that for any powerful enough system either there is at least one (hence many) A & ~A, or there are A for which neither A nor ~A can be proved by the system.
That is almost correct. But “powerful enough” isn’t clear at all at your statement.
“Powerful enough” means those infinite sets with arithmetics.
It’s not “almost correct”, it’s actually correct. I wasn’t trying to make it complete :-). Yes, the “powerful enough” criterion is about being able to represent arithmetic. But if you want an actually-complete statement then “with arithmetics” and “contain the ‘standard calculus’” really aren’t any better than “powerful enough”. (I can never remember the exact requirements and am too lazy to look them up for this sort of discussion, and it’s not like the details matter very much here.)
Not only arithmetic, but an arithmetic with arbitrary large numbers is required for the Godel’s Incompleteness Theorem.
You need at least “all integers”, that is an infinite set.
Maybe some incompleteness can be proved for the finite sets as well. But it’s not known to be so. Nobody proved that yet.
Finite sets of numbers have obvious problems, well explored in computer science (if you integer has be be expressed in 4 bytes, you can express only a finite set of integers. Or real numbers, for that matter).
Trivially, if your finite set ends at n, what is the sum of (n-1) + (n-2)? A plausible answer is “you can’t do that”, but that’s also an answer to any paradox whatsoever.
How do you feel about imaginary numbers, by the way?
No, this is not true. The only answer we have to paradoxes is—change you axiom system!
There are no paradoxes there. At least no paradoxes which wouldn’t be present in the real numbers already.
SQRT(-1)=i is no paradox whatsoever. People might often say—“oh, well that’s paradoxical”.
No, it’s not. No A & ~ A here. It might be weird to somebody, that you can calculate square roots from negative numbers. I really don’t see why, but it might.
Weirdness and paradoxicality are two different things.
I don’t mind weirdness, I hate paradoxicality.
You can eliminate paradoxes by declaring operations which lead to them “illegal” (or their outcomes to be undefined).
Is i a number? If not, what is it? Does it have a sign?
Precisely! You narrow your axiomatic system, hoping that one of contradictors will fall out. They’ve eliminated self containing sets from the Naive Set Theory and Russell’s paradox vanished. Hopefully.
It is not “what is it”, it is “how does it behave”. Whatever behaves as a number, is a number.
If it behaves as it had a sign … And it does. So yes, it has a sign!
Which is what?
Everything which behaves like a sign.