why is 1+1=2? There is no logic sense.

Question

sgvxfcbxvxv · Answer

but y did we design it like that?

sgvxfcbxvxv · Answer

it is just so confusing

akshay_budhkar · Answer

it is an internationally acllaimed logic..
you learn it for now
you can design yours for later..

anonymous · Answer

hahahahahaha......

anonymous · Answer

empty set, {empty set},{empty set {empty set}},....

is one way of looking at it, you can then supply names for these as u like before implementing any other axiomatic properties such as successor.

The third one could be an ordered pair...

anonymous · Answer

This is the most basic set of axioms you need for a number system. http://en.wikipedia.org/wiki/Peano_axioms

anonymous · Answer

That isn't strictly true, there are number systems with less axioms...

anonymous · Answer

It is true if u want to insist on being able to do the vast majority of what we would today call mathematics..

anonymous · Answer

Sorry I meant to say the natural numbers.

anonymous · Answer

Here is an example...

anonymous · Answer

http://en.wikipedia.org/wiki/Presburger_arithmetic

anonymous · Answer

Yes but any number system that is so simple that it doesn't fall into Godel's category of either inconsistent or incomplete is too simple for meaningful mathematics.

anonymous · Answer

Or http://en.wikipedia.org/wiki/Robinson_arithmetic

akshay_budhkar · Answer

this question was a simple query!
its so complicated now!
lol
you guys rock estudier n alchemista

anonymous · Answer

It was a well founded question that goes to the heart of mathematics.

akshay_budhkar · Answer

right!

anonymous · Answer

What axioms should be included and what not is somewhat debatable.

Modern mathematics bases on ZFC where the C part is the Axiom of choice.

This together with infinity axioms allows eg the Banach Tarski "paradox" (cutting up a ball and rearranging it into 2 balls equal to the original).

anonymous · Answer

The Banach Tarski paradox is really just about whether Lebesgue unmeasurable sets exist or not.

anonymous · Answer

If you include the axiom of choice unmeasurable sets exist, otherwise they don't

anonymous · Answer

It is impossible to prove or disprove the axiom of choice based on the ZF axioms. Therefore the axiom of choice can be either taken or not depending on your preference.

anonymous · Answer

Godel has shown that it is impossible to prove or disprove AC based on ZF axioms with his incompleteness theorem.

anonymous · Answer

If it is an Axiom u don't have to prove it.

anonymous · Answer

What u mean is that both systems are consistent.

anonymous · Answer

I am talking about whether or not it is independent from ZF

anonymous · Answer

Anyways, there are important theories based on AC. Like every vector space has a basis (Zora's lemma) and the product of compact spaces is compact.

anonymous · Answer

True but many (most) of these same conclusions can be reached, say, in constructive mathematics (albeit in a more roundabout fashion).

At the end of the day, mathematicians as a group decide what they think is "useful" and go on accordingly.

Whether it should be like that continues to be a subject of debate.

anonymous · Answer

Since ZF together with AC isn't inconsistent and there are useful theories implied by AC, there doesn't seem to be a reason to reject it. It is useful and that should be enough.

anonymous · Answer

On that theory, u should go ahead and add large cardinals, lots of people think they are "useful" (that's twice now I have put "useful" in quotes).

anonymous · Answer

For myself, I think there is the sort of math that people who are not professional mathematicians need and then the other.