# Thread: how to prove 1+1=2?

1. how can we prove 1+1=2?we all know that one is one,and two is two,so what we should prove is either"+" or "=".
i choose to accept the "=",and prove the"+".
So if we define add as to keep parts as individuals and also to see them as a whole in the same time,will this prove?  2.

3. Originally Posted by allenyuang
how can we prove 1+1=2?we all know that one is one,and two is two,so what we should prove is either"+" or "=".
i choose to accept the "=",and prove the"+".
So if we define add as to keep parts as individuals and also to see them as a whole in the same time,will this prove?
1+1=2

turns out to be a definition. It is basically the first step in the definition of "+" that occurs when constructing the arithmetic of the integers from the Peano Axioms.  4. I can confirm that I have verified the result using a calculator.  5. I think 2 is not in itself an independent entity. What really is meant by 2 is a unit and another unit (1+1). Similarly 3 means a unit, another unit and yet another unit (1+1+1). All numbers are basically meant to be human instruments that describe our concepts of collections of unities. Therefore, we dint find that 1+1 and 2 are of the same measure, but that, we named 1+1 to be 2, or in other words, we defined 2 as 1 + 1. So 1+1=2 is not a result but a definition. If it is a definition, the question of proving it doesnt make any sense.  6. Originally Posted by J Rahman
I think 2 is not in itself an independent entity. What really is meant by 2 is a unit and another unit (1+1). Similarly 3 means a unit, another unit and yet another unit (1+1+1). All numbers are basically meant to be human instruments that describe our concepts of collections of unities. Therefore, we dint find that 1+1 and 2 are of the same measure, but that, we named 1+1 to be 2, or in other words, we defined 2 as 1 + 1. So 1+1=2 is not a result but a definition. If it is a definition, the question of proving it doesnt make any sense.
That is correct.

It is also a definition that 2+1 = 3.

But when you go through the development of arithmetic starting from the Peano ?Axioms there is actually something to prove that 1+2 = 3 (i.e. that (1+1)+1 = 1+ (1+1) ).  7. you can prove this through algebra. 1+ x = 2. 2-1=1. 1 = 1. x = 1. 1+1=2.  8. No you can't. Algebra is built on the definitions that 1+1=2, 1+2=3, etc. So trying to prove it by algebra would be circular.  9. Originally Posted by MagiMaster
No you can't. Algebra is built on the definitions that 1+1=2, 1+2=3, etc. So trying to prove it by algebra would be circular.
absolutely right.

In fact the algebraic properties of commutivity and associativity are a result of the construction of the number systems, so you have to to have that development in place before you can even talke sensibly about algebra.  10. if you don't know the definition of 1 or 2, isn't that effectively trying to prove x+x=y?  11. Originally Posted by schiz0yd
if you don't know the definition of 1 or 2, isn't that effectively trying to prove x+x=y?
no

If you don't know the definition of 1 or 2 this is gibberish.  12. Sorry DrRocket, I respect your knowledge and all, but I'm pretty sure the main concept of a variable is something that is has not been defined.  13. Originally Posted by schiz0yd
Sorry DrRocket, I respect your knowledge and all, but I'm pretty sure the main concept of a variable is something that is has not been defined.
Then you are certainly wrong.

A variable is nothing more and nothing less than a symbol that stands for some member of some set that has been defined at the outset of the definition of a problem.

If you don't have a definition for whatever variable that you are using the you are talking nonsense.  14. Before I go research the Peano Axioms, I would like to guess at the nature of this conversation. From my latest studies and using the context of the conversation I guess that;

2 IS NOT = 2

2 = 1+1

1 = 1

(1+1)(1) = 1+1

(1+1+1)1 = 1+1+1

(1+1)(1+1) = 1+1+1+1

(1+1)(1+1)+1 = 1+1+1+1+1

(1+1+1)(1+1) = 1+1+1+1+1+1

1-1x = x-1

x(1+1+1) = 3x

Either, I have a deep understanding of a very basic concept, or I am a fool making scratches in dirt. I really don't have the formal math training to make that judgement. If I've "created" something, I call it "Very Discreet Math".

Thank you for your consideration. 8)  15. Right track, but the details are a little off. It goes more like:

2 = 1+1
3 = 1+2 = 1+(1+1)
4 = 1+3 = 1+(1+(1+1))
.
.
.

Getting to 4 = 2+2 requires some more work.  16. Originally Posted by DrRocket Originally Posted by schiz0yd
Sorry DrRocket, I respect your knowledge and all, but I'm pretty sure the main concept of a variable is something that is has not been defined.
Then you are certainly wrong.

A variable is nothing more and nothing less than a symbol that stands for some member of some set that has been defined at the outset of the definition of a problem.

If you don't have a definition for whatever variable that you are using the you are talking nonsense.
so i think i understand now. are you saying that in 2+x = 3, the definition of x would be x=3-2? i was thinking that variables that you would have to solve for would be considered undefined, but now i'm assuming that the rest of the problem would act as the definition?  17. Originally Posted by MagiMaster
Right track, but the details are a little off. It goes more like:

2 = 1+1
3 = 1+2 = 1+(1+1)
4 = 1+3 = 1+(1+(1+1))
.
.
.

Getting to 4 = 2+2 requires some more work.
4=1+(1+(1+1)=((1+1)+(1+1))=(1+1)+(1+1)=2+2  18. You're using associativity without proving it exists. If you're starting that far back, it's not a given.  19. Ah yes, I have mucked up Mr Peano's axioms with algebra. But isn't the use of the distributive property in conjunction with caveperson hashmarks interesting? Okay, it stops being interesting when expressing any amount greater than ten or so. Although it is interesting to view all numbers above one as set's of polyreplicates of a single entity, it is very inefficient.

Now, are there proofs for the concepts of successor(addition by 1), set, and the number zero, assumed in Mr Peano's axioms? Is there a mathematical proof for the concept of 1, or have we then wandered inextricably into a realm of circular philosophy? 8)  20. Originally Posted by GiantEvil
Now, are there proofs for the concepts of successor(addition by 1), set, and the number zero, assumed in Mr Peano's axioms? Is there a mathematical proof for the concept of 1, or have we then wandered inextricably into a realm of circular philosophy? 8)
The Peano Axioms, or more formally the Zermelo Fraenkel axioms are axioms, hence the name. There is no proof of axioms. Since they allow one to construct arithmetic, one can apply Godel's theorems and also conclude that it is not possible to prove, using only the axioms and formal logic, that they are consistent.

Moreover, one does not prove concepts. One proves theorems, and there is a big difference. A concept is usually a definition, and definitions doe not require proof. Only statements regarding concepts require proof. Such statements are called theorems.  21. Also, I should add that when doing things like calculus there's an implicit "assuming the Peano axioms" (or some equivalent set of axioms) at the beginning of every proof. It's just that you rarely see it explicitly since you wouldn't be able to do much of anything without it.  22. Originally Posted by MagiMaster
Also, I should add that when doing things like calculus there's an implicit "assuming the Peano axioms" (or some equivalent set of axioms) at the beginning of every proof. It's just that you rarely see it explicitly since you wouldn't be able to do much of anything without it.
Essentially ALL of mathematics, unless explicity stated otherwise, assumes the Zermelo-Fraenkel axioms plus the axiom of choice. The Peano axioms are less formal, but the content is included in ZF.

You never see this stated in most cases, but it is correct, and the reason that you never see it stated is that ZF plus choice is universally accepted in the mathematics community.

The only exceptions are some arcane studies in symbolic logic, and the constructivist school of Errett Biship, which assumes ZF but does not assume the axiom of choice. But any tme you see anyone using even ordinary first-grade arithmetic, you can be sure that the Peano axioms or something equivalent have been assumed. You can't do addition of natural numbers, or even counting, without them.  23. Originally Posted by DrRocket
Essentially ALL of mathematics, unless explicity stated otherwise, assumes the Zermelo-Fraenkel axioms plus the axiom of choice. The Peano axioms are less formal, but the content is included in ZF.
What is the axiom of choice? What does that mean? Please enlighten me.  24. Originally Posted by J Rahman Originally Posted by DrRocket
Essentially ALL of mathematics, unless explicity stated otherwise, assumes the Zermelo-Fraenkel axioms plus the axiom of choice. The Peano axioms are less formal, but the content is included in ZF.
What is the axiom of choice? What does that mean? Please enlighten me.
It is the axiom that given any set of non-empty sets there is a function that assigned to each set a member of that set.

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

To most people this appears to be obvious, but it has implications that are not all obvious, and in fact seem paradoxical.  