1. ## 1+1=0?

Within the following axiomatic system :

1) a+b=b+a......................................a.b=b .a

2) (a+b)+c=a+(b+c)................................... .(a.b).c=a(b.c)

3) a+0=a............................................. .......1.a=a

4) a+(-a)=0.................................... $a\neq 0\Longrightarrow \frac{1}{a}.a=1$

5) a(b+c)= a.b+a.c

6) $1\neq 0$

is it possible to have 1+1 =0 ??

2. ## Re: 1+1=0?

GF(2) obeys all these properties, i.e. binary arithmetic.

Note that in GF(2) $-x=x$

3. ## Re: 1+1=0?

If you are working in the Z2 set I think it's possible...

4. ## Re: 1+1=0?

Yes, any field of characteristic 2 will satisfy those properties. This includes $GF_2$ (the field with two elements) which is also the same as $\Bbb Z_2$, the integers modulo 2 (arithmetic of parity: oddness and evenness), as well as any extension field of $GF_2$ (for example, the field of rational functions with coefficients that are integers (mod 2), or $\Bbb Z_2(x)$).

While it may seem strange, if we think of 1 as acting on a light switch via the instruction: flip the switch, and 0 acting on a switch by: do nothing, and "adding" these functions is interpreted as: do one thing, then the next, we have:

1 + 1 = 0,

as flipping the switch twice, does the same thing as not flipping it at all.

We can view multiplication as: do one thing, or the other, whichever makes the most difference. This ensures the other axioms are satisfied as well.

5. ## Re: 1+1=0?

Originally Posted by romsek
GF(2) obeys all these properties, i.e. binary arithmetic.

Note that in GF(2) $-x=x$
Note the above axiomatic system is the one mentioned in nearly every high school book and sufficient enough to cover all the calculations between Nos .

One of the important theorems is the theorem : -(a+b) = -a-b ,provided that we define : a-b =a+(-b).

However ,since (a+b) = -( a+b) ,how should we simplify the following :

-{-[(a+b)-(b-a)]}

6. ## Re: 1+1=0?

Originally Posted by psolaki
Note the above axiomatic system is the one mentioned in nearly every high school book and sufficient enough to cover all the calculations between Nos .

One of the important theorems is the theorem : -(a+b) = -a-b ,provided that we define : a-b =a+(-b).

However ,since (a+b) = -( a+b) ,how should we simplify the following :

-{-[(a+b)-(b-a)]}
(a+b)-(b-a) = (a+b) + (-(b-a)) = (a+b) + (-(b+(-a))) = (a+b) + ((-b) -(-a)) = (a+b) + ((-b) + a) = ((a+b) + (-b)) + a = (a + (b +(-b))) + a = (a + 0) + a = a + a.

Since -(-c) = c, taking c = (a+b)-(b-a), we have: -{-[(a+b) - (b-a)]} = (a+b) - (b-a) = a+a.

The above is true is any field. In a field of characteristic 2, we have a + a = a(1) + a(1) = a(1 + 1) = a(0) = 0.

The axioms you describe are called FIELD AXIOMS, and while they are enough to do ALGEBRA, they are NOT sufficient to cover ALL calculations between ALL numbers.

For example, using ONLY the axioms listed, given that our field F = the rational numbers, we CANNOT solve:

(x)(x) = 2.

7. ## Re: 1+1=0?

Originally Posted by Deveno
(a+b)-(b-a) = (a+b) + (-(b-a)) = (a+b) + (-(b+(-a))) = (a+b) + ((-b) -(-a)) = (a+b) + ((-b) + a) = ((a+b) + (-b)) + a = (a + (b +(-b))) + a = (a + 0) + a = a + a.

Since -(-c) = c, taking c = (a+b)-(b-a), we have: -{-[(a+b) - (b-a)]} = (a+b) - (b-a) = a+a.

The above is true is any field. .
Thus IN ANY FIELD (which is the field in the OP) can we prove that : $a\neq -a$??

8. ## Re: 1+1=0?

Originally Posted by psolaki
Thus IN ANY FIELD (which is the field in the OP) can we prove that : $a\neq -a$??
No, Deveno, Prove It, and romsek all gave you examples of fields where any number $a$ in the field satisfies $a = -a$. Deveno even gave you more information that it is true for ANY field of characteristic 2. Additionally, in EVERY field, $0 = -0$ is always true, so for at least one element, you can NEVER prove that $a \neq -a$.

Edit: You can prove that given any field $F$ with $\text{char}(F) \neq 2$, then for all $a\in F\setminus\{0\}$, you have $a \neq -a$.

9. ## Re: 1+1=0?

Originally Posted by psolaki
Thus IN ANY FIELD (which is the field in the OP) can we prove that : $a\neq -a$??
I think you haven't had much exposure to finite fields. My guess is that you've only had experience with fields of characteristic 0, that is:

1+1+...+1 (n times) is not equal to 0 for ANY n.

This is the case for the rational numbers, the algebraic numbers, the real numbers and the complex numbers; all of which are fields of characteristic 0. In all of THESE fields, 1+1 is more commonly denoted as "2".

It turns out that any FINITE field (these are also called "galois fields") has characteristic $p$, where $p$ is a positive prime integer. Since the only prime where $\dfrac{p}{2}$ is also an integer is $p = 2$, fields of characteristic 2 are the ONLY fields for which:

$a + a = 0$, for any $a$ (which is the same thing as saying $a = -a$).

I will give you a small example of a field with 4 elements:

$F = \{0,1,a,a+1\}$. The addition is defined like so:

$0 + 0 = 0$
$0 + 1 = 1$
$0 + a = a$
$0 + (a + 1) = a + 1$
$1 + 1 = 0$
$1 + a = a + 1$
$1 + (a + 1) = a$
$a + a = 0$
$a + (a + 1) = 1$
$(a + 1) + (a + 1) = 0$ with all other sums defined by commutativity: $b + c = c + b$. It is a daunting task to manually show that all 256 pairs of triple sums are associative, but it can be done.

The multiplication is defined like so:

$0 \cdot 0 = 0$
$0 \cdot 1 = 0$
$0 \cdot a = 0$
$0 \cdot (a + 1) = 0$
$1 \cdot 1 = 1$
$1 \cdot a = a$
$1 \cdot (a + 1) = a + 1$
$a \cdot a = a + 1$
$a \cdot (a + 1) = 1$
$(a + 1)\cdot (a + 1) = a$, and products not listed are defined by commutativity. This is also associative, though showing it manually is just as exhaustive as proving associativity for addition.

The distributive law is likewise tedious to prove just from this definition, but can be done. This field is known as ${}GF_4$, and even forms a finite vector space, which you could view as the four corners of a square; where, when you get to a corner, you "keep going" by travelling on the "back side" of the square (the actual surface we embed this "square" in, is called the projective plane, which is impossible to physically display, it's sort of like a mobius strip with "no edges").

It's likely you may never have heard of these exotic structures. "Numbers" probably aren't what you think they are.

10. ## Re: 1+1=0?

Originally Posted by SlipEternal
No, Deveno, Prove It, and romsek all gave you examples of fields where any number $a$ in the field satisfies $a = -a$. Deveno even gave you more information that it is true for ANY field of characteristic 2. Additionally, in EVERY field, $0 = -0$ is always true, so for at least one element, you can NEVER prove that $a \neq -a$.

.
Hence the equation 2.x=0 has no solution within the field of the OP

11. ## Re: 1+1=0?

Originally Posted by psolaki
Hence the equation 2.x=0 has no solution within the field of the OP
The OP is a general set of axioms for ANY field. "The field of the OP" is not defined as a single field. EVERY field (including all fields of characteristic 2) satisfy those axioms.

Edit: Also, $2x=0$ has the solution $x=0$ in EVERY field (so it is incorrect to say it has no solutions in any field).

12. ## Re: 1+1=0?

Originally Posted by Deveno
"Numbers" probably aren't what you think they are.
this is a great line

13. ## Re: 1+1=0?

The OP makes no provision for the introduction of integers, or systems containing integers or using integers in their description. It just specifies symbols, including 0, and 1, and certain axioms.

Birkhoff, for example, proves from the OP axiom system that (-a)(-b)=ab. There is no mention of integers or numbers.

For example, 1='0 -> 1+1 =' 0. NO. Because you can't conclude 1+1 =' 0 from the axiom system.

14. ## Re: 1+1=0?

Originally Posted by Deveno
(a+b)-(b-a) = (a+b) + (-(b-a)) = (a+b) + (-(b+(-a))) = (a+b) + ((-b) -(-a)) = (a+b) + ((-b) + a) = ((a+b) + (-b)) + a = (a + (b +(-b))) + a = (a + 0) + a = a + a.

Since -(-c) = c, taking c = (a+b)-(b-a), we have: -{-[(a+b) - (b-a)]} = (a+b) - (b-a) = a+a.
a+a belongs to the system. a+a=c

15. ## Re: 1+1=0?

Back to a public computer, since I was locked out of the internet after I left this site the last time. Took a while to get back on. Happened before. Scary.

The OP is very interesting. I assumed it was trivial: assume 1+1=0, -> a= -a all a, and that's only possible if a=0. Yet I couldn't prove it for the given axiom system. Would love to see a proof.

As for the other systems mentioned in this thread,
1) Is multiplication and division defined?
2) Are all the OP axioms satisfied?

EDIT: I see Deveno did attempt to do so in post #9. I don't see 1/a defined, or proof that (1/a)xa=1, nor do I see subtraction.
Assuming it can be done, can you assume that if you can come up with a specific system that satisfies all the axioms, the proof is true for any system that satisfies all the axioms, intuitively it looks like you can.

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