is there any base where 1+1=0? is this a legit proof?

Question

i think not, and i want to prove that there isn't such base.

is this a legit way to prove:

$$ \begin{split} 1_x+1_x=0_x &\implies 1 \cdot x^0+1\cdot x^0=0\cdot x^0 \quad \text{conversion to decimal base}\\ &\implies 1+1=0 \\ &\iff 1=-1 \quad \text{this is false} \end{split} $$

Answer 1

This is ok for any base > 2.

• In unary base, $1=0$ is true because there is only one number.
• In $\mathbb{Z}_2$, you have $1+1 = 0$ since $2 \equiv 0 \pmod{2}$.

UPDATE

In this context, unary means a field with one element $0$ with $0+0=0=0 \cdot 0$.

Answer 2

The set of integers $N$ obtained from Peano's axioms is defined independently of a base, and in this set, $1+1 \neq 0$.

Writing in base 2, $1+1=10\neq0$

Now, the boolean field $Z/2Z$ is defined in such a way that equality is a modulo 2 comparison.

Answer 3

No, there is no base where $1+1=0$.

Alluding to other "interpretations" to your question, in Binary $1+1=10$ not $0$.

For a proof verification, we need your axioms. \

Answer 4

It depends on what your question actually is.

Question 1: Is there any base, $b$ representation of the integers, $\mathbb Z$ in which $1_b + 1_b = 0$.

Answer: No.

And your proof is fine.

But $1_b$ is always equal to $1*b^0 = ONE$ and $0_b= 0*b^0 = ZERO$ and $ONE + ONE = TWO \ne ZERO$ no matter what representation you use to express them in.

And it doesn't matter what base or language we use to represent the numbers. That never changes what the numbers are. The $1 + 1 =2$ whether it is "une" + "une" = "deux" or $1_2 + 1_2 = 10_2$ or $I + I = II$.

Question 2: Is there any system in which $1 + 1 = 0$.

Answer: Yes. But not the integers and not with integer addition.

Simplest system is boolean algebra where you have two values total. $0$ and $1$ and two binary operations[1] $+$ and $\cdot$ and the following are given: $0+0=0; 0+1=1; 1+0=1; 1+1=0$ and $0\cdot 0=0; 0\cdot 1=0; 1\cdot 0=0; 1\cdot 1=1$. (This is often called $\mathbb Z_2$ or $\mathbb Z/2\mathbb Z$.)

[1](a "binary" operation means you combine two elments; not that you have two elements, BTW)

Another system would be the algebraic group $\mathbb Z_2\times \mathbb Z_2$ where you have for elements: $0,1, 10, 11$ and an operation $+$ and it is defined that $a+b = b+a$ and $0+0=0; 0+1=1; 0+10=10; 0+11 = 11; 1+1=0;1+10=11; 1+11= 10; 10+10=0; 10+11=1; 11+11 =0$.

(Another way to think of this is $0=(0,0), 1=(0,1), 10=(1,0), 11=(1,1)$ and $(a,b)+(c,d) = (a+b, c+d)$ where $0+m = m+0 = m$ and $1+1 = 0$.)

But these are different algebraic systems and are not "numbers" as you know them.

Answer 5

When thinking about base representations of numbers you have to be careful to distinguish between the actual number (a mathematical abstraction) and the various ways we choose to write or name the number.

Let's suppose for the moment that the abstract numbers are collections of stars, with the symbol "$Z$" for "no stars". Then in everyday notation

$************$ is $12$

While in base $***$

$************$ is $110$

The digits "$0$", "$1$", "$2$" and so on are just symbols. We need as many of them as the base we decide on and the usual conventions for the columns, corresponding to units, multiples of the base, the square of the base and so on.

In arithmetic addition means just "combine the groups of stars" so we always have $$ * + * = ** $$ and we never have $$ * + * = Z . $$

So your assertion is correct, but your proof is not. There is really nothing to prove.