Why is $1 + 1 = 0$ in $(\{0, 1\}, +, x) \subset \mathbb{R}$, which is the smallest field in $\mathbb{R}$.

  • $\begingroup$ The smallest subset of any set is $\emptyset$. Are you actually asking about fields? $\endgroup$ – Henry Oct 24 '11 at 21:34
  • $\begingroup$ Yes, sorry, fields is what I meant. Better? $\endgroup$ – Lenar Hoyt Oct 24 '11 at 21:36
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    $\begingroup$ $\mathbb K$ is a field, and $0$ is the neutral element with respect to the addition.. I think there are no many places in which $1+1$ can be sent. Indeed $1+1=1=1+0$ would imply $1=0$ but in a field this is not allowed. $\endgroup$ – uforoboa Oct 24 '11 at 21:44
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    $\begingroup$ But note, this $\{0,1\}$ is a field, but it is not a field "in $\mathbb R$". $\endgroup$ – GEdgar Oct 25 '11 at 0:39
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    $\begingroup$ @GEdgar {0, 1} could well come as a subset of R, given 0 as a real number, and 1 as a real number. "+" just isn't real number addition. So, ({0, 1}, +, x) could qualify as a field where the set comes as a subset of the set of all real numbers. $\endgroup$ – Doug Spoonwood Oct 25 '11 at 3:02

What you seem to ask is why $\{0,1\}$ with addition in which $1+1=0$ is a field.

Well, to verify that an object is a field we only need to verify a few axioms. First let us write the complete table of addition and multiplication (this can be skipped if we use an alternative definition, such as $\mathbb Z/2\mathbb Z$):

$$\begin{array}{ c | c |} x + y & 0 & 1 \\\hline 0 & 0 & 1\\ 1 & 1 & 0\\ \end{array} \begin{array}{ c | c |} x \times y & 0 & 1 \\\hline 0 & 0 & 0\\ 1 & 0 & 1\\ \end{array} $$

It is a matter of writing the possibilities down, but it is not very hard to show that addition and multiplication are:

  1. Commutative, that is $x+y=y+x$ and $x\times y=y\times x$.
  2. Associative, that is $x+(y+z) = (x+y)+z$, and similarly for the multiplication.
  3. We have that $0$ is the neutral element in addition, $x+0=x$, and that $1$ is neutral for multiplication so $x\times 1=x$.
  4. We can see that $x\times (y+z) = (x\times y)+(x\times z)$.

All these can be shown by checking all the possible cases, which is not too much work by hand.

From this we have that indeed $\{0,1\}$ behave as a field. And indeed $1+1=0$.

As for why it cannot be any other way? Well, if $\mathbb F$ is a field, and $x\in\mathbb F$ then the function $f(y)=y+x$ is injective. From this we have that since $0+1=1$, it cannot be that $1+1=1$ as well, so it has to be $0$.

  • $\begingroup$ Your multiplication table has the entries in the bottom row switched. $\endgroup$ – Quinn Culver Oct 24 '11 at 23:34
  • $\begingroup$ @Quinn: Thanks! I knew it looked strange :-) $\endgroup$ – Asaf Karagila Oct 24 '11 at 23:35
  • $\begingroup$ Is $0 + 1 = 1 \Rightarrow 1 + 1 ≠ 1$ the only proof? $\endgroup$ – Lenar Hoyt Oct 25 '11 at 17:00
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    $\begingroup$ @mcb: No, there can be other proofs. For example, $a+a=a\Rightarrow a=0$, since $1\neq 0$ it means $1+1\neq 1$ so $1+1=0$. $\endgroup$ – Asaf Karagila Oct 25 '11 at 17:41

The smallest field doesn't necessarily have to have elements with the names $0$ and $1$. Any field of two elements under an operation $\oplus$, say $(\{a,b\}, \oplus)$ will have a unique identity $e$ of $\oplus$ so that $a \oplus e = a$ and $b \oplus e = b$. Now $e$ must be either $a$ or $b$; suppose it's $a$.

Then $b$ must have the property that $b+b=a$, because if it didn't, then you would have the formula $b+b=b$, and you would conclude that $b=e$, an impossibility because the identity element is unique.

So in the particular case of $(\{0,1\},+)$, $1+1=0$ because there's nothing else for it to equal due to the uniqueness of the identity element $0$.

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    $\begingroup$ But the two elements are the $0$ and the $1$ of the field no matter what we called them outside the field context. $\endgroup$ – hmakholm left over Monica Oct 24 '11 at 22:57
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    $\begingroup$ To say something is "the $0$" and "the $1$" are informal statements that makes a metaphor with numbers to make it easier to understand the field. To use the terminology of field theory, you should say "$a$ is the additive identity" and "$b$ is the multiplicative identity" to express these ideas for the field in my answer. $\endgroup$ – tomcuchta Oct 24 '11 at 23:01
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    $\begingroup$ Perhaps. It depends on whether your field axioms mention $0$ and $1$ as constants (nullary operations). Some axiom sets do, others don't. $\endgroup$ – hmakholm left over Monica Oct 24 '11 at 23:13

It isn't. The smallest subset of $\mathbb R$ is $\varnothing$, not $\{0,1\}$. And as long as you're thinking of subsets of $\mathbb R$, then $1+1$ is $2$ and nothing but $2$.

On the other hand $1+1=0$ in the ring $\mathbb Z/2$ of residue classes modulo 2. That is because $(1+1)-0=2$ is a multiple of $2$. But that has nothing in particular to do with $\mathbb R$.

  • $\begingroup$ He was speaking about fields, not about subset $\endgroup$ – Evgeny Savinov Oct 24 '11 at 22:15
  • $\begingroup$ Sorry, I didnt see edit. $\endgroup$ – Evgeny Savinov Oct 24 '11 at 22:18

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