Yog79 Posted May 3, 2020 Share Posted May 3, 2020 Since $\mathbb{R}$ is not first order definable, the statement $\forall x \in \mathbb{R}(x = x)$ is not a first order statement and thus not provable in ZFC. Does that mean we can assume $\exists x \in \mathbb{R}(x \neq x)$? If so, would this provide us with the basis for a field with one element? Link to comment Share on other sites More sharing options...

John Cuthber Posted May 3, 2020 Share Posted May 3, 2020 Am I the only one who thought of this? Link to comment Share on other sites More sharing options...

taeto Posted May 3, 2020 Share Posted May 3, 2020 The truth of the predicate \(x=x\) follows from logic, i.e., it is how equality is defined. It has nothing to do with \(\mathbb{R}\). The field \( ( \{0\}, +, \cdot ) \) with one element is kind of okay as a field, as it satisfies the important axioms. Yet most authors exclude it explicitly, because it is a cumbersome exception for some theorems. In this 'field' you can always divide by zero, for example.so it is a little too exceptional. Link to comment Share on other sites More sharing options...

Martin Rattigan Posted July 28, 2020 Share Posted July 28, 2020 When I learned it a field was required to contain a genuine herd. Link to comment Share on other sites More sharing options...

HallsofIvy Posted September 3, 2020 Share Posted September 3, 2020 Would "a field of daisys" constitute a herd? Link to comment Share on other sites More sharing options...

ahmet Posted October 25, 2020 Share Posted October 25, 2020 (edited) On 5/3/2020 at 7:22 PM, taeto said: ({0},+,⋅) this is an ideal (also it is one of trivial ideal) but not a field. Could you explain how you have defined . and + operations here? what is more ,I can additioally say that there was one theorem claiming that a matehmatical structure containing just 2 element also would not be field.(i.e. not only just one element but just two element is also insufficient to be field.) Edited October 25, 2020 by ahmet Link to comment Share on other sites More sharing options...

studiot Posted October 25, 2020 Share Posted October 25, 2020 1 hour ago, ahmet said: this is an ideal (also it is one of trivial ideal) but not a field. Could you explain how you have defined . and + operations here? what is more ,I can additioally say that there was one theorem claiming that a matehmatical structure containing just 2 element also would not be field.(i.e. not only just one element but just two element is also insufficient to be field.) I agree that the OP proposition does not constitute a field, since it does not contain the required inverses. However the smallest field contains just two elements, along with appropriate rules for addition and multiplication. An example would be {1, 0} with a De Morgan/Cayley table. Quote Wikipedia https://en.wikipedia.org/wiki/Field_(mathematics) Classic definition Formally, a field is a set F together with two binary operations on F called addition and multiplication.^{[1]} A binary operation on F is a mapping F × F → F, that is, a correspondence that associates with each ordered pair of elements of F a uniquely determined element of F.^{[2]}^{[3]} The result of the addition of a and b is called the sum of a and b, and is denoted a + b. Similarly, the result of the multiplication of a and b is called the product of a and b, and is denoted ab or a ⋅ b. These operations are required to satisfy the following properties, referred to as field axioms. In these axioms, a, b, and c are arbitrary elements of the field F. Associativity of addition and multiplication: a + (b + c) = (a + b) + c, and a · (b · c) = (a · b) · c. Commutativity of addition and multiplication: a + b = b + a, and a · b = b · a. Additive and multiplicative identity: there exist two different elements 0 and 1 in F such that a + 0 = a and a · 1 = a. Additive inverses: for every a in F, there exists an element in F, denoted −a, called the additive inverse of a, such that a + (−a) = 0. Multiplicative inverses: for every a ≠ 0 in F, there exists an element in F, denoted by a^{−1} or 1/a, called the multiplicative inverse of a, such that a · a^{−1} = 1. Distributivity of multiplication over addition: a · (b + c) = (a · b) + (a · c). Link to comment Share on other sites More sharing options...

ahmet Posted October 25, 2020 Share Posted October 25, 2020 (edited) 1 hour ago, studiot said: I agree that the OP proposition does not constitute a field, since it does not contain the required inverses. However the smallest field contains just two elements, along with appropriate rules for addition and multiplication. An example would be {1, 0} with a De Morgan/Cayley table. I think there should be three element at least. lets see 1+(-1)=0 but -1 is not element of {1,0} set. Thus we can say that additive inverse does not exist for "1" element in {1,0} in fact, the second operation should meet/satisfy group criteria to be field, and to be ring; first operation should also satisfy group criteria. Edited October 25, 2020 by ahmet Link to comment Share on other sites More sharing options...

studiot Posted October 25, 2020 Share Posted October 25, 2020 3 hours ago, ahmet said: I think there should be three element at least. lets see 1+(-1)=0 but -1 is not element of {1,0} set. Thus we can say that additive inverse does not exist for "1" element in {1,0} in fact, the second operation should meet/satisfy group criteria to be field, and to be ring; first operation should also satisfy group criteria. Read Here https://en.wikipedia.org/wiki/GF(2) Link to comment Share on other sites More sharing options...

ahmet Posted October 25, 2020 Share Posted October 25, 2020 (edited) 1 hour ago, studiot said: Read Here https://en.wikipedia.org/wiki/GF(2) no, this is not a group for multiplication operation.To be honest Galois's theorem does not specify the case for 2 ,which is prime number. But this prime number is unique prime number that is even number. There is one a very basic theorem that will not accept any set which has had just two elements to be a field. That basic theorem is given as below: Theorem: a ring having just two elements cannot be a group for/across multiplication operation.(Thus <H,+,.> cannot be a field) Proof: <H,+,.> is a ring. if <H,.> is a group, we should have at least x element in H such that for 0_{H}≠ b , a.x=b should be satisfied. when we care one other theorem ,which states 0_{H.}x=0H , we have b=0_{H and this is a contradiction. } thus, as any ring that has just two elements cannot be a group in regard to multiplication operation, it will also not be accepted as a field. I think Galois theorem is invalid specifically for 2. meanwhile, I think wikipedia might be not suitable for mathematical sciences ,where the issue is upper level than MSc or equivalent. (for Z_{2} there 0 is not invertible. ) Edited October 25, 2020 by ahmet Link to comment Share on other sites More sharing options...

wtf Posted October 25, 2020 Share Posted October 25, 2020 (edited) 1 hour ago, ahmet said: (for Z_{2} there 0 is not invertible. ) 0 isn't invertible in the real numbers either, but the reals are a field. 0 never has a multiplicative inverse. Write down the addition and multiplication tables for [math]\mathbb Z_2[/math] and you'll see it's a perfectly good field. Remember that 1 = -1, that's a misunderstanding you had earlier. 1 + 1 = 0 so 1 is its own additive inverse. Edited October 25, 2020 by wtf Link to comment Share on other sites More sharing options...

ahmet Posted October 25, 2020 Share Posted October 25, 2020 (edited) 49 minutes ago, wtf said: 0 isn't invertible in the real numbers either, but the reals are a field. 0 never has a multiplicative inverse. This is okay. ✔️ but what about the theorem appearing above. In Regard to the multiplication operation, Z_{2} - {0_{z2}} should satisfy group criteria.(Should be a group for multiplication operation) 2 hours ago, ahmet said: <H,+,.> is a ring. if <H,.> is a group, we should have at least x element in H such that for 0_{H}≠ b , a.x=b should be satisfied. is this also being satisfied? Here H=Z_{2} maybe you might be right, not sure. Because I am sure that my hodja had made me note that theorem. But already galois thorem is not specified for specific prime number (to be exempted),therefore you might be right. Edited October 25, 2020 by ahmet Link to comment Share on other sites More sharing options...

wtf Posted October 25, 2020 Share Posted October 25, 2020 18 minutes ago, ahmet said: is this also being satisfied? 1 x 1 = 1. That's the multiplication table for the nonzero elements. What's the problem? Link to comment Share on other sites More sharing options...

ahmet Posted October 25, 2020 Share Posted October 25, 2020 (edited) 22 minutes ago, wtf said: 1 x 1 = 1. That's the multiplication table for the nonzero elements. What's the problem? I need to contact my hodja for help. Could you wait for a while for the clarification. There should be some unclarified details. (And are we confident/sure that Z_{2} contains just two elements. Because this ring is defined by congruence or equivalence (I could not find the correct word to explain very well) but there are such details in our hand .... 23≡1 22≡0 .... (infinity) so, is this group (or claimed so) finite? as I said,I shall contact my hodja and make suitable clarification. Edited October 25, 2020 by ahmet Link to comment Share on other sites More sharing options...

studiot Posted October 26, 2020 Share Posted October 26, 2020 (edited) 20 hours ago, ahmet said: no, this is not a group for multiplication operation.To be honest Galois's theorem does not specify the case for 2 ,which is prime number. But this prime number is unique prime number that is even number. So I have finally conquered the MathML here for tables. Here are the Cayley Tables I referred to earlier. [math]\begin{array}{*{20}{c}} + & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array}[/math] So the rules are 0 + 0 = 0 0 + 1 = 1 + 0 = 1 1 + 1 = 1 These demonstrate the commutative and additive inverse requirements for both 1 and 0. [math]\begin{array}{*{20}{c}} X & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array}[/math] 0 X 0 = 0 0 X1 = 1 X 0 = 0 1 X 1 = 1 These demonstrate the commutative requirements for 1 and 0 and multiplicative inverse requiremtnts for 1 0 is not required to have a multiplicative inverse. Edited October 26, 2020 by studiot Link to comment Share on other sites More sharing options...

ahmet Posted October 26, 2020 Share Posted October 26, 2020 (edited) hi, my lovely hodja replied today. But probably.. as I understand she implies that if except unitary element (of first operation) any ring contains just two elements ,then this will not satisfy group criteria according to second operation. To be honest her e-mail is a bit mixed. I am not sure whether I should recontact her. But I see "loves" conclusional word instead something more formal (e.g. regards, best regards also sincerely etc. Generally I had seen "sincerely" if someone would act across me rather formal ,especially when they do not know me, but loves is a bit intimate (not like warmly "kind regards" (I mean this last one seems like both positive and negative or at the centre of these approaches ) she also accepts that just {1} set is a group according to multiplication operation and it is obvious. mmm,yes I reread the taken e-mail and decided that she would express that if except the unitary element of first operation(I mean the ring's zero) ,if we have just two element,then this (according to the second operation) will not meet/satisfy group criteria. Z_{2} satisfies group criteria , because this ring has just one element except ring's zero. Edited October 26, 2020 by ahmet Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now