[HN Gopher] When 1+1+1 Equals 1
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       When 1+1+1 Equals 1
        
       Author : surprisetalk
       Score  : 27 points
       Date   : 2025-11-13 20:13 UTC (5 days ago)
        
 (HTM) web link (mathenchant.wordpress.com)
 (TXT) w3m dump (mathenchant.wordpress.com)
        
       | pyuser583 wrote:
       | Interesting. I always associate "1 + 1 = 1" with idempotency.
       | Here, "1 + 1 + 1 = 1", but "1 + 1 = 0".
       | 
       | I'm not a math whiz, so I'm just stuck with "1 + 1 = 2."
        
         | aatd86 wrote:
         | you're wrong. 1 + 1 = 10
        
           | lisper wrote:
           | Which of course is 0 mod 2.
        
         | harperlee wrote:
         | XOR is a simple operation that shows that behavior.
        
       | khannn wrote:
       | Terrence Howard has entered the chat
        
         | kevin_thibedeau wrote:
         | Shhh! Don't give them any ideas.
        
       | susam wrote:
       | A simple example where 1 + 1 + 1 = 1 is Z2, the group of integers
       | modulo 2 under addition.
       | 
       | In fact, in any group with binary operation, say +, the identity
       | element 0, and a non-identity element a, we have a + a + a = a if
       | and only if a + a = 0 (i.e. a has order 2).
       | 
       | There are plenty of groups with elements a satisfying a + a = 0.
       | Z2 as mentioned above has its unique non-zero element of order 2.
       | The Klein group V4 has three non-identity elements, each of order
       | 2. Dihedral groups D2[?] (the symmetry groups of regular n-gons)
       | contain reflections, all of which have order 2. Symmetric groups
       | S[?] (n >= 2) contain transpositions, each of which has order 2.
       | 
       | For example, in the dihedral group D8, if we let a be a
       | reflection of the square, then a + a = 0 and a + a + a = a. But
       | this is conventionally written in multiplicative notation as a2 =
       | the identity element, so a3 = a.
       | 
       | Similarly, in the symmetric group S3 under the binary operation
       | of composition, if a denotes the transposition (12), then
       | (12)(12) is the identity element and (12)(12)(12) = (12). In
       | other words, applying a transposition three times is the same as
       | applying it once.
       | 
       | In the last two examples, it is conventional to use product
       | notation instead of +, although whether we use + or * for the
       | binary operation does not matter mathematically. It is
       | conventional to use + in some subjects (coding theory, additive
       | groups of integers modulo n, etc.) and * in others (permutation
       | groups, dihedral groups, etc.). Often + is used for the binary
       | operation in abelian groups and * in non-abelian ones. I'm sure
       | none of this is particularly insightful to someone who has
       | studied group theory, but still I wanted to share a few concrete
       | examples here.
        
         | HWR_14 wrote:
         | [My post below is wrong]
         | 
         | > In fact, in any group with binary operation +, identity
         | element 0, and a non-identity element a, we have a + a + a = a
         | if and only if a + a = 0 (i.e. a has order 2).
         | 
         | The "if" is correct. The "only if" is not. (I assume that '+'
         | and '0' are used as shorthand for "any binary operation" and
         | "the identity of that binary operation", as I don't recall
         | cases where "+" and "*" are used for specific types of binary
         | operations).
        
           | susam wrote:
           | > The "if" is correct. The "only if" is not.
           | 
           | Both "if" and "only if" are correct.
           | 
           | Let a + a + a = a. Adding the inverse of a to both sides, we
           | get a + a = 0.
           | 
           | Let a + a = 0. Adding a to both sides, we get a + a + a = a.
           | 
           | > I assume that '+' and '0' are used as shorthand for "any
           | binary operation" and "the identity of that binary operation"
           | 
           | Yes. As I mentioned in my previous comment, "In the last two
           | examples, it is conventional to use product notation instead
           | of +, although whether we use + or * for the binary operation
           | does not matter mathematically."
           | 
           | In multiplicative notation, the statement becomes: a*a*a = a
           | holds if and only if a*a = e, where e denotes the identity
           | element.
        
             | HWR_14 wrote:
             | > mentioned this in my previous comment
             | 
             | You did. I'm sorry I glossed over the ending to your
             | comment. I was focused on a counterexample I was working on
             | and went only on my memory of group theory.
             | 
             | > Adding the additive inverse of a, i.e., -a from both
             | sides, we get a + a = 0.
             | 
             | That assumes associativity, but that's a nitpick, not a
             | real objection.
             | 
             | In reality, I got a bit tired and mentally shifted the
             | question to a + a + a = 0, not a + a + a = a. That of
             | course has numerous examples. But is irrelevant.
             | 
             | Thanks for taking the time for the thoughtful, and non-
             | snarky, response. Sorry if I was abrupt before.
        
               | susam wrote:
               | > That assumes associativity, but that's a nitpick, not a
               | real objection.
               | 
               | I don't think that is a valid nitpick. My earlier
               | comments assume associativity because a group operation
               | is associative by definition. If we do not allow
               | associativity, then the algebraic structure we are
               | working with is no longer a group at all. It would just
               | be a loop (which is a quasigroup which in turn is magma).
               | 
               | > Thanks for taking the time for the thoughtful, and non-
               | snarky, response. Sorry if I was abrupt before.
               | 
               | No worries at all. I'm glad to have a place on the
               | Internet where I can talk about these things now and
               | then. Thank you for engaging in the discussion.
        
               | HWR_14 wrote:
               | You are again right. I misrecalled a group as a loop.
               | 
               | Thank you again. It's been too long since I've had to use
               | this knowledge and am happy to have the opportunity to
               | (try to) use it.
        
           | patrickthebold wrote:
           | I'd be good to give an example of where the 'only if' doesn't
           | apply. If only to clear up the confusion.
        
             | HWR_14 wrote:
             | Sorry, I had a mental skip. I was thinking of solutions to
             | a+a+a=0, not a+a+a=a.
        
         | thaumasiotes wrote:
         | > The Klein group V4 has three non-identity elements, each of
         | order 2.
         | 
         | Unrelated, but this calls out for a link to the classic song
         | _Finite Simple Group (of Order Two)_ by the Klein Four:
         | https://www.youtube.com/watch?v=BipvGD-LCjU
        
       | kazinator wrote:
       | 1 + 1 + 1 [?] 1 (modulo 2)
       | 
       | In the modulo 2 congruence, 1 + 1 + 1 is the same element as 1.
        
         | discoinverno wrote:
         | The piece is about something else, cases when 1 + 1 + 1 [?] 1
         | but 1 + 1 != 0
        
       | voxleone wrote:
       | Beautiful math tricks. For things like these I think math
       | education should start with sets and groups instead of numbers.
       | 
       | https://d1gesto.blogspot.com/2025/11/math-education-what-if-...
        
       | CyberDildonics wrote:
       | I'm not a mathematician but I don't think that's right.
        
       | anthk wrote:
       | 1 as a boolean? true
        
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