[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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