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 D₂ₙ (the symmetry groups of regular n-gons) contain reflections, all of which have order 2. Symmetric groups Sₙ (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.
> A simple example where 1 + 1 + 1 = 1 is ℤ₂, the group of integers modulo 2 under addition.
That’s a good example of an algebra where 1 + 1 + 1 = 1, but the article is specifically about systems where in addition to that condition, this second condition is also true: 1 + 1 != 0 (not equal!). ℤ₂ is not an example of that.
> 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).
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.
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.
> 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.
In tropical geometry, tropical multiplication (⊙) is replaced by standard addition (+), and tropical addition (⊕) is replaced by the minimum (min) function, so 1⊕1⊕1=min{1,1,1}=1.
>there are other operations that aren’t quite involutions – “near-involutions”, one might call them2 – that nonetheless have the property that thrice is the same as once, four times is the same as twice, etc. ... Unlike mod-two arithmetic, which is about counting “zero, one, zero, one, zero, one, …,” the kind of counting that governs these operations goes “zero, one, two, one two, one two, …”
Interesting! Earlier I had a shower thought about "what would be an variant of idempotence?" That's where an operation has the same effect whether done one or many times.
One variant would be "has the same effect whether one two or many times". Another would be "can be in any one of two possible states after done one or many times" (as opposed to one possible state for idempotence). This looks like the latter!
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 D₂ₙ (the symmetry groups of regular n-gons) contain reflections, all of which have order 2. Symmetric groups Sₙ (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.
> A simple example where 1 + 1 + 1 = 1 is ℤ₂, the group of integers modulo 2 under addition.
That’s a good example of an algebra where 1 + 1 + 1 = 1, but the article is specifically about systems where in addition to that condition, this second condition is also true: 1 + 1 != 0 (not equal!). ℤ₂ is not an example of that.
[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).
> 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.
> 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.
> 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.
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.
I'd be good to give an example of where the 'only if' doesn't apply. If only to clear up the confusion.
Sorry, I had a mental skip. I was thinking of solutions to a+a+a=0, not a+a+a=a.
> 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
The intuitionistic negation example is so mind blowing. I can only wrap my head around it when I think about it in terms of functions and types:
Now it all just turns into function application :)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-...
I like Chapter 1 of Evan Chen's An Infinitely Large Napkin for some more theory.
https://web.evanchen.cc/napkin.html
In tropical geometry, tropical multiplication (⊙) is replaced by standard addition (+), and tropical addition (⊕) is replaced by the minimum (min) function, so 1⊕1⊕1=min{1,1,1}=1.
1 + 1 + 1 ≡ 1 (modulo 2)
In the modulo 2 congruence, 1 + 1 + 1 is the same element as 1.
The piece is about something else, cases when 1 + 1 + 1 ≡ 1 but 1 + 1 != 0
>there are other operations that aren’t quite involutions – “near-involutions”, one might call them2 – that nonetheless have the property that thrice is the same as once, four times is the same as twice, etc. ... Unlike mod-two arithmetic, which is about counting “zero, one, zero, one, zero, one, …,” the kind of counting that governs these operations goes “zero, one, two, one two, one two, …”
Interesting! Earlier I had a shower thought about "what would be an variant of idempotence?" That's where an operation has the same effect whether done one or many times.
One variant would be "has the same effect whether one two or many times". Another would be "can be in any one of two possible states after done one or many times" (as opposed to one possible state for idempotence). This looks like the latter!
1 as a boolean? true
modulo 2
I'm not a mathematician but I don't think that's right.
Terrence Howard has entered the chat
Shhh! Don't give them any ideas.
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."
XOR is a simple operation that shows that behavior.
you're wrong. 1 + 1 = 10
My son once asked me what 1+1 was equal to. I said, "two!"
He said, "no, it's eleven." And that's when I realized he's going to be a JavaScript coder.
Which of course is 0 mod 2.
And 0^0 x 2