Enough high school crap. It R Harderer

Let S be a set with two binary operations called addition and multiplication, satisfying
(1) S is an abelian group under addition
(2) S \ {0} is a group under multiplication (where 0 is the additive identity of S)
(3) x(y+z) = xy+xz for all x,y,z \in SSuppose that for some integer n > 0, 0 = 1 + 1 + ... + 1 (n times). Prove that for all x \in S, (1)x = x.
Proposed by T. J. Kaczynski, Evergreen Park, Illinois. Am. Math. Month. 71:689, 1964.

So prove (1)x=x for Integers mod n assuming only left distributivity?

Could be $\mathbb{N}$ as well.

I'm going to assume $\mathbb{N}$ is natural numbers, which don't form a group.

Shit, you're right. I must be rusty. Still, doesn't much matter what group it is.

Meh, sets forming groups are overrated. I'm much more interested in using a pipewrench to beat solutions out of
studentslearners than I am in correcting group definitions. Also, more fun to berate the teacher than answer the question.

I may even have read a proof of this in uni, I know we did quite a bunch of abelian groups. But this is one of the things I remember as a basic property, so I'm not sure which other basic properties are proven yet at this point.

Before I post anything here: How do I get the ****ing spoiler tags to work?

It's been a few years since my first semester (the first and last time I had to solve anything like that). I found a proof that looks ok, but I (think I) didn't have to use
for some integer n > 0, 0 = 1 + 1 + ... + 1 (n times).
so I suspect I made a mistake somewhere.
spoiler x 0 for any x \in S, since:
x y = x (y + 0) [0 is the additive identity]
x y = x y + x 0 [(3)]
therefore: (1) 0 = 0 > 0 + (1) 0 = 0 > (1) 0 = 0
(ii) (1) 1 = 1 [due to (2), since 1 is the multiplicative identity]
(iii) the rest follows via induction (assumption: (1) i = i):
(1) (i + 1) = (1) i + (1) 1 [(3)]
(1) (i + 1) = i + 1 [assumption and (ii)]
therefore: (i + 1) + (1) (i + 1) = i + 1 + i + 1
= (i + i) + (1 + 1) = 0 [using (1)][/spoiler]Where's the problem?
Also: Fuck you too, Discourse.

but I (think I) didn't have to use
@Forgotmylogin1 said:for some integer n > 0, 0 = 1 + 1 + ... + 1 (n times).
Oh, wait, I did, [spoiler]otherwise I couldn't have used induction at all.[/spoiler]Edit: My proof still doesn't work, though, since I buttumed S is a subset of N, which is not stated anywhere.

but I (think I) didn't have to use
(Spoilered for very very minor potential hint)
[spoiler]For the record, not all the hypotheses are necessary, but OP is how Prof. Unabomber posed the question (edited slightly to modernize the language)[/spoiler]

For any set at all satisfying the assumptions.

since 1 is the multiplicative identity
I wonder if that should follow by definition (if it does, the addendum to 2 is rather pointless), or whether it should be derived from (3) and (4).
Fairly trivial to do (5) from x(y + z) = xy + xz follows x(y1 + y2 + y3 + ... + yn) = xy1 + xy2 + xy3 + ... + xyn, as x(y1 + (y2 + y3)) = xy1 + x(y2 + y3) = xy1 + xy2 + xy3 and so forth with induction (left as an exercise to the reader),(6) since (4), for some integer n, x(1 + 1 + 1 + ... + 1 (n times)) = x*0 = 0. Thus, for x(1 + 1 + 1 + ... + 1 (n+1 times)) = x(0 (from (4) again)+ 1) = (from 3) 0 + x = (from (2)) x.Actually, I'm not sure (5) is necessary, but it's 7AM and I need to get to work.Oh hey, is that circular logic I see here? Screw that, will think further later.

By common convention, "1" is used to represent the identity of a group whose operation is written multiplicatively. Same with "0" and "+".

That'd be too easy (see my failed attempt).

[spoiler]For the record, not all the hypotheses are necessary[/spoiler]
[spoiler]That's just evil then. One of the first things I learned in college was to check whether I used everything that was given. One of the best sanity checks for proofs.[/spoiler]

@asdf said:
since 1 is the multiplicative identity
I wonder if that should follow by definition (if it does, the addendum to 2 is rather pointless), or whether it should be derived from (3) and (4).Neither (4) nor the hypothesis to be proven have any meaning if you do not assume that 1 denotes the multiplicative identity. So yeah, the comment in my proof was completely pointless.

(2) doesn't have sense either if you don't assume 0 is an additive identity, and yet, it deserves a mention.
Anyway, the solution dragged out from Am. Math. Month:
[spoiler]if z denotes 1, then z+z+zz = z(1+1+z) = z, so zz = 1. Now z(x + zx) = zx + x = x + zx, so either x + zx = 0, or z = 1. In either case (huh?  me), zx = x[/spoiler]
Elegant proof, but I'm still confused  [spoiler]if z = 1, then zx is most definitely not x. I assume what was meant is that z = 1 leads to contradiction with what we've asserted before?[/spoiler]

[spoiler]if z denotes 1, then z+z+zz = z(1+1+z) = z, so zz = 1.[/spoiler]
Ah, I noticed that proving that would make everything really easy, but didn't manage to prove it.
[spoiler]I assume what was meant is that z = 1 leads to contradiction with what we've asserted before?[/spoiler]
[spoiler]1 = 1 is definitely possible in an arbitrary group (e.g. mod 2), so there's no contradiction there.[/spoiler]

[spoiler]1 = 1 is definitely possible in an arbitrary group (e.g. mod 2), so there's no contradiction there.[/spoiler]
Ah, right. But [spoiler]how does it follow that if 1 = 1, then still zx = x? As stated in the proof, if z = 1 = 1, then x + zx might not be 0, and as such x != zx.[/spoiler]

OR != XOR
(I still don't get why we can draw the same conclusion in that case, though.)

OR != XOR
That's why
might not be 0
I assume it's provable that [spoiler]1 = 1 also implies (1)x = x[/spoiler], but it doesn't seem trivial.

Oops, I misread what you wrote.
I assume it's provable that [spoiler]1 = 1 also implies (1)x = x[/spoiler], but it doesn't seem trivial.
Maybe the OP can enlighten us?

(2) doesn't have sense either if you don't assume 0 is an additive identity, and yet, it deserves a mention.
I actually added that myself in an attempt to prevent such confusion (and then left it off for "1" since I thought the point was made). My bad lol.

[spoiler]1 = 1 iff 0 = 2, so 0 = x0 = x2 = x*(1+1) = x+x = x+1*x = x+(1)*x.[/spoiler]


[spoiler]z=0, so z * (1+ 1 + z) = z is really 0 * (1 + 1 * 0) = 0 which is 0 = 0[/spoiler]

[spoiler]I thought z=1? When did z become 0?[/spoiler]

[spoiler]Because i'm tryiung to mess with your head and am not attempting to solve the problem not comment on any proposed solution[/spoiler]

Seems that [spoiler]they just substitute z = 1 in the parentheses and end up with z(1 + 1 + 1) = z*1 = z[/spoiler].
I admit, it was a bit of a leap of logic to me too.


I admit, it was a bit of a leap of logic to me too.
[spoiler]I think the fact that there exists a value that acts like 1 is implicit in it being an Abelian group under addition (i.e., every member other than the additive identity has an inverse). Because we know that such a value exists, we can denote it as z and use the additive rules to establish that 1 + 1 + z = 1 (since 1 + z = 0 by the definition of inverses, and 1 + 0 = 1 again by definition). Which is quite enough to get that z(1 + 1 + z) = z*1 = z under the multiplicative rules.[/spoiler]
You have to watch out for words like “Abelian”.

But how does it follow that if 1 = 1, then still zx = x?
If 1 = 1, that implies that [spoiler]there are only two elements in the group, {0, 1}[/spoiler]. Working from the definition of 1, you then find that it must indeed be equal to 1.

Meaning the last assumption is actually:
[spoiler]Suppose that forsomeall even integers n > 0, 0 = 1 + 1 + ... + 1 (n times).[/spoiler](Ooh! Look at the pretty spoilers!)

[spoiler]Algebra over finite groups[/spoiler] is fun! It looks just like the ordinary thing, but isn't…

Suppose that for some integer n > 0, 0 = 1 + 1 + ... + 1 (n times).
Is that right? Shouldn't that read
Suppose that for some integer n > 0, n = 1 + 1 + ... + 1 (n times).


Oh! Some n, not any n. Misread that.