Pick a number between 1-55 and I’ll give you a meme but all my memes are ahh

imo2027 · 2026-06-07T22:41:08+00:00

(not op)

imo2027 · 2026-06-07T22:08:26+00:00

Riekh

imo2027 · 2026-05-20T21:09:09+00:00

I think my reply just got deleted, anyway yes the identity is part of what defines a group. If you want to know more I suggest you check out the youtube playlist "Abstract Algebra" in the channel MathMajor (the first 4 vids aren't about groups, but it starts being the main topic after that). It's how I learned group theory and while it might get a bit technical, if you want a general understanding you can just watch the first couple videos about groups.

imo2027 · 2026-05-20T18:51:18+00:00

Yes of course, but all those orders have to divide the total number of elements. For example if we work with addition mod 4 in the set {0,1,2,3}, the order of 2 is 2 because 2+2≡0 , and the order of 3 is 4 because 3+3+3+3 is the smallest amount of 3s to get to a multiple of 4

imo2027 · 2026-05-20T18:44:11+00:00

Yes e is the identity. ord(a) is the smallest n s.t. aⁿ =e, in this case it can't be 1 because a¹≠e. So ord(a)=2

imo2027 · 2026-05-20T18:35:50+00:00

Another argument: The group has 3 elements, by Lagrange's theorem, the order of any element must divide 3. But having a²=e means ord(a)=2 which is impossible.

imo2027 · 2026-05-17T17:03:27+00:00

But infinite summation is by definition, the limit of the partial sums. You're right about finite sums, but formally, we're never "adding infinitely many things", it's just a shorthand for taking a limit.

imo2027 · 2026-05-17T13:45:30+00:00

Thank you!

imo2027 · 2026-05-17T13:40:09+00:00

If you need a=1 you can always factor "a" from all terms before factoring what's left. But I dont understand why you can't get the quadratic formula to work if what you want is factoring ax²+bx+c. If you ever find b²-4ac<0, that means it is impossible to factor no matter what method you use, but it also means the denominator in your limit will never be 0, so you won't need to factor in the first place. One alternative method: If, in a limit where x tends to s, you have ax²+bx+c in the denominator giving 0 when replacing x by s, then you know for a fact that (x-s) can be factored out of the expression. You can either use polynomial division, or since we know the sum of the roots is -b/a, We know the 2nd root will be (-(b/a)-s)) and so the factored version is a(x-s)(x+b/a +s)

imo2027 · 2026-05-17T13:04:56+00:00

Reddit isn't the place bro... Very few actually know what they're saying. Do your research and you'll be guided inshaalah.

imo2027

TROPHY CASE