Is there a solution to x=x+1 (x is not a complex) maybe x is

>>8846894
Dumb anime poster. State your domain for x

>>8846900
If we allow for any domain whatsoever.... I suppose... I suppose you could define `mod 0` in such a way that `x` is congruent to `x+1 mod 0`. Kinda dumb though.

You can choose any domain i find the trivial domain of modulo 1 but is there other domain ?

>>8846894
>>>8846900
>If we allow for any domain whatsoever.... I suppose... I suppose you could define `mod 0` in such a way that `x` is congruent to `x+1 mod 0`. Kinda dumb though.
>>8846916
I don't think mod 0 is define cause mod is the reste by the euclidian division and you can't divise by 0

>>8846894
Define {1} with + and * as a group where 1+1=1 and 1*1=1

>>8846923
You are mostly correct, but I think it's possible to flex out definitions sufficiently to make it possible.

Consider the set of integers as a group. Taking cosets of a given subgroup would induce equivalence classes for which we may define the modulo operator. In this way, ........

wait no, then n mod 0 would just be n, not 0.

I can't think of a way to define mod 0 to always equal 0. I thought I could. OH WELL.

>>8846894
it's easy really.

x-1 = x = x+1

so x-1-1 = x = x+1
so x - 1 - 1 - 1 = x = x+1
and so on

so x = x + 1 + 1 + 1 + 1+ ...

Therefore x is anything and 1 + 1 + 1 + 1 + ... = 0

Numberphile can contact me any time for a video btw.

>>8846940
>>8846940
Maybe what you want to say is the trivial mod as
>>8846917 said
>>

>>8846958
mod 1 won't work. Even numbers are congruent to 0 mod 1; odd numbers are congruent to 1 mod 1.

Unless you define a domain with a broken field, an operation of +1 will toggle between even and odd.

>>8846967
mod 1 implies at least 2 elements
You want mod 0

>>8846971
I challenge you to define mod in such a way that n mod 0 = 0.

>>8846894
subtract x from both sides
0 = 1
nobel prize when

>>8846916
obviously for [math]\mathbb{Z}_1[/math] as >>8846917 said its sole element is a solution.

You also get 2 solutions with the extended reals and infinitely many with ordinal numbers

>>8846976
>>8846939 already did
The only thing I'd change is that where he uses the symbol '1', I'd use the symbol '0' instead.

Here are a few tricks you can use that I've learned from this thread:

>Define 1 as 0
>Define x as your mom
>Define = as my dick

and there you go

>>8846967
How are odd numbers congruent to 1 mod 1?

In the most basic sense where we define the congruence relation by remainder after division, we have that any integer mod 1 is 0 since 1 always divides into that number.

Mod 2 is the one that associates to odds 1 and to evens 0.

x, where x is defined as the answer to the question.

Another cheap solution is with boolean algebra

>>8846894
x=0
now where's my nobel?

>>8846894
Well, infinity is a solution since inf+1=inf, but I don't think that's a satisfying answer.

>>8846894
There are two examples where this comes up, and neither is particularly interesting.
Firstly, there's the zero ring, which is an algebraic structure with only one element. In a ring, there is no restriction saying that zero may not be equal to one. However, it does follow that any other element x = (1)x = (0)x = (0+0)x, so x+x = x, and cancelling out an x gives x = 0. In other words, the only element in the ring is 0. It's sort of a trivial structure similar to the empty set.
The other example is infinity, and all the different variants of infinity. For example, you can extend the reals (or the naturals, or the integers) with an element infinity such that infinity plus anything is infinity, infinity times anything non-zero is infinity, infinity times 0 is 0, infinity is bigger than everything else. Etc. You can set up negative infinity in the same way. The problem with this addition is that what you have is no longer a ring since the additive operation is no longer cancellable. 0 + infinity = 1 + infinity, but 0 is not equal to 1.
To give a less contrived example of the same idea, you can define addition on equivalence class of sets defined by A ~ B if and only if there is a one to one correspondence between elements of A and elements of B (the sets have the same size) to be [A] + [B] is the equivalence class of [X union Y] where X is in [A], Y is in [B] and X and Y are distinct. Under this construction, and associating the non-negative integers with the equivalence class of sets with that many elements, there are a large number of distinct "infinities" (sets which cannot be put into one to one correspondence with any natural number) for which X + 1 = X. That is, adjoining a single element to an infinite set does not change its size. Examples of infinite sets which are not the "same size" via this definition include the natural numbers, the real numbers, and the set of all sequences of real numbers.

[-1/2] + [1] = [-1/2 + 1] = [1/2]
[-1/2] = [1/2]

in some projective space, probably