Mathematics

1:56 AM

What are non-real complex numbers called?

@Bennett i Think pure imaginary is used

@KasmirKhaan I thought that was when the real part $0$. So So $i+1$ would non-real and not purely imaginary at the same time.

pure imaginary means a real multiple of i

nonreal complex number is called just that

@Jacksoja you have ax=x, let xa=y, using associativty axa=y or ay=y

so now you have both left identity and right identity, so a is your "e"

@Bennett I would trust anon if I were you :D he is very good :D

@anon Anon ! can you check my other answer is good or not ? :D

your other answer?

2:02 AM

to jacksoja@

just few lines up ><

am also curious about that question , looks like a handsome Q

Let G be a finte set, closed under an associative operation, such that , ax =ay forces x=y and ua=wa forces u=w ( ie left and right cancellation laws)

prove that G is a group

@anon What can one do with unordered fields? Are there examples of unordered fields where addition and multiplication never return elements belonging to some ordered field?

you can do anything with an unordered field you can do with an ordered field that doesn't use the ordering

also, fields are closed under addition and multiplication

as are pretty much any algebraic structure

@KasmirKhaan I don't understand how Jacksoja "has" ax=x...

in any case, you absolutely need to use the fact the set is finite (else it may not be a group), so yeah pigeonhole principle

well hmm

I did solve it this way

f : G--->G , f(x) = ax

this map is injective and since G is finite it is also surjective

but without finding an identity sepratly i could not really go further

I had to have an identity to find inverse for each element

right. that's part of the answer. (dunno how jacksoja got it without essentially using that the set is finite is what I'm getting at)

:D

@anon okay I understand why you would say that =p fair point

btw if we want to complete the proof now

since the map left multiplication by a

is a 1-1 correspondance

e in G

ax=e for any x in G

hence we are done yeah ? :D

2:11 AM

@anon That they are closed algebraically means they return elements within the field, right? I suppose my question was asking whether there are unordered fields that don't contain elements within the field belonging to an ordered subfield. I don't know a lot of advanced maths, so my questions may not even make sense.

I should really work on such questions more ._. do you have a hard question of this flavor?

@KasmirKhaan I still don't see how you've established the existence of an identity element. once you have ax=x for some a and x, that doesn't establish a is an identity in G, because you need ag=g to be true for all g

Xander Henderson

I have no questions of that flavour, but I do have some fabulous chocolate flavoured exercises

@Bennett well, for example, finite fields can't be ordered, so any element of a finite field cannot be contained in a turn-into-an-ordered-field-able subfield

@anon well since G is finite , and we can safely assume that all elements are unique, so fixing one element , and taking all the left Products with that element would also yields unique elements ( because of cancelation law ) thus one of them must be equal to a, hence ax=a for some x in G

now doing the same thing but with right multiplication by a , so xa = y for some y in G

using associativity , ax=x, xa=y, axa=y or ay=y

wait wait wait -.-

i see that I got it backwards

i should end up with ya=y

2:18 AM

for a given choice of a, you've concluded that ax=a for some element x. for x to be an identity you now need gx=g and xg=g for all g in G

hmm yes yes :D

working on it now !

hint: show you can left-multiply ax=a by some element b to get gx=g (where g is any other element of G)

@anon Nice, thanks. It's hard to first see how unordered fields can be useful when there's no $>$ or $<$.

you add/subtract/multiply/divide polynomials and rational functions right?

we order monomials, but not really polynomials themselves

and integers mod n are also a common thing

computers use finite fields as number systems too

and finite fields have no ordering

@anon This bit is going over my head a bit.

The bit about monomials and polynomials.

Integers mod n is an amazing example. I can't believe I haven't thought about that.

2:30 AM

like {1,x,x^2,x^3,...} we can think of as being ordered, and you can put orderings on monomials with even more variables (like "lexicographic ordering"), but then it gets weird and arbitrary thinking of one multivariable polynomial as "bigger" than another

ax=a , axg=ag but by the cancelation law xg=g @anon

even better, good

yeeey :D

now go from xg=g (for all g) to gx=g