When your teacher asks you to solve x2 + x + 1 = 0 and you answer x = 0 because they didn't specify where to solve it...

They talk to me with memes, I answer with memes too

>just google it, monoid
FTFY.

>For The First Year

That's some 9geg-tier meme user.

>For The Fourteenth Year

Your reasoning is flawed. The trivial ring contains one element, 0.
This can't be solved in the trivial ring because +1 in the left hand side of the equation is not defined.

Except this is (almost) correct.

Sure, OP is dumb. But if he is working in the trivial ring then he could simply denote the element of the trivial ring as '1' and then he could solve it with x=1

Symbols are just symbols after all.

I have a question regards imaginary numbers,

x^2 + 1 = 0

At the time our teach always solved with the quadratic formula for some reason. Went online to jewtube and found a lot of people do the same for some reason.

Why can't I just say x := sqrt(-1)

Thus,

(sqrt(-1))^2 + 1 = 0

sqrt(-1) => i

i^2 + 1 = 0

i^2 => -1

-1 + 1 = 0

0 = 0

Is that okay vs the stupid quadratic formula? Am I being a nigger by just introducing the squareroot of a negative?

Not him, but it would be unusual for the OP to use both 0 and 1 in his equation if he were to define 0 = 1.

Taking a square root like that gives two solutions: +i and -i. I have no idea why you would choose to use the quadratic formula. What you're talking about is exactly the definition of i that most people begin with.

When you have no b part in your quadratic, then you can use that method. b part is the real part of your complex number.

Quadratic formula is the easiest method to solve those equations so that's why people use it

purplemath.com/modules/complex.htm

Was based off that ^

I think he just really liked the quadratic formula for some reason and seemed to use it to demonstrate a lot of things. *shrug*

The real problem with introducing squareroots of negative numbers is that the usual properties of exponents no longer hold quite like you want them. You have to be extra careful.

For example, before you had (ab)^{1/2} = a^{1/2}b^{1/2} whenever everything existed (i.e., a & b positive). But if a^{1/2} exists for all real a, then (ab)^{1/2} is no longer always equal to a^{1/2}b^{1/2}. (Take a = -1 & b = -1 to get a contradiction.)

However, by using the letter i to be a ``number'' with the property i^2 = -1. Then you can get around a number of these issues.

Defining a proper square root (or generally any arbitrary real power) for negative reals (or complex numbers in general) requires you to be very careful so you don't generate lots of contradictions.

Are imaginary numbers actually legitimate though or are we just artificially introducing something to solve these issue we have with exponents and negative sqrts?

numberphile confirmed this

also, illuminaughty did this

X^2 + X +1 = 0 has no real solutions

>Are imaginary numbers actually legitimate though
yes

No because there's already a 0 on the right hand side.

what are pragmatic inferences

1 is defined as for all x in the ring, 1 * x = x * 1 = x.
0 in the trivial ring agrees with this definition so 1 = 0 in the trivial ring.

Complex numbers are just couples of real numbers, for which the operations + and * are defined like:
(a,b)+(c,d)=(a+c,b+d), and
(a,b)*(c,d)=(ac-bd,ad+bc)
Now notice that the complex numbers of the form (a,0) behave exactly in the same way as the real numbers (between themselves) so you just write them a instead of (a,0)
And numbers of the form (0,b) can be written as:
(0,1)*(b,0), so if we put i = (0,1), (b,0)=b, then (0,b)=ib
Finally any number (a,b) can be written as: (a,0)+(0,b) which is exactly a+bi.
As a coincidence (it's not really one, it's more of a design choice): try calculating i*i, which is (0,1)*(0,1)=(-1,0)=-1.

>which is exactly a+bi
a+ib sorry

Just heading out to dinner, will be back later. Thank you for the response that actually clears up their behavior for me a lot.

>(a,b)+(c,d)=(a+c,b+d), and
>(a,b)*(c,d)=(ac-bd,ad+bc)

So the operations of + and * are redefined correct? Will this hold for all real numbers?

Exactly, they're redefined.
And yes, all (a,0) complex number behaves like 'a' real number as in:
(a,0)+(a',0)=(a+a',0), basically a+a'
and
(a,0)*(a',0)=(a*a',0), basically a*a'
This 'behaves like' thing is called isomorphism.

No that's bullshit. You can only have one symbol, not two, defining the same element. OP thought he was being witty, but in fact he can't grasp basic algebra.

Identity element for + is denoted 0.
Identity element for * is denoted 1.
OP (me) is right.

Girls are smelly

>Identity element for * is denoted 1
>>When your teacher asks you to solve x2 + x + 1 = 0 ...
>x2 + x + 1 = 0
>+ 1
>+

Then why do you add here? Sorry OP but you're not making sense.

Btw, for the trivial ring, 0 is both the identity element for multiplication and for addition.

>Are imaginary numbers actually legitimate though or are we just artificially introducing something to solve these issue we have with exponents and negative sqrts?

If there is anything I learned from my set theory class is that numbers don't exists, we invented all of them from basically nothing.

The only things that exist in the universe are logic and sets.

>Btw, for the trivial ring, 0 is both the identity element for multiplication and for addition.
>Identity element for * is denoted 1
You just said that 1=0 in the trivial ring, therefore I can write 1 instead of 0 in expressions, but that doesn't make not behave like 0, because again, in the trivial ring, 1=0.
How am I not making sense?

>How am I not making sense?
Ambiguous notation. There's no reason to mix 0 and 1.
It wouldn't be ambiguous like this: [math]x^2+x+1_F=0_F[/math]

Should be R instead of F of course

Girls are yucky.

To add to what he's saying, as far as I know, that's the only way you can define a operation for multiplication of couples of real numbers. Intuitively and rather naively, the complex numbers are the only way you can have 2 dimensional real numbers.

I love plowing in their yucky smelly and tasty hole.

>Ambiguous.
As in it pushes you to think that 0 and 1 are different elements? Hmm, maybe.
Anyway you know that this is just a joke and I solved it in C like a faggot, but the teacher should've specified where he wants you to solve it, because in R it has no solutions, and maths teachers should be rigorous in their notation.

Holes*

>I'd love to plow
Fixed.

:'(

Thank you again for the response, this has clarified a lot for me. Much appreciated

Yeh this is really what's beginning to dawn on me but you've actually articulated what I've been trying to grasp perfectly.

Brainlet here, where do I go from here?
[math]x^2+x+1=0[/math]
[math]x^2+x=-1[/math]
[math]x+\sqrt(x)=i[/math]
[math]???[/math]

you go kys

dumb rudeposter

where [math] ax^2 + bx + c = 0, \\ x = (-b±√(b^2-4ac))/2a

Fug I never remember to close math tags. I keep assuming they work like spoiler tags [math] ax^2 + bx + c = 0, \\ x = (-b±√(b^2-4ac))/2a[/math]

What you are using is the idiot's square root identity. To solve this problem the smart way, do this: x^2 + x + 1 = 0, x^2 + (0.5 + 0.5)x + 1 = 0,
x^2 + (0.5 + 0.5)*x + 0.5*0.5 + 0.75 = 0, (x+0.5)^2 + 0.75 = 0
(x+0.5)^2 = -0.75, z^2 = -0.75. z = r*cos(t) + i*r*sin(t), z^2 = (r*cos(t) + i*r*sin(t))^2 = (r^2)*cos(2*t) + i*(r^2)*sin(2*t) = -0.75. Clearly, 2*t = pi + 2*pi*n, where n is any integer, so t = pi/2 + pi*n, where n is any integer, and r^2 = 0.75 = 3/4, so r = sqrt(3)/2. z = (sqrt(3)/2)*(cos(pi/2 + pi*n) + i*sin(pi/2 + pi*n)). However, this gives two possible solutions, z = i*(sqrt(3)/2) or -i*(sqrt(3)/2). Finally, solving for x, we get -0.5 + i*(sqrt(3)/2), or -0.5 - i*(sqrt(3)/2).

you cant distribute powers on sums m8

>x2 + x + 1 = 0
x=-1/3

So are guys. Just depends on the person and their hygiene habits

haha gayyy

[math]x_{2} + x +1 = 0[/math]
[math]x=-(x_{2}+1)[/math]
SOLVED

REEEEEEEEEEEEEEE

My analysis book had an exercise showing that the sum of [math]9/10^n[/math] actually converges to 1.

It not only converges to 1, user, it's the definition of 1 actually.
Real numbers are defined to be sets of Cauchy sequences over the rationals, so the sequence with general term [math]9/10^n[/math] represents the set 1 of [math]\Bbb{R}[/math]

>Reminder: Veeky Forums is for discussing topics pertaining to science and mathematics
So leave, roastie

...

I prefer the analysis approach to the algebraic one

I didn't study maths since high school user, but I think that the one I stated is the analysis one.
Is that right? If so, link to algebraic one? (if understandable with high school level).

I was talking mostly about terminology/vision.

Algebraic "terms" are that it's the definition of one.
Analysis "terms" is that the set of real numbers is a given, and then we have

0.99.. = 0.9 + 0.99...*1/10
0.99...(1-0.1) = 0.9
0.99... = 0.9/0.9
0.99... = 1

I'm the retarded one here. You were right.

Yeah the sequence converges to 1 of course, and it's nice to see 0.999... as just an ugly way to write 1.
Analysis is really fascinating, I've bought some books and I'm planning to study some university level Algebra and Analysis next holidays.

[math]x^2+x+1=0, \\ x^2+x=-1, \\ 4x^2+4x=-4, \\ 4x^2+4x+1=-3, \\ (2x+1)^2=-3, \\ 2x+1=±i√3, \\ 2x=±i√3-1, \\ x=(±i√3-1)/2[/math]