/sci/ - Science & Math

[Anonymous No.10468691]

[math] [/math]

[Anonymous No.10468693]

>>10468691
1

[Anonymous No.10468695]

>>10468693
> |1| = -1

[Anonymous No.10468699]

|x| is non invertible function. Therefore solving for x is not possible.

[Anonymous No.10468701]

>>10468691
what "number" base system ? there are a lot of possible solutions to this problem

[Anonymous No.10468710]

>>10468701
Base 10.

[Anonymous No.10468716]

>>10468691
No solution. Next.

[Anonymous No.10468718]

>>10468710
Absolute or relative ?

[Anonymous No.10468719]

>>10468691
in order to answer this question we introduce the supernegative numbers indicated by an = where |=x|=-x. Therefore
5+|x|=4=>|x|=-1=>x==1

[Anonymous No.10468720]

no solution exists

[Anonymous No.10468722]

>>10468719
Love your creativity, I guess thats where dark energy comes from, having difficulties curve fitting the galaxy rotation curve ?

[Anonymous No.10468733]

>>10468691
x=i

next

[Anonymous No.10468734]

>>10468691
x is non-invertible. No solution for the codomain of abs(x) does not contain negative numbers.

[Anonymous No.10468749]

>>10468691
5 + |x| = 4
5 + |x| - 5 = 4 - 5
|x| = -1
(|x|)^2 = (-1)^2
x^2 = 1
x = +-1

[Anonymous No.10468753]

>>10468691
The real question is do we break maths by assuming a number has a negative absolute value, or would that create valid new numbers similarly to how assuming a number has negative square creates imaginary numbers?

[Anonymous No.10468797]

>the answer is literally i

[Anonymous No.10468811]

>>10468699
??????????????????????????????????????????????????????????????????????????????

[Sage No.10468821]

This thread is making me physically sick

[Anonymous No.10468852]

Dismissing the problem as impossible is not a valid mindset.
The square root of a negative number is impossible but look at what's taught in school.

[Anonymous No.10468854]

What are those lines next to x?

[Anonymous No.10468856]

No real or imgainary solutions.

[Anonymous No.10468876]

Why does wolfram say that |i| = 1?

i = Sqrt(-1)
isn't this supposed to be a distinct number?
If |i| = 1 then |Sqrt(-1)| = 1 and that means i = -1


looks like someone forgot to tie up all the loose ends to their dumb made up math terminology.
Why is higher math such a fucking joke?

[Anonymous No.10468882]

>>10468876
Absolute value is just distance from 0 on the number plane. Eg. |Sqrt(i)| = 1 because its complex constituencies add up to 1 when treated as positive numbers. (1/sqrt(2))^2=1/2 2/2=1 so absolute value is 1.

[Anonymous No.10468891]

How many times does -2 go into 8?
show your work.

[Anonymous No.10468897]

>>10468876
abs=magnitude

[Anonymous No.10468903]

>>10468691
|x|=-1
x^2=1
x=+- 1
x=-1,x=1

[Anonymous No.10468952]

>>10468897
I can't build abs.

[Anonymous No.10468962]

The answer is i

[Anonymous No.10469094]

>>10468891
infinitely many
Proof: The proof is trivial.

[Anonymous No.10469108]

>>10468691
abs(x) = -1
obviously this won't work for any real number, so we need to make a new type of number. Call it the absolute numbers, such that abs(x) = -r, where r is a real number. Also let's call the answer to this solution to be alpha-theta-epsilon, or as shorthand, the sterling number of the fourth kind, because why not?

[Anonymous No.10469111]

>>10469094
>professor says proof to the answer is trivial
>finds he can't solve it easily
>looks up the source of the original proof of the solution
>it's his book
>the book says the proof is trivial.

[Anonymous No.10469120]

>>10469111
Take -2 from 8. You have left 8 - (-2) = 10. Clearly you can do this infinitely many times.

[Anonymous No.10469155]

>>10468691
[math] | x | \equiv 9 \mod 10 [/math]

;^)

[Anonymous No.10469174]

>>10468753
yes

[Anonymous No.10469281]

>>10468854
basically you just take the positive value
so -5 => 5
and 5 => 5 also

[Anonymous No.10469306]

Still a distance thing

[Anonymous No.10469357]

I'm only in calculus 2, the abs function can be viewed as the distance from zero, so a negative distance doesn't make sense.
I mean i don't know shit about math but i don't think there's a meaning to negative distance (if there's any tell me becayse that sounds cool)

[Anonymous No.10469409]

>>10468699
its literally just the square root of x^2, you can reverse it like anything else

[Anonymous No.10469466]

>>10469409
For real numbers thats true but for anything else you're a brainlet. Its embarassing how many of you miss this.
For real numbers you can define absolute value many ways take this peicewise function as an example.

If x>=0 f(x)=x if x<0 f(x)=-x

This gives you the same as square into root, the proper definition of abs is more complex.

[Anonymous No.10469474]

>>10469466
Specifically for complex numbe
|+-i*a+-b| = |sqrt(a^2+b^2)|
Brainlet

[Anonymous No.10469482]

>>10469466
>>10469474
>retard acting smug because he knows middle school math

[Anonymous No.10469510]

>>10469482
>complex numbers
>middle school math
Come back when you get out of highschool.

[Anonymous No.10469518]

>>10469466
Shut the fuck up, retard.

[Anonymous No.10469963]

>>10468876
[math]\left|z\right| = \sqrt{\Re{\left(z\right)}^2 + \Im{\left(z\right)}^2}[/math] for all [math]z\in\mathbb{C}[/math].
For [math]\Im{\left(z\right)} = 0[/math] (that is, for all [math]z\in\mathbb{R}[/math]), this evaluates to [math]\left|z\right| = \sqrt{z^2}[/math], which I'm sure you'll find agreeable.

[Anonymous No.10470052]

How is this not 1?

[Anonymous No.10470072]

Fucking laughable brainlets don't know the correct order of operations.

absolute |x| comes LAST so there is no solution

[Anonymous No.10470123]

>>10469155
9 mod 10 = 9 retard

[Anonymous No.10470128]

>>10468897
in complex number domain abs is absolute square, which is xx*

[Anonymous No.10470134]

>>10468691
5 + abs(x) = 4
abs(x) = - 1

[Anonymous No.10470137]

>>10469111
calling stuff trivial or giving proofs as exercise without a solution should be illegal

[Anonymous No.10470139]

Imagine being so dumb you dont know the answer is -2

[Anonymous No.10470188]

>>10468691
call j the number such as 1 + |j| = 0.

I just created a new form of complex numbers.

suck my dick mathfags

[Anonymous No.10470243]

>>10469482
>retard greentexting because he forgot middle school math

[Anonymous No.10470276]

>>10470188
Now explore its properties

[Anonymous No.10470318]

>>10468876
The absolute value means magnitude, distance from zero

[Anonymous No.10470335]

1 + |j| = 0
1+ |j + 1 + |j + 1 + |j + 1 +...||| = 0

[Anonymous No.10470460]

>>10470243
Are you confused?

[Anonymous No.10470981]

>>10468749
i want you to plug +1 back into that equation

[Anonymous No.10470982]

>>10470052
5+1=4?

[Anonymous No.10471161]

>>10468691
it's 4294967295

[Anonymous No.10471187]

This proof is absolutely trivial. Absolute value of a number is the reference distance aka a distance from a point to another point. How can you have negative distance? Use a path integral and move your reference point.
Pretend you are standing on a 1 d number line. Now pretend you are at point 0, you want to get to move 5 steps so that you are actually 4 steps away. But wait, you were at 0? Move 1 step in the opposite direction, and then move 4 steps toward that direction and bam you are at x=4. Proof couldn't be more trivial

[Anonymous No.10471944]

>>10468962
|i|=1

[Anonymous No.10471982]

>>10471187
Nope. You'd be at x = 3. Path integrals don't care about distance so they won't be helpful here.

[Anonymous No.10472185]

>>10468876
complex numbers are represented in two dimensions as a+bi where a represents the x-axis and b represents the y-axis. The absolute value is found by the formula sqrt(x^2 + y^2) which is just the distance formula where the first point is always the origin

[Anonymous No.10472188]

>>10468691

Can I get a real answer please ?
Is this unsolvable in the sense that "it should be solvable but we don't know how yet" or "it's impossible to solve and will always be impossible" ?

[Anonymous No.10472194]

>>10472188
The latter.

[Anonymous No.10472240]

>>10469409
>its literally just the square root of x^2
No, it's really not just that. I think you meant the POSITIVE root of x^2, and even that is problematic.

[Anonymous No.10472467]

>>10468733
>abs(i) = -1
>oof

[Anonymous No.10472698]

>>10472467
>>oof
>>>/reddit/

[Anonymous No.10472714]

>>10468852
>The square root of a negative number is impossible
No, it isn't. [math] \sqrt(-1) [/math] doesn't break anything. This, on the other hand, leads to a contradiction an breaks all of math.

[Anonymous No.10472787]

>>10470981
He didn't put 1 anywhere

[Anonymous No.10472914]

5+|x|=4
|x|=x, |x|=-x
5+|x|-5=4=5
|x|=-1
x=1, x=-1
4≤5+|x|≤6, x∈N

[Anonymous No.10472917]

>>10472914
>5+|x|-5=4=5
>5+|x|-5=4-5

[Anonymous No.10472925]

>>10469111
Kek

[Anonymous No.10472937]

[math]\text{The function }|\cdot| \text{can be write by } |x|=\sqrt{<x,x>}[/math]. Inner product is always non-negative. Then, \sqrt{<x,x>}=|x| is non negative.

[Anonymous No.10472971]

>>10472467
>>10471944
>>10469963
>>10468797
>>10468733

[math]|x|=\sqrt{x^2}[/math]
[math]|i|=\sqrt{i^2}=\sqrt{-1}=i[/math]
If [math]5+|x|=4[/math] then [math]x\neq i[/math] because [math]5+i\neq 4[/math]

[Anonymous No.10472973]

>>10472467
>>10471944
>>10469963
>>10468797
>>10468733

[math]|x|=\sqrt{x^2}[/math]
[math]|i|=\sqrt{i^2}=\sqrt{-1}=i[/math]

If [math]5+|x|=4[/math] then
[math]x \neq i[/math] because [math]5+i \neq 4[/math]

[Anonymous No.10472987]

>>10472467
>>10471944
>>10469963
>>10468797
>>10468733

[eqn]|x|=\sqrt{x^2}[/eqn]
[eqn]|i|=\sqrt{i^2}=\sqrt{-1}=i[/eqn]
If [math]5+|x|=4[/math] then [math]x \neq i[/math] because [math] 5+i \neq 4[/math]

[Anonymous No.10473006]

If that true?
>>10468733
>>10468962
[eqn] \left| x \right| =\sqrt{x^2}[/eqn]
[eqn] \left| i \right|=\sqrt{i^2}=\sqrt{-1}=i[/eqn]

If [math] 5+ \left| x \right|=4 [/math] then [math]x \neq i[/math] because [math] 5+i \neq 4[/math]

But these are better definitions:
>>10472937
>>10469963

In that case:
[math] \left| i\ right| = 1[/math]

[Anonymous No.10473010]

[eqn] \left| x \right| =\sqrt{x^2}[/eqn]

[eqn] \left| i \right|=\sqrt{i^2}=\sqrt{-1}=i[/eqn]

[Anonymous No.10473015]

[eqn] \left| x \right| = \sqrt{x^2} [/eqn]

[math] \left| i \right|=\sqrt{i^2}=\sqrt{-1}=i [/math]

[Anonymous No.10473017]

>>10468691
someone explain absolute value to a brainlet like me

[Anonymous No.10473035]

>>10469482
>waiting until middle school to learn about complex numbers

[Anonymous No.10473040]

>>10469963
wow how did you draw that fancy R?

[Anonymous No.10473047]

>>10473015
[math]
\left | i \right | = \sqrt{0^2+1^2} = 1
[/math]

[Anonymous No.10473049]

>>10473040
\left|z\right| = \sqrt{\Re{\left(z\right)}^2 + \Im{\left(z\right)}^2}

[Anonymous No.10473064]

>>10468691
undefined, abs(x) can't equal -1.

[Anonymous No.10473067]

>>10469510
>>10473035
Where do they teach complex numbers before middle school?

[Anonymous No.10473161]

Even if you could generate a set to solve |x| = -1, it would be pointless because |•| is useful as it is intuitively the size of a number which we all know can't be negative. Seriously, |•|≥0 is a very important property where |x|=0 iff x=0, so if we can find |x|<0, then our absolute value is useless!

[Anonymous No.10473584]

>>10468719
But are they lower or higher than 0?

[Anonymous No.10473605]

>>10469963
What the hell is J

[Anonymous No.10473614]

>>10468691
5 - 4 = -|x| you fucking retards
I'm not even a math guy and I can figure it out.

[Anonymous No.10473647]

>>10468691
>OP made a thread about [math]|\omega|=-1[/math] and got told
>now he makes this one
It's like mathematical shitposting.
I guess it's on topic

[Anonymous No.10473663]

>>10473584
You would feel dirty if I told you.

[Prof Sai No.10473756]

>>10468733

>x=i
That's what I thought, but real mathematicians have beat us to it. |x| for complex numbers means distance to 0,0. So |i|=1.

Oh well. I think >>10468719 is correct. There is no answer unless you define one.

[Prof Sai No.10473764]

>>10473067

In photoshop land.

[Anonymous No.10473871]

>>10473647
That thread came after this one, retard

[Anonymous No.10473887]

>>10473647
>It's like

[Anonymous No.10473955]

>>10473614
5-4 < 0

[Anonymous No.10474012]

>>10468691
>No set indicated
>Assuming this is four-bit two's compliment decimal representation
The answer is the absolute value of 10, overflow happens after 7 to -8, then 7 more is -1.

[Anonymous No.10474110]

>>10468691
This is easy. You just need to find the number that has an absolute value of -1.

[Anonymous No.10474144]

x∈O

[Anonymous No.10474159]

>>10468691
Not possible without smartassery

[Anonymous No.10474173]

>>10468854
Absolute value, so if a>0, |a|=a and |-a|=a, if a<0, |a|=-a and |-a|=-a. If a=0, |a|=|-a|=a

[Anonymous No.10474183]

>>10468854
AKA √x2

[Anonymous No.10474199]

>>10468722
galaxy rotation curves is evidence for dark matter, not dark energy

[Anonymous No.10474445]

>>10468691
if you're allowed to just make shit up like i^2 = -1 for the sake of solving equations then let t be a number such that abs(t) = -1.

[Anonymous No.10476218]

>>10468691
5+x=4 (-5)
x=4-5
x=-1

[Anonymous No.10476626]

>>10468691
given: 5+|x|=4
let u =ix then u/i=x
5+|u/i|=4
5+sqrt((u/i)^2)=4
5+sqrt(-u)=4
5+isqrt(u)=4
isqrt(u)=-1
sqrtu=-1/i
u=(-1/i)^2
u=1/-1
u=-1
ix=-1
x=-1/i

5+|x|=5+sqrt((-1/i)^2)=5+sqrt(-1/-1)=5+1=6

therefore 5=4 Q.E.D.

Where's my fields medal?

[Anonymous No.10476844]

>>10468691
Question's assumptions: there exists an x in the real numbers such that 5 + |x| = 4,
|x| = -1
x2 = 1 ∵ x2 >= 0 for all real x
x = ±1. By substitution,
5 + |x| = 4 5 + |±1| = 4 5+1 = 4 2 = 0. By contradiction, there are no real solutions.

[Anonymous No.10476846]

>>10476844
Nevermind; character encoding errors.

[Anonymous No.10477003]

>>10468691
Solving this equation makes time travel possible

[Anonymous No.10477350]

[math]x = [[1, 0], [0, -1]] [/math]

[Anonymous No.10477358]

This is just the rehashed version of the imaginary triangle meme

[Anonymous No.10477360]

>>10468691
5+x=4
x=4-5
x=-1

[Anonymous No.10477364]

>>10473584
superlower

[Anonymous No.10477531]

>>10468691
5 + abs(x) = 4
abs(x) = -1
abs(x) = i^2

[Anonymous No.10477582]

>>10477360
It's an absolute value, no such thing as a negative.