college math

[Anonymous No.10070929]

A teacher in my college showed us this equation. Anybody got an idea why is it true/wrong?

[Anonymous No.10070938]

>>10070929
first equation is wrong because the third root of negative one is not negative one.

[Anonymous No.10070945]

>>10070938
how come? (-2)^3 is -8

[Anonymous No.10070953]

nvm, got it

[Anonymous No.10070954]

>>10070929
The last step is where the wrong sixth root was used.
He could have just written -2=sqrt(4)=2.

[Anonymous No.10070956]

>>10070945
good point, then i have no clue

[Anonymous No.10070957]

>>10070954
I dont get this

[Anonymous No.10070964]

>>10070957
There are 2 different square roots: positive and negative.
There are 6 different sixth roots, positive and negative being two of them.
In the last step, the positive sixth roots was arbitrarily chosen instead of the negative.

[Anonymous No.10070970]

>>10070964
ok thx. what are the other 6. roots ?

[Anonymous No.10070974]

>>10070970
(+/-) 1/2 (+/-) i*sqrt(3)/2

[Anonymous No.10070976]

>>10070974
Those are the sixth roots of 1.
(+/-) 1 (+/-) i*sqrt(3) are the other sixth roots of 64 apart from (+/-) 2.

[Anonymous No.10070986]

>>10070945
because it's a complex number
https://www.wolframalpha.com/input/?i=(-8)%5E(1%2F3)

[Anonymous No.10071006]

ez for college math

[Anonymous No.10071195]

>>10070929
Any kind of exponential operation is not a bijective function and therefore can produce additional (and fake) solutions, but the original solution is never lost though.

(2)^6 = 64 (principle root & additional solution)
(1+1.73i)^6 = 64 (additional solution)
(-1+1.73i)^6 = 64 (additional solution)
(-2)^6 = 64 (original solution)
(-1-1.73i)^6 = 64 (additional solution)
(1-1.73i)^6 = 64 (additional solution)

The simplest demonstration of this is:

x = 2
x^2 = 4
sqrt x^2 = sqrt 4
abs(x) = 2
x = {2; -2} while x was 2 originally

[Anonymous No.10071199]

>>10071195
the last line should be x (member symbol) {2; -2}

[Anonymous No.10071327]

>>10070929
[math]\sqrt[6]{64}=\pm2[/math]
[math]\pm[/math] implies plus OR minus, in this case you would chose minus.

[Anonymous No.10071374]

>>10071327
It doesn't imply plus or minus.
It's defined as |64^1/6|.
Therefore, it's only plus or minus if there's an equivalence relation between the function and another value.

[Anonymous No.10071395]

Isn't -8^(2/6) an invalid operation in real numbers?

[Anonymous No.10071422]

>>10071395
Yes you cant have negatives under even roots or you get a complex number

[Anonymous No.10071447]

>>10070945
-2 * -2 = 4

4 * -2 = -8

[Anonymous No.10071466]

>>10070929
z^(1/n), with z not positive and n some integer, is a "multi-valued function"
the equation almost certainly confuses one of those values for the other

[Anonymous No.10071470]

>>10070929
Because

-2^6 = 64
2^6 = 64

Retard

[Anonymous No.10071473]

>>10070929
The third equality is where the mistake is. LHS is -2, RHS is 2

[Anonymous No.10071486]

>>10070929
yes, but it doesnt means that -2=2 since he uses a definition of square root that isnt a function

[Anonymous No.10071493]

The sixth root of 64 is also -2

[Anonymous No.10071499]

>>10071470
This is embarrassing but -2^6=-64

[Anonymous No.10071510]

>>10071499
[math](((((((((((((((((((([/math]-2[math]))))))))))))))))))))[/math]^6 = 64
is that better?

[Anonymous No.10071512]

>>10071510
>as an engineer, i like to add an extra buffer of protection for the numbers i use in my math

[Anonymous No.10071530]

>>10070929
last equation is supposed to be (64)^(1/6) = +/- 2

[Anonymous No.10071542]

>>10070929
The last step is dishonest.

It should be plus or minus 2. Even roots have two solutions.

[Anonymous No.10071589]

>>10071530
>>10071542
Wrong.
Why bother posting if you don't read the thread?

[Anonymous No.10071593]

>>10071589
>not understanding working in different number spaces
their answers are perfectly fine when restricted to reals

[Anonymous No.10071607]

>>10071593
No it's not.
>there are two solutions...
Nothing is being solved, only evaluated.

Definition of the square root function is perfectly clear.

[Anonymous No.10071622]

>>10071607
oh, you're being THAT picky
i literally don't give a fuck about that, they obviously understand the concept of a fractional power being ambiguous
but yes, you're right
but i'm annoyed that you care

[Anonymous No.10071678]

of which they speak, if they are speaking, the error begins since it equals 1/3, it is a cubic raita therefore it is adding 2 more solutions to the problem, and when multiplying it by 2 up and down it adds 3 more solutions, therefore it would be 6 solutions to that question, that's why 2 and -2 are solutions to the sixth root question of 64, but it does not mean that both solutions are correct, so you tell me that the root of -1 in reals does not exist but inside of the imaginary if y is the root of -1 = i

[Anonymous No.10071721]

>>10070929
The error is the third to fourth part. If x \ge 0, then x^{a/b} = (x^a)^{1/b}. However, this fails when x < 0.

This is, or course, assuming that we're using the usual principle branch of the argument (and that we've fixed a branch at all so roots aren't multivalued). If you're doing multivalued roots, then both the first and the last equalities are incorrect.

[Anonymous No.10071735]

>>10070929
Roughly, let x=(-8)^2/6. If x is 2 or -2, the statement is true.
Identity for roots doesn't mean exactly what identity usually means.

[Anonymous No.10071738]

>>10071607
>>10071622
PS i didn't mean to sound that offended. sometimes i just can't control my emotions. i apologize

[Anonymous No.10071752]

>>10070929
From second to the third part. The exponential is not the same thing. Cube root accept negative number but sixth doesn't.

[Anonymous No.10072038]

>>10070929
Well it's already wrong because how can -2 = 2, but the problem arises when he takes a 6 root of a negative problem, would spit out an i, and when he squares a -2 it would be come positive. So 3/4 steps

[Anonymous No.10072050]

People that can't solve this problem should not be on /sci/, aka, a large portion of posters here.

[Anonymous No.10072064]

>>10070929
It goes through an even power. Once you start fucking around with even powers and negative signs, you can bullshit up some way to make a negative equal to a positive.

[Anonymous No.10072100]

The only thing wrong with the equation is the first and last numbers.
-2=2. Using = implies that the only answer that is correct is -2. -2 = -2. There is six answers to sixrt(64), because it has six roots. But the only root that is correct is -2.

[Anonymous No.10072115]

>>10070954
This guy has it right. By squaring negative 8 in that last step he made the negative number positive. This is the case in any equation like this, such as
(-3)^2=9
9^(1/2)=3
But everyone knows that negative 3 does not equal 3.
He’s just trying to fuck with you by making the equation longer, hoping that it’s enough to confuse you.

[Anonymous No.10072176]

>>10070929
It's really just a roundabout way of saying sqrt((-n)^2) = n. You can't square -8 before taking the sixth root. It nullifies its negativity.

[Anonymous No.10072209]

>>10071510
The Jewish -2

[Anonymous No.10072237]

>>10072176
You can though. Positive 64 still has six roots, two of them being -2 and 2. I believe you're thinking about taking sqrt of negative #s which leaves an imaginary.

the six roots would look like

-2 x -2 x -2 x -2 x -2 x -2
2 x 2 x 2 x 2 x 2 x 2
2i * 2i * -2 * -2 * -2 *-2
2i * 2i * 2i * 2i * -2 * -2
2i * 2i * 2i * 2i * 2i * 2i
2i * 2i * -2 * -2 * 2 * 2

[Anonymous No.10072366]

>>10070929
LMAO SERIOUSLY CASIO?? JUST WANTED TO BAIT BUT IT ACTUALLY SAYS THIS....

[Anonymous No.10072484]

>>10072237
What? It isn't the same number multiplied by itself 6 times?

[Anonymous No.10072504]

>>10072484
Ya but writing the notation properly in latex is cancer.
Just solve for (re^ix)^6=64.

[Anonymous No.10072607]

>>10071199
I got triggered a bit by this. I use the € for that, it's kinda similar.

[Anonymous No.10072619]

>>10072484
This explains it best

[Anonymous No.10072665]

>>10071622
FWIW, the chair of the math department asks this same question to all grad students applying.

He gave me a big rant about it, so it has stuck with me.

[Anonymous No.10072678]

>>10070929
because you can't square root negative numbers. My guess is that this is some imaginary i example or something

[Anonymous No.10072683]

>>10072678
they never try to sqrt a negative number. (-8)^2 is 64 hence positive. The problem at the end is that the sixrt of 64 has six possible answers, two of them being -2 and 2. He chose 2, but really he should have chose -2, because that is the correct root according to the initial -2 of the whole equation. -2 = 2. That's not a true statement.

[Anonymous No.10073189]

>>10072683
Since nobody can read, I'll say it again:
the equation x^6 = 64 has 6 solutions.

The sixth root of 64 evaluates to 1 value, which is the absolute value of 2.

[Anonymous No.10073199]

>>10070986
We're talking about the real domain here. Over reals, each odd root of a negative number has one, negative but real, value, and non-odd roots of negative numbers are undefined.
>>10070929
Underage brainlets >>>/out/

[Anonymous No.10073223]

>>10070929
-2 is not 2.

By rooting positive number (n^x) if x//2=0, you always get two results, positive and negative with same distance from zero.

[Anonymous No.10073275]

>>10071195
this.
>>10070929
>hurr look at me I can use half-assed tricks to make a paradox
t. brainlet prof

[Anonymous No.10073311]

>>10070964

This is the correct answer

[Anonymous No.10075042]

>>10070964
I find your lack of "self" in math being disturbing.

There are just odd and even exponents and their roots that differ... 6th root doesn't have 6 solutions, because root n(x) is what number you n times multiply to get number.

If you have more numbers different numbers in a set of which you multiply all of elements in to get x there it's factorization not root.

[Anonymous No.10075214]

>>10070929
[math]\left(e^{i\frac{1}{3}\pi}2\right)^3 = e^{i\pi}8 = -8[/math]
[math]-2 \neq e^{i\frac{1}{3}\pi}2 = (-8)^{1/3}[/math]

[Anonymous No.10075218]

>>10075042
how far are you into your studies?