Cubic root (-8) = 2 ?



#29

What do you think of this :

All seems perfect and nothing seems violate math rules... but....


Edited: 4 Aug 2013, 6:22 p.m.


#30

At the first step the equality is not true, the "problem" is similar like SQRT(x^2) is not equal with x.

When you are expanded the 1/3 to 2/6, you are make SQRT(x^2)==x mistake.


#31

The first step is perfectly fine. It is the third step that is the problem.
Edit: Third expression, not third step.


Edited: 5 Aug 2013, 4:53 a.m.


#32

Quote:
The first step is perfectly fine.


BS
#33

...well, you're wrong, but my hungarian thinking gives me an other point of view when i'll try to solve similar problems...

For example:

(-1)^1     = -1, but
(-1)^(2/2) = SQRT((-1)^2) = +1

The mathematical symbol

(...)^c == (...)^(a/b)
, where
a/b = c
contains the mistake not the mathematical operation (which have not even been performed).

You do not have to make error to know that a mistake.
So, the first step is wrong.

Csaba


#34

Quote:
So, the first step is wrong.

So you want to say that 1/3 = 2/6 is wrong??

LOL, what madhouse is this here?

Franz


#35

:D

Of course,

1/3 = 2/6
is true, BUT
(-8)^(1/3) = (-8)^(2/6)
is NOT true.

(-8)^(1/3) == ((-8)^(1)) ^ (1/3) == (-8) ^ (1/3) = -2
(-8)^(2/6) == ((-8)^(2)) ^ (1/6) == (64) ^ (1/6) = +2

My english is not very good, so one another example for clarification:

A little BASIC program:

10 REM This program will be calculate logarithm of (-5)
20 A=-5
30 PRINT LOG(A);

a.) The error as Gilles seems: "Everithing is fine, I do not understand why Line 30 is wrong because of LOG() ?!??!"
b.) The error as an interpreter seems: "Line 30 is wrong because of LOG(neg)"
c.) The error as Harald, Reth, fhub seems: "Line 30 is wrong because of LOG(neg)"
d.) The error as a smart compiler seems: "Line 20 is wrong because of A is negative and this will cause error in Line 30 as LOG(neg)"
e.) The error as I see: "Line 10 is wrong. This idea causes all trouble below." ;)

Csaba


Edited: 6 Aug 2013, 3:20 a.m.


#36

Quote:
Of course,
1/3 = 2/6
is true, BUT
(-8)^(1/3) = (-8)^(2/6)
is NOT true.

That's pure nonsense!

The error is NOT in replacing 1/3 by 2/6, but in what you do with this 2/6 later. Of course it's wrong to put this numerator 2 to the base (-8) first and then do the exponentiation ^(1/6), i.e. (-8)^(2/6) is NOT the same as ((-8)^2)^(1/6), simply because the rule a^(b*c)=(a^b)^c is NOT valid for ALL real numbers!

So the 'error' in the starting post is indeed in the 3rd expression (or in the 2nd step) - besides that there's still another error in having forgotten a pair of important parentheses (because exponentiation is right-associative as already mentioned by an other member).

Franz


Edited: 6 Aug 2013, 4:05 a.m.


#37

Quote:
...rule a^(b*c)=(a^b)^c is NOT valid for ALL real numbers

Yes, when the base is negative. The rational exponent method cannot be used with negative bases because it does not have continuity.

Quote:
...but in what you do with this 2/6 later
yes, see above!



Then again, see above, if the base is negative then, the rational exponent method cannot be used so 1/3 should not be changed to 2/6!


Edited: 12 Aug 2013, 11:55 a.m.


#38

Quote:
if the base is negative then, the rational exponent method cannot be used

Correct!
Quote:
so 1/3 should not be changed to 2/6!

Well, it should not be changed (simply because it doesn't make any sense at all), but it definitely can be changed (if you want), because 1/3=2/6=5/15=.....=70/210=..... etc., no matter in which expression or formula this subexpression is contained.

So as I've stated already a few times before: replacing 1/3 by 2/6 is NOT any error (it's just useless), the error is how the original calculation proceeded with this 2/6!

And even if the base would have been positive (i.e. 8 instead of -8), an expression 8^(2/6) has to be calculated as 8^(1/3) but NOT as (8^2)^(1/6). Although it's not really wrong in this case, but the expression (2/6) has definitely priority over 8^... (ever heard of PEMDAS?), because 2/6 CAN be simplified to 1/3.

Franz


#39

You mean we should use fractions that have been reduced! Ok, I can buy that.

Still, the Rational Exponent Method does not apply to a negative base so the entire argument is axiomatic. And,

Quote:
simply because it doesn't make any sense at all

#40

Quote:
Ok, I can buy that.

I'm glad you finally got it - although it took quite some time! ;-)

So peace again from now on ... :-)

Franz

Edited: 12 Aug 2013, 12:55 p.m.


#41

I have always got it. Disagreement does not translate to ignorance; however, acceptance of differences leads to greatness. Putting aside the argument of application, I still disagree with the entire concept of teaching math with a nursery rhyme, i.e. PEMDAS, because I feel that leads to only teaching the manipulation of numbers and not the understanding of math. Similarly, I think this thread is somewhat a product of that same type of problem.


Peace from now on is fine by me.

#42

...
f/ There is no error because LOG(-5)=(LOG(5)+i*PI)/LOG(10) ;)

Edited: 6 Aug 2013, 4:18 a.m.

#43

[HP-Prime CAS] approx(evalc((-8)^(1/3))) [Enter] return 1+73...i OK (principal complex root)

but

exact(1+73...i) [Enter] .... /=/ 2 (-1)^(1/3) =(

approx((-8)^(1/3)) [Enter] return -2 real root

Request for HP-Prime rename evalc to evalCplx

Thanks

more info

http://www.wolframalpha.com/input/?i=%28-8%29%5E%281%2F3%29

#44

Watch it: 641/6 = ±2 !


#45

So we can say that

sqrt(4)=±2


#46

I hate opening that box once again, but there's a "convention" declaring SQRT(4) = +2 only. OTOH, 41/2 = ± 2 since it inverts (±2)2 = 4.

#47

The sixth roots of 64 are: -2, 2, square_root(3)*i+1,-square_root(3)*i+1, square_root(3)*i-1, -square_root(3)*i-1

#48

Exponentiation is right-associative, thus:

Cheers

Thomas


#49

;)

and it's <> 64^(1/6)

#50

So much ado just to come to terms with the fact that square root is a multi-value function... that's all there is to it, nothing mysterious or bizarre. Sure enough not a matter of "conventions" or "dual rules" or "exceptions", but plain and simple math.

Cheers,
'AM


#51

I still don't get it. The cubic root has three solutions, two complex and one real. What transformation in Gilles' process intoduces +2 as a fourth solution? Expanding 1/3 to 2/6?


I guess it's just been too long since I learned that stuff...)


Kind regards, Victor


#52

(-8)^(1/3) is taking finding the three roots of -8

(-8)^(2/6) is taking the the six roots of (-8)^2, or 64


#53

Quote:
(-8)^(2/6) is taking the the six roots of (-8)^2, or 64

Mathematics: fail

#54

Both with 50G and Prime CAS :

a^b^c is not (a^b)^c but a^(b^c)

Exponentiation is a curious beast : right-associative as wrote Thomas.


#55

Quote:
a^b^c is not (a^b)^c but a^(b^c)

Yes, of course I know this, but what has this to do with your (wrong!) assumption that (-8)^(2/6) has to be (or should be) calculated as ((-8)^2)^(1/6) ???

Your previous statement "(-8)^(2/6) is taking the the six roots of (-8)^2, or 64" would require the following transformation:

(-8)^(2/6) = (-8)^(2*(1/6)) != ((-8)^2)^(1/6)

But the used exponentiation rule a^(b*c) = (a^b)^c is NOT true in this case (i.e. when the base a is negative)!

Franz


Edited: 12 Aug 2013, 4:25 a.m.


#56

Quote:
Yes, of course I know this, but what has this to do with your (wrong!) assumption that (-8)^(2/6) has to be (or should be) calculated as ((-8)^2)^(1/6) ???

I think we said the same things ...

The idea of my post was (like a puzzle) to show that ^ is right associative (as far I know it's the only operator like this ?)

(-8)^(2/6)= (-8)^2^(1/6) is true

but

(-8)^(2/6)=((-8)^2)^(1/6) is false

And of course

(-8)^2^1/6 <> 64^(1/6).

Here is the 'error' in the initial post


edit : typo and quote error

Edited: 12 Aug 2013, 6:02 a.m.


#57

Quote:
(-8)^(2/6)= (-8)^2^(1/6) is true

No, it's not!

The RHS would mean to calculate the 6th root of 2 first, and that's something completely different!

Franz


#58

Oh ! I see it now !

Thanks

#59

Quote:
to show that ^ is right associative

Otherwise abc = abc, thus we wouldn't gain much by this convention.

Quote:
as far I know it's the only operator like this ?

Any assignment operators are also typically right-associative.

#60

Also, with the NTHROOT functions, the calculator strives to give the principal root. If you want all the roots and have a calculator with CAS abilities, we will need to use a cSolve function, like this:

CSolve(x^3-8=0,x)


#61

Here are all (?) the ways to do this with the 50G

You can notice than even in Real Approx mode, the 50G returns a complex number (it's a legacy of the 48 series where there was no 'Complex mode') and because (-8)^0.33333333 is a complex number (no real root).It should result an error (no solution in |R)
Also note the different results in |C~ mode between
-8 3 XROOT
and
-8 1 3 / ^ it's logic, if you consider the second way is (-8)^0.3333333333...

However, returning (-2) in '|C~' mode with XROOT is not logic (It must retunr the principal root); in '|R=' mode, XROOT and ^ should return (-2) and not a warning to switch in |C


Edited: 6 Aug 2013, 7:08 p.m.


#62

What about [ 1 0 0 8 ] PROOT to get the roots of the polynomial x^3 + 8 = 0 ?


#63

Quote:
What about [ 1 0 0 8 ] PROOT to get the roots of the polynomial x^3 + 8 = 0 ?

[(1.000,-1.732) (1.000,1.732) (-2.000,0,000)]
#64

Just google Rational Exponent Method with Negative Base and it will be found that it does not apply (that way it is not me stating why). So 1/3 cannot be changed to 2/6 with the Rational Exponent Method.


#65

Quote:
Just google Rational Exponent Method with Negative Base and it will be found that it does not apply (that way it is not me stating why). So 1/3 cannot be changed to 2/6 with the Rational Exponent Method.

I think:

If the base is negative and the exponent is even, the result is positive, and if the exponent is odd, the result is negative.

So, this expression is true only:



ABS((-8)^(1/3)) = ABS((-8)^(2/6)) or ABS(((-8)^2)^(1/6))

Because SQRT(X^2)= +-X as Csaba Tizedes wrote.



Edit:

(-8)^(1/3) = (-8)^(2/6) but only ABS((-8)^(1/3)) = ABS(((-8)^2)^(1/6))

Edited: 12 Aug 2013, 3:55 p.m.


#66

The rational exponent method cannot be used with negative bases because it does not have continuity


Quote:
think:
If the base is negative and the exponent is even, the result is positive, and if the exponent is odd, the result is negative.
so it is not continuous!


f(q)=bq is continuous for b>0 but it is not continuous for b<0 and more importantly, when b<0 it is not continuous for the rational set of q for which it is defined.

Anyway, I thought I was not going to explain this ....

John

Edited: 12 Aug 2013, 10:34 p.m.


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