silly? - math - question about fractional powers of negative base

asked 2020-10-10 11:25:16 +0200

newbie-02
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updated 2020-10-11 02:03:48 +0200

hello @all,

there are some questions and bugs around about the evaluation of fractional powers (roots) from negative bases, e.g. -8^(1/3) is the third root from -8 and thus -2, while -8^(1/2) results in an imaginary number 'i * squareroot(8)' [math: i=squareroot(-1)], and is not representable in spreadsheets ...

a simple marginal question of these problems: fractions can be 'extended', so (-1)^(1/3) would / should be the same as (-1)^(2/6)?

but if you now evaluate this according to school maths (base powered to a fraction is 'denominator root' from 'base to power of numerator'):

you get once ((-1)^1)^(1/3) - or cube root of '-1' - so '-1',

and once ((-1)^2)^(1/6) - or 6th root of '-1 squared', thus 6th root of 1, thus 1,

that can't be true for both?

probably there are some rules or definition restrictions or additional solutions in 'math'? but i don't remember if we were told this in school, and i couldn't find instant answer on the web, i hope someone here knows about ...

or do I have a cardinal error in thinking somewhere?

(why am i asking this question here ... imho calc should try to produce mathematical correct results, and for that such questions are important)

tia fah,

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Please remove tag math because it advertises issues with the formula editor. Instead use the unabbreviated mathematics.

ajlittoz ( 2020-10-10 11:48:11 +0200 )edit

@ajlittoz: thanks, i'd try to 'retag' math by mathematics, found editing the headline, the question, but nothing for the tags :-(

newbie-02 ( 2020-10-10 13:20:00 +0200 )edit

Retagging is a bit whimsical. When you have entered/replaced a tag, you must hit Enter twice.

ajlittoz ( 2020-10-10 16:15:48 +0200 )edit

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answered 2020-10-11 08:34:37 +0200

newbie-02
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updated 2020-10-25 23:07:21 +0200

<<< posted as an answer as it nearly solves the question, and is too long for a comment >>>

@Mike Kaganski in tdf#69293:

'And it's not correct to spaculate that "(-8)^(1/3)=(-8)^(2/6)=((-8)^2))^(1/6)=64^(1/6)=2". This speculation is a sort of sophism that masks the fact that there are more than one nth roots. When you make any transformations of powers you must take into account the possible loss of some roots or introduction of new roots. The correct result of "sqrt(x)" should be a multitude of two (possibly complex) numbers, not a single number. As we cannot return such a result in a spreadsheet we agree to see only positive answer, but it doesn't mean that we don't deserve to get correct negative real number as a cube root of a negative number.'

(i now remember that i did! read that in the past, just didn't find it, neither remembered exact content),

it doesn't teach which concept forbids extend or shortening of fractional exponents, but neatly explains why the results vary ... :-) at Mike,

would be nice if somebody can point to the math rules about it ...

and as a 'calc question' it's left if it's 'more correct' that calc delivers -2 for -2^2/6 when it's constructed by itself (and not by expanding the exponent of -2^1/3), (as calc does and is correct for a 'shortened' exponent), or if +2 would be better as users would expect acc. manual calculation?

mathematically one can justify both solutions? '-2' holds for a re-conversion 'second root of -2 to the power of 6', just like '2' doe's, but slides down to the 'odd roots' while the original term formulates an 'even root'?

'parentheses dictate' isn't applicable imho because the way of calculation prescribes the resolution of the parentheses according to mathematical calculation rules,

in other words: if 'This speculation is a sort of sophism that masks the fact that there are more than one nth roots. When you make any transformations of powers you must take into account the possible loss of some roots or introduction of new roots.' 'forbids' the expansion of 1/3 to 2/6, the conversion of 2/6 to 1/3 should also be avoided for fractional exponents, and thus +2 would be the correct result for 2/6 as exponent?

or in other other words: -2 violates 'As we cannot return such a result in a spreadsheet we agree to see only positive answer', the specified exception refers to 'odd roots',

still a little confused ...

[edit]
P.S. got it?

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

squareroot and 6-th root are 'even roots', and thus have two solutions, +/-x, thus calculating from 2/6 you get directly to 'also -2', and the reduced exponent 1/3 just reduces the ... (more)

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I don't know if this is a sophism, but in maths a^expr as a value is not supposed to be "decomposed" as a set of calculations. It stands for a^{value of expr}, whatever the expr.

In the case of a^b/c = a^{(b×k)/(c×k)} is still an identity but (a^b×k)^1/(c×k) is no longer guaranteed to yield the same value because of the k factor. When dealing with fraction exponent, the only safe computation involves irreducible fractions. With negative base, an added k factor in numerator and denominator is susceptible to change the result.

The use of the positive root is a long standing tradition in maths when they were not as elaborate as nowadays. Negative numbers were formalised only in the VIIth century. Algebra as we know it appeared in late XVIth century.

Nevertheless, this is not an excuse ...(more)

ajlittoz ( 2020-10-11 09:26:03 +0200 )edit

@ajlittoz: but how do you calculate a^b/c to a value? or e.g. a^0,333

newbie-02 ( 2020-10-11 23:59:01 +0200 )edit

If you can't ascertain that the exponent is a fraction m/n in lowest terms, it can't be done reliably for a<0 because a^x is not continuous. Even enlarging the problem to C (complex numbers) which sometimes eases processing does not help here because steps involve multi-valued functions (in factors of i 2 pi k).

a^x is well-defined only for a>0 in computer. The case a<0 must be specifically handled based on extra information in your possession.

ajlittoz ( 2020-10-12 08:23:44 +0200 )edit

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answered 2020-10-10 11:57:30 +0200

ajlittoz
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There is a flaw in your derivation because it does not take into account precedence of operators. Unfortunately this is implicit in maths classes and should be emphasised.

When you write (-1) ^ /(2 / 6), the parentheses are very important because they delimit the range of operators. The expression is made of 3 parts:

(-1) which applies unary operator - only to number 1

Incidentally, the APL language makes a distinction between the unary - operator and the negative sign by using a different character. This clears a common ambiguity.
^ the dyadic exponentiation operator
(2 / 6 ) an expression which is the right-hand operand to the exponentiation

The parentheses also mandates that the expression be computed before the exponentiation. Therefore ((-1)^2)^(1/6) is not equivalent to the original expression because you didn't compute (2 / 6 ) first.

In principle, this kind of question is off-topic (not LO specific) and is discarded.

To show the community your question has been answered, click the ✓ next to the correct answer, and "upvote" by clicking on the ^ arrow of any helpful answers. These are the mechanisms for communicating the quality of the Q&A on this site. Thanks!

In case you need clarification, edit your question (not an answer which is reserved for solutions) or comment the relevant answer.

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@ajlittoz: parentheses: they were not meant as 'formula expression', but to 'group operands'? i'm just unsure, 2^2/6 would be 4/6 regarding precedence of '^' over '/' and / or 'left to right' rule, thus 2 powered 2/6 needs to be written as 2^(2/6), and in school math if you write 2/6 with a horizontal breakline that preceedes '^'?
but to calculate the expression, 2/6 is 'decomposed' into numerator and denominator (instead of calculating it to 0.3-period) and calculated in two successive steps, base 'up' numerator and then 'root' denominator ... and there are different results for different representations of the same value ... (1/3, 2/6, 0,33333333...),(logically, not in calc, it seems to break everything down to 1/3)
it becomes a calc question when evaluating terms like '=-2^(3333333333333330/1E+016)', positive result reg. even enumerator in exponent, or negative result reg ...(more)

newbie-02 ( 2020-10-10 13:28:04 +0200 )edit

Mathematics notation is very ambiguous on first sight because it relies on really implicit rules. It is a 2D writing systems with agreed-upon precedence rules. When you write 2^2/6, it is implied that the division in the exponent should be computed first to reduce the expression to a single number. Otherwise you get the inconsistencies you pointed out.

You also made a mistake in your identity. In math, a^b × a ^c = a^b+c. Note the +,operator in the exponent, not ×. That is, you cannot decompose a^b×c into a^b × a^c.

In other words, a multiplication in the base expressions translate into an addition in the exponents. This is used in the logarithms to compute multiplications with additions.

ajlittoz ( 2020-10-10 16:13:33 +0200 )edit

@ajlittoz: nice,
i don't know how to format in 'superscript',
i didn't want to convert 'a to the power of b times a to the power of c' to 'a to the power of (b+c)', or vice versa, but
'a to the power of (b (2) diveded by c (6))' which is - afaik the 'c-th root' of a to the power of b
a^(b*1/c) = (a^b)^(1/c) = c-th root of a^b
isn't that correct?
'to convert the exponent first' is ... difficult ... the only recipe i know to handle fractional exponents is computing them separated in enumerator and denominator, powers like e.g. 2^0,333 are not defined afaik, only power 333 and root 1000 ... or am i wrong?

newbie-02 ( 2020-10-10 16:34:41 +0200 )edit

or am i wrong?

yes you are wrong: a^(b*1/c) does not equal (a^b)^(1/c)

Opaque ( 2020-10-10 17:03:55 +0200 )edit

a^b-c = a^b × a^-c = a^b / a^c

a^b×c =(a^b)^c: you can't factor it in the base. This is where you made a mistake.

PS: to format superscript, I used the HTML <sup> element.

ajlittoz ( 2020-10-10 17:05:01 +0200 )edit

@Opaque: but it holds for plenty cases ... just tried =3^(4*1/5) and =(3^4)^(1/5), both result in 2,40822468528069000000 - within the limitations of calc - and it's what i've learned in school ... for sure there are limitations with c being 0 or similar, but as common rule it holds? just to check an exotic case: =21^(1/14*-6/5) ... holds ... i know! that that's not a mathematical 'q.e.d.', but a proof that i'm not totally wrong?

newbie-02 ( 2020-10-10 20:27:57 +0200 )edit

@ajlittoz: i wanted to 'convert' a^b/c to a^(b*1/c) and that to (a^b)^1/c with the values a:-1, b:1, c:3, and a:-1, b:2, c:6, (-1¹)^1/3 is definitively -1 while (-1²)^1/6 results in 1 ... while 2/6 is at least very similar to 1/3, and thus the results should match ... what they do in calc, but which is the rule that forbids expanding 1/3 to 2/6?
hmmm ... coming nearer ... ^1/6 is something like a square root, and thus has two solutions, =1^1/2 is [1;-1]

newbie-02 ( 2020-10-10 20:37:45 +0200 )edit

In math, with b and c integers, a^(b/c) is not always (a^b)^1/c. The identity holds only of b/c is an irreducible fraction. Otherwise, as you notice, the common factors in the numerator and denominator play nasty tricks. With common factors, the operations are no longer distributive.

ajlittoz ( 2020-10-10 20:52:00 +0200 )edit

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