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What do you think of this :
All seems perfect and nothing seems violate math rules... but....
Edited: 4 Aug 2013, 6:22 p.m.
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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.
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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.
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Quote:
The first step is perfectly fine.
BS
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...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
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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
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: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.
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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 rightassociative as already mentioned by an other member).
Franz
Edited: 6 Aug 2013, 4:05 a.m.
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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.
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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
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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
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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.
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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.
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...
f/ There is no error because LOG(5)=(LOG(5)+i*PI)/LOG(10) ;)
Edited: 6 Aug 2013, 4:18 a.m.
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[HPPrime 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 HPPrime rename evalc to evalCplx
Thanks
more info
http://www.wolframalpha.com/input/?i=%288%29%5E%281%2F3%29
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Watch it: 64^{1/6} = ±2 !
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So we can say that
sqrt(4)=±2
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I hate opening that box once again, but there's a "convention" declaring SQRT(4) = +2 only. OTOH, 4^{1/2} = ± 2 since it inverts (±2)^{2} = 4.
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The sixth roots of 64 are: 2, 2, square_root(3)*i+1,square_root(3)*i+1, square_root(3)*i1, square_root(3)*i1
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Exponentiation is rightassociative, thus:
Cheers
Thomas
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So much ado just to come to terms with the fact that square root is a multivalue 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
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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
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(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
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Quote:
(8)^(2/6) is taking the the six roots of (8)^2, or 64
Mathematics: fail
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Both with 50G and Prime CAS :
a^b^c is not (a^b)^c but a^(b^c)
Exponentiation is a curious beast : rightassociative as wrote Thomas.
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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.
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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.
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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
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Oh ! I see it now !
Thanks
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Quote:
to show that ^ is right associative
Otherwise a^{bc} = a^{bc}, 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 rightassociative.
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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^38=0,x)
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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.
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What about [ 1 0 0 8 ] PROOT to get the roots of the polynomial x^3 + 8 = 0 ?
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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)]
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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.
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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.
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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)=b ^{q} 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.
