# Unclear situation involving roots.

1. May 10, 2010

### QED-Kasper

Let's say we have:
A=B3
r=6B2
and we want to rewrite A in terms of r.
My way to do it is:

B2=r/6 => B=+-$$\sqrt{(r/6)}$$ => B3=+-(r/6)$$\sqrt{(r/6)}$$ => A=+-(r/6)$$\sqrt{(r/6)}$$

In a book I found the following solution:

B2=r/6 => B3=(r/6)1.5 => A=(r/6)1.5

As you can see the book doesn't put a +-. Doesn't this mean that their answer is wrong as it only gives the positive answer? I think this can be brought back to a simpler problem:

$$\sqrt{4}$$ = 2
not +-2.
+-$$\sqrt{4}$$ = +-2

Writing the 4 within the root symbol doesn't by definition imply that its either +-2, right?. It means that this is a positive root of 4. In order to include both the positive and negative root of 4 one must write +- in front of it, right? So does this mean that the books answer only includes the positive value, thus making it an incomplete and wrong answer?

I would like to add that in the book itself, the problem is stated as follows: Rewrite this system of equations as A(r) = and rearrange (its from a dutch book, I'm not sure if rearrange is the proper word in English) as far as possible.
So if A is a function of r, it means that it cant have both + or - values for the same r. So is the books answer valid in that case as only the principal value is taken to be a valid solution of the function? Or is it still wrong as the ± sign has to be used, because it symbolizes there are two functions and the answer actually includes two separate functions. Or maybe then the ± sign has to be put in front of the A aswell? (to distinguish between two functions, because A(r) by definition only defines one function so either another F(r) has to be put there for the negative equation, or one can just write ±A(r) where -A(r) = F(r))

Thank you.

Last edited: May 10, 2010
2. May 10, 2010

### tiny-tim

Hi QED-Kasper!

(have a square-root: √ and a ± )

Yes, I agree with you …

for example, if A = B = -1 and r = 6, then A = minus (6/6)1.5

3. May 10, 2010

### QED-Kasper

hi tiny-tim :)

Thank you! So (r/6)1.5 doesn't by definition include both the positive and negative value? I have a friend who claims otherwise. He said that he wouldn't put a +- sign unless he would calculate out this expression for example if he had a value for r. (He also claims that $$\sqrt{4}$$ inherently means -2 or +2.) He says that just raising both sides of the equation to the power of 1.5 is enough and that no +- has to be written in front.
I somehow struggled to explain why I thought this is wrong. Could you elaborate a bit further on this, so I could logically and mathematically be able to explain to my friend why this is wrong?

4. May 10, 2010

### tiny-tim

Hi QED-Kasper!

(what happened to that √ i gave you? )
Certanly not.

Any expression like that, or √, or sin-1 etc, is deemed to give the principal value.

The principal value is established by convention … in this case, as the positive value.

(btw, try asking your friend whether he would say that (r/6)1.49 and (r/6)1.51 also include a negative value )

5. May 10, 2010

### QED-Kasper

hi again :D

√ ± i like these signs, thx :) how do you make them?

OK, to be a bit more extreme, is there any "law book" that I could use as a proof, that the "principal value" rule (when writing just √4 it means its equal to 2) , has indeed been established?

"(btw, try asking your friend whether he would say that (r/6)1.49 and (r/6)1.51 also include a negative value ) " I'm sorry, could you explain what you mean?

I would like to add that in the book itself, the problem is stated as follows: Rewrite this system of equations as A(r) = and rearrange (its from a dutch book, I'm not sure if rearrange is the proper word in English) as far as possible.
So if A is a function of r, it means that it cant have both + or - values for the same r. So is the books answer valid in that case as only the principal value is taken to be a valid solution of the function? Or is it still wrong as the ± sign has to be used, because it symbolizes there are two functions and the answer actually includes two separate functions. Or maybe then the ± sign has to be put in front of the A aswell? (to distinguish between two functions, because A(r) by definition only defines one function so either another F(r) has to be put there for the negative equation, or one can just write ±A(r) where -A(r) = F(r))

Last edited: May 10, 2010
6. May 10, 2010

### tiny-tim

Hi QED-Kasper!
I make them by just typing on my standard Mac keyboard (they've been standard every since Macs started).

If you don't have a Mac, you will have to copy them from my post above.
Not that I know of.

It's just a bit obvious that every function is supposed to have a unique meaning (ie, be single-valued).

If eg √ could be ±, it wouldn't be a function.

Well, I expect he'd say no to that …

in which case why does he have a special rule for 1.5 which he doesn't have for 1.49 and 1.51?
That seems to be begging the question, ie it's assuming, for no apparent reason, that A is a function of r.

(and of course a function must be single-valued)

7. May 10, 2010

### D H

Staff Emeritus
No. The book is correct.

There is a big difference between $a^{1/2}$ and $\sqrt a$. The first denotes both the positive and negative solutions. The latter denotes the principal value only. In general, $a^b$ is a multi-valued function.

8. May 10, 2010

### Redbelly98

Staff Emeritus
Sometimes a textbook will state that all values are nonnegative for a set of problems involving exponents, in order to avoid this confusion. I wonder if that could be the case here.

9. May 10, 2010

### QED-Kasper

@D H
So tiny-tim is wrong?

From http://mathworld.wolfram.com/SquareRoot.html
There appears to be no difference in meaning between both notations according to wolfram. :s

Last edited: May 10, 2010
10. May 10, 2010

### Martin Rattigan

How many values then would $2^\pi$ represent? And could you define $2^\pi$ as the limit of $2^q$ as $q$ approaches $\pi$ through rational values? If not, how would you define it and would the same definition apply to $2^{1/2}$ ?

And would you say the derivative of $2^x$ is undefined in $\mathbb{R}^+$?

Last edited: May 10, 2010
11. May 10, 2010

### HallsofIvy

I have a problem with this. You have to be very careful with your notation. Normally when we write "$x= \pm a$" we mean that either of those two values will satisfy whatever properties x is supposed to satisfy.

That is not the case here. If B= 1, then r= 6 and so $$\pm\frac{r}{6}\sqrt{\frac{r}{6}}= \pm 1$$.

But, in fact, A= B3= 1 only.

If B= -1, on the other hand, r= 6 still and $$\pm\frac{r}{6}\sqrt{\frac{r}{6}}= \pm 1$$ but now A= -1 only.

With $A= B^3$ and $r= 6B^2$ we simply cannot "write A in terms of r only". Calculating r loses information about B (its sign) that is essential for calculating A.

We can only say that if $A= B^3$ and $r= 6B^2$ then A is either $$\frac{r}{6}\sqrt{\frac{r}{6}}$$ or $$-\frac{r}{6}\sqrt{\frac{r}{6}}$$ but we don't know which.

Last edited by a moderator: May 10, 2010
12. May 10, 2010

### QED-Kasper

Both answers are valid because both satisfy all equations so the ± is justified. The reason why I must use the ± sign is exactly because we dont know the value of B, it could be negative or positive. But we do know the value of r, which is positive. I'm not sure what you mean though. I'm just confused why the book doesn't have a ± in front of its answer. Doesn't the answer unjustly only include the positive value?

13. May 10, 2010

### tiny-tim

I still think I'm right.

If a1/2 a3/2 etc are double-valued, then presumably z1/6 is 6-valued, and so is z3/6

so z3/6 ≠ z1/2.

I do accept there could be a convention that makes that last equation true, but I've never come across it.

(and while I'm not unhappy about such a convention for z9/6 and z3/2, I woulnd't like it for exponents written non-fractionally, as in z1.5)

However, even if that convention applies, it doesn't make the book correct …

the statement A = (r/6)1.5 would give A two values, while in any particular case it clearly only has one value.

HallsofIvy makes a similar point about ± …

… A = ±(r/6)1.5 is just as wrong as A = (r/6)1.5 would be under the two-valued convention, and for the same reason.

14. May 10, 2010

### D H

Staff Emeritus
The first supposition is correct; the latter is not. z3/6 is double valued. Think about it in terms of the primitive roots of unity.

15. May 10, 2010

### tiny-tim

oops!

oops!

16. May 10, 2010

### Martin Rattigan

I believe when defining rational powers the convention is rather to make the last (in)equation false. I quote from Wikipaedia:

________________________________________________________________________
When one speaks of the n-th root of a positive real number a, one usually means the principal n-th root.

Rational powers

A power of a positive real number $a$ with a rational exponent $m/n$ in lowest terms satisfies

$a^{m/n}=(a^m)^{1/n}=\sqrt[n]{a^m}$

where $m$ is an integer and $n$ is a positive integer.
_______________________________________________________________________

Note the reference to lowest terms.

But since this definition would generally be restricted to the positive reals, and the definition $z^w=e^{wLog(z)}$ (with $Log$ meaning the principal $log$) used when complex numbers are involved, there would only be two values to consider for $x^{1/6}$ in this case, even for people who wish to take $x^y$ to be multivalued.

It is clear that Wikipaedia at any rate agrees with you that the definition of $x^y$ is normally taken to be single valued.

The questions I posed in my previous post were intended to suggest that any multivalued definition (if that could properly be called a definition) would be unworkable.

17. May 10, 2010

### D H

Staff Emeritus
Wikipedia is not the best source for resolving a dispute. You must admit that there certainly is some utility to having a multifunction that represents both solutions of x2-a=0 , and also quite a bit of utility to having a function that represents the principal value.

In any case, we have wandered a bit from the original post. As Halls pointed out, there is only one solution, not two. Using ± is incorrect. A will be positive if B is positive; there is no negative solution. Similarly, will be negative if B is negative; there is no positive solution in this case. If the problem specified that B is positive (we don't know) then the book's solution correct assuming the book is using (r/6)1.5 to denote the principal value.

18. May 10, 2010

### Martin Rattigan

I have to admit to failing to understand the distinction. I would count two solutions.

If a circle of radius 2 be drawn with centre (0,1) I would be quite happy to say that it intersects the x axis when x = ±√3.

It will only intersect the x axis at -√3 when the angle $$\theta$$ between the radius and the positive direction of the x axis is $$\pi$$/6 and it will only intersect the x axis at +√3 when $$\theta$$ is 5$$\pi$$/6. But I don't see how the value of $$\theta$$ is relevant to the statement in the previous paragraph.

Similarly for the value of B in the original question.

But I would certainly agree with you that if the book specified that B (or A) were positive the answer it gave would be correct.

Last edited: May 10, 2010
19. May 10, 2010

### QED-Kasper

All we know is that B squared is equal to some positive value. So B could be negative and positive. Thus the final answer could be negative and positive. So how can the book be correct?

20. May 10, 2010

### Martin Rattigan

Sorry, I just noticed that you already said in #12 that you weren't given the sign of B, in which case I'd agree with Tiny Tim that the book is incorrect and the expression you originally gave is correct.

But as you can see there would also be dissenting views.