Colin Foster

Posted on Sunday, 28 March, 2004 - 02:56 pm:

When we write $\sqrt{64}$ , the convention is that the positive square root is meant. Is this the same when we write $64^{1 / 2}$ ? If you want to indicate both roots, do you have to put $\pm 64^{1 / 2}$ ?
What is the convention about fractional powers that are unsimplified; e.g., $(- 8)^{2 / 6}$ ?
Do you simplify the fraction first to get $(- 8)^{1 / 3} = - 2$ or do you work out $((- 8)^{2})^{1 / 6} = 64^{1 / 6} = + 2$ ?
When you have a power to the power of something else, such as '' $\sqrt{2}$ to the power of $\sqrt{2}$ to the power of $\sqrt{2}$ '' is there a convention for whether you mean $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{2} = 2$ or $\sqrt{2}^{\sqrt{2}^{\sqrt{2}}} =$ I'm not sure what! (How do you work out something like this?)

When we write

e^{x^{2}}

we don't mean

(e^{x})^{2} = e^{2 x}

, so I think it must be the latter, but why?

The rules of BIDMAS don't seem sufficient for dealing with this.

Many thanks.

David Chen

Posted on Sunday, 28 March, 2004 - 07:12 pm:

Interesting Colin, I'm not sure about 2. But I agree with you for 3, it should be the latter one. Since

e^{x^{2}} = e^{(x^{2})}

, it differs from

(e^{x})^{2}

For 1, in my own opinion, you should indicate both roots whenever it needs. I think sqrt(64) only refers to positive sqrt(64), so you can just write 64^1/2 . But I found that quite a lot people when solving problems, they wrote sqrt(64)=+(-)8, which I think they are wrong.

Marcos

Posted on Monday, 29 March, 2004 - 09:48 am:

When I was doing O-level maths my teacher, when I asked her, said that

64^{1 / 2}

was the same thing as

\sqrt{64}

but I think that this isn't actually the case.

64^{1 / 2}

means and so can refer to either square root. When you meet complex numbers (if you haven't already) something like

a^{1 / n}

is multivalued and we can't order complex numbers (in any useful way) like we can with reals so we can't restrict to ''the positive root(s)''.

Marcos

Ian Short

Posted on Monday, 29 March, 2004 - 11:38 am:

I think David and Marcos have answered all of your queries now as Marcos' message explains (2).

Ben Challenor

Posted on Monday, 29 March, 2004 - 05:16 pm:

My question is pretty similar so I'll post it here.

When substituting values into an expression with more than one root in it, eg:

LaTeX Image

Can you take one root to be positive and one to be negative? Would the above expression have four values or two?

Marcos

Posted on Monday, 29 March, 2004 - 05:42 pm:

I assume you're defining

\sqrt{x}

to mean either square root (or else your question doesn't make sense). I think there situations where this may be a problem. For example (if you've met complex numbers) a famous paradox is:

$1 = \sqrt{1} = \sqrt{- 1 \times - 1} = (- 1)^{1 / 2} (- 1)^{1 / 2} = [(- 1)^{1 / 2}]^{2} = - 1$

The paradox is at once resolved if we understand exactly what is meant by the step $(- 1 \times - 1)^{1 / 2} = (- 1)^{1 / 2} (- 1)^{1 / 2}$ . For the correct selection of square roots the equality does hold, for example when we choose one $(- 1)^{1 / 2}$ to be $i$ and the other to be $- i$ and the LHS is 1 (as we've started off with it being 1).

Your mapping is one to many. Usually we aim to deal with many to one (or ideally one to one functions, so that they have inverses) and this can be achieved with your expression if we define $\sqrt{x}$ to be the positive square root.

Perhaps someone can correct me as I'm not entirely sure of my answer,
Marcos

Ben Challenor

Posted on Monday, 29 March, 2004 - 06:36 pm:

Using my expression above as an example..

Say x=64. Can you substitute in +8 for the denominator and -8 for the numerator (or vice versa) ? Or do you take root(x) to be the same value throughout?

This would 'create' two more solutions. Actually, the above is a bad example, as two of them will be the same. You'd get -9/9 , 7/-7 , -9/-7 , 7/9.