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Cubes and the Imagination

by Mentallic
Tags: cubes, imagination
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Mentallic
#1
May8-09, 02:07 AM
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This problem had arose in a question I was trying to answer in which I could take two approaches to solve it, but one method would never work, even though I was almost certain there were no algebraic errors made.

Simplified, a section of the problem was as so:

[tex](\sqrt{-a})^3[/tex]

Now, I would go about trying to cube the inside of the root, and cubing a negative gives a negative so I end with:

[tex]\sqrt{-a^3}[/tex]

but after vigorous searching for algebraic errors and finding none, I took another look at this part of the question and (not trusting negatives in roots) went about it like this:

[tex](\sqrt{-a})^3=(i\sqrt{a})^3=i^3\sqrt{a^3}=-i\sqrt{a^3}=-\sqrt{-a^3}[/tex]

which gave me the correct answer. But I'm stumped as to why the first method didn't work
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deferro
#2
May8-09, 03:07 AM
P: 49
Because i is a universal inner measure or a kind of polarizing function that resolves or inflects the cube.
Trying to use integers doesn't work, but can you see why?
HallsofIvy
#3
May8-09, 06:41 AM
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Quote Quote by deferro View Post
Because i is a universal inner measure or a kind of polarizing function that resolves or inflects the cube.
Trying to use integers doesn't work, but can you see why?
I really, really hope this is some kind of silly joke!

HallsofIvy
#4
May8-09, 06:43 AM
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Cubes and the Imagination

Quote Quote by Mentallic View Post
This problem had arose in a question I was trying to answer in which I could take two approaches to solve it, but one method would never work, even though I was almost certain there were no algebraic errors made.

Simplified, a section of the problem was as so:

[tex](\sqrt{-a})^3[/tex]

Now, I would go about trying to cube the inside of the root, and cubing a negative gives a negative so I end with:

[tex]\sqrt{-a^3}[/tex]

but after vigorous searching for algebraic errors and finding none, I took another look at this part of the question and (not trusting negatives in roots) went about it like this:

[tex](\sqrt{-a})^3=(i\sqrt{a})^3=i^3\sqrt{a^3}=-i\sqrt{a^3}=-\sqrt{-a^3}[/tex]

which gave me the correct answer. But I'm stumped as to why the first method didn't work
The "exponent rule" (ax)y= axy is true for real numbers but not for general complex numbers.

In particular, [itex]\left(\sqrt{a}\right)^x= \sqrt{a^x}[/itex] is not true for complex numbers.
deferro
#5
May8-09, 07:44 PM
P: 49
Halls of Ivy, would you like to discuss the 2-ribbon problem, or the derivation of root(-1)?
Mentallic
#6
May8-09, 09:48 PM
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Quote Quote by deferro View Post
Because i is a universal inner measure or a kind of polarizing function that resolves or inflects the cube.
Trying to use integers doesn't work, but can you see why?
Sorry, I don't understand this.

Quote Quote by HallsofIvy View Post
The "exponent rule" (ax)y= axy is true for real numbers but not for general complex numbers.

In particular, [itex]\left(\sqrt{a}\right)^x= \sqrt{a^x}[/itex] is not true for complex numbers.
But [tex]\sqrt{-a}^3[/tex] would still be defined for a<0. For a>0, being complex, am I not allowed to go about the simplification as I did before?

Let a=2, then [tex]\sqrt{-2}^3\neq\sqrt{-8}[/tex]?

[tex]\sqrt{-8}=2\sqrt{2}i[/tex]

Taking the other approach: [tex]\sqrt{-2}^3=(i\sqrt{2})^3=i^3\sqrt{8}=-2\sqrt{2}i[/tex] which is incorrect?

From what I know about complex numbers, taking the root of a non-positive number (negative or complex) yields both positive and negative values.

e.g. [tex]\sqrt{-1}=\pm i[/tex] or [tex]\sqrt{i}=\pm\frac{\sqrt{2}}{2}(1+i)[/tex] or am I wrong here?
Russell Berty
#7
May9-09, 12:15 AM
P: 105
The square root symbol is defined to return only the principal square root of the number. In the case where the radicand is a nonnegative real number, the principal root is the positive root.
Thus,

[tex]2=\sqrt{(-2)(-2)}\neq\sqrt{-2}\sqrt{-2}=-2[/tex]


So compare the following and notice we cannot “break up” the roots of -2 as above.

[tex]\left(\sqrt{-2}\right)^{3}=\sqrt{-2}\sqrt{-2}\sqrt{-2}=-2\sqrt{-2}[/tex]

But,

[tex]\sqrt{\left(-2\right)^{3}}=\sqrt{(-2)(-2)(-2)}=\sqrt{4(-2)}=2\sqrt{-2}[/tex]
jbunniii
#8
May9-09, 12:50 AM
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The problem is that the square root is a multivalued function: for any complex [tex]a \neq 0[/tex], there are two distinct complex numbers, [tex]b[/tex] and [tex]-b[/tex], such that

[tex]b^2 = a[/tex] and [tex](-b)^2 = a[/tex]

Thus there are also two distinct square roots of [tex](-a)[/tex], namely [tex]ib[/tex] and [tex]-ib[/tex], such that

[tex](ib)^2 = -a[/tex] and [tex](-ib)^2 = -a[/tex]

Then if I cube both of these square roots, I again get two distinct answers that are both equally valid:

[tex](ib)^3 = -i b^3[/tex] and [tex] (-ib)^3 = i b^3 [/tex]

Your "paradox" is that you did the calculation two different ways, each of which implicitly assumed that there was only one answer. One way you ended up with the first valid solution, [tex]-ib^3[/tex], and the other way you ended up with the second valid solution, [tex]ib^3[/tex], and you expected them to be equal even though there's no reason for them to be.
Mentallic
#9
May9-09, 03:24 AM
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Quote Quote by Russell Berty View Post
The square root symbol is defined to return only the principal square root of the number. In the case where the radicand is a nonnegative real number, the principal root is the positive root.
Thus,

[tex]2=\sqrt{(-2)(-2)}\neq\sqrt{-2}\sqrt{-2}=-2[/tex]


So compare the following and notice we cannot “break up” the roots of -2 as above.

[tex]\left(\sqrt{-2}\right)^{3}=\sqrt{-2}\sqrt{-2}\sqrt{-2}=-2\sqrt{-2}[/tex]

But,

[tex]\sqrt{\left(-2\right)^{3}}=\sqrt{(-2)(-2)(-2)}=\sqrt{4(-2)}=2\sqrt{-2}[/tex]
Thanks for making this very clear

Quote Quote by jbunniii View Post
Your "paradox" is that you did the calculation two different ways, each of which implicitly assumed that there was only one answer. One way you ended up with the first valid solution, [tex]-ib^3[/tex], and the other way you ended up with the second valid solution, [tex]ib^3[/tex], and you expected them to be equal even though there's no reason for them to be.
If I did the answer two different ways which are both correct, then the question I was attempting to answer should have two solutions, which the book does not express (I might have to post the question + my working). The multivalued function of the square root as you have said doesn't help to explain why one of the solutions to my problem is invalid.
deferro
#10
May9-09, 03:25 AM
P: 49
And, no-one has explained why -1 exists at all. Or how we know it does and why we do; or as I stated earlier why i is an inner polarization operator for number valued (and any kind of) functions.

It is, because without being able to induce its existence we would not have computers.
(Because I know /giggle)
Russell Berty
#11
May9-09, 04:13 AM
P: 105
I believe, historically, i was introduced to name a solution to x^2 = -1.

I dunno if this is what you want, but here is a “real world” interpretation of i.

In the real world, we have the idea of marking a central point on a line (call it 0.) Then we can mark evenly spaced units along a direction (call them 1,2,3…) Then, make subdivisions (fractions.) For irrationals, we need to fill in all “gaps” (for this we name a marker at the least upper bound of any bounded set of our numbers so far. For example, the square root of two is the name of the marker that lies just above all numbers x such that x^2 < 2 and below all number y such that y^2 > 2.

Multiplying by a positive number is then visualized as stretching or compressing by some amount.
However, also in the real world, we have a notion of rotation (and reverse direction, but here rotation is key.) Picture multiplication by a negative number as flipping 180 degrees about 0 on the real number line (and stretching/compressing.) So, we put a mark where we end up from 1 after rotating 180 degrees, and call it -1.
So, multiplying by 1, leaves you where you are. Multiplying by -1 is rotating 180 degrees. Hence, starting at -1 and multiplying by -1 rotates you 180 degrees to 1;
(-1)*(-1) = 1. Hence, -1 is a square root of 1.
What about rotating by 90 degrees (say, couter-clockwise)? If we do it twice starting at one, we are at -1 (180 degree rotation.) So, whatever this 90 degree rotation is, it is a square root of -1. Starting at 1 and rotating 90 degrees counter-clockwise, we put a marker and call it i. Then i*i = -1. Also, -i is 180 degrees from i about 0.
deferro
#12
May9-09, 04:35 AM
P: 49
There was this bloke called George, a cobbler whose logic strongly implies or 'proves' that propositions require the root of -1.

in fact if you couldn't reciprocate there would be no propositions.
Dragonfall
#13
May9-09, 11:57 AM
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Quote Quote by deferro View Post
i is an inner polarization operator for number valued (and any kind of) functions.
I'm almost certain that's from "Fashionable Nonsense" by Alan Sokal.
HallsofIvy
#14
May9-09, 12:46 PM
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Thank you, Dragonfall, now I understand.
deferro
#15
May9-09, 12:50 PM
P: 49
I'm almost convinced you don't know why -1 exists.
If it didn't, what would you do? Do you know how to explain why you believe it does?
Given there is no other way to prove it, than by "using" subtraction, which is reciprocation.

Math often does assume things, like 'continuous', you don't prove a formula has continuity from first principles or prove the circle has a circumference of 2pi with an exhaustive analysis.

however, -1 is different, it exists, obviously and we accept this. I bet you will have a hard time demonstrating exhaustively that the proposition: "-1 does not exist" is false (that means, not true).

p.s. it's not like I care or anything dumb like that...
p.p.s has anyone dismissing the proposition "i is a polarization operator" done any real math? by which I mean Boolean logic of course.
Dragonfall
#16
May9-09, 05:06 PM
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You know what else is a polarization operator?

[tex]\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&1\\1&-1\end{array}\right)[/tex]

It turns [tex]\left| 0\right>[/tex] into [tex]\frac{\left| 0\right> +\left| 1\right>}{\sqrt{2}}[/tex].
deferro
#17
May10-09, 02:05 AM
P: 49
Yay! the Hadamard, a recursive, or 'the recursive' function that is involute. Which is generally available.

But why does H imply i?


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