The Players Corner Archive

A formal proof will be most appreciated.

6 = squart(36)
squart(36) = squart(4*9)
squart(4*9) = squart(-4*-9)
squart(-4*-9) = squart(-4) * squart(-9)
squart(-4) * squart(-9) = 2i * 3i
2i * 3i = 2*3*i^2
2*3*i^2 = 6*i^2
6*i^2 = -6
therefor 6 = -6

- Lord Kranar, human Archwizard reg

The identity i isn't negative or positive, it's imaginary... however since i = squart(-1), then i^2 = -1.

- Lord Kranar, human Archwizard reg

Anywho, an old math trick. Why isn't this right? (Yes, I know why.)

Given A = B
A*A = A*B
A^2 = AB
A^2 - B^2 = AB - B^2
(A+B)*(A-B) = B(A-B)
A + B = B
B + B = B (Since A = B is given.)
2B = B
2 = 1

[This message has been edited by Gnomad (edited 11-29-2001).] reg

In anycase, back to my problem... I did request a PROOF that my statement was wrong.

You said that squart(-6)*squart(-9), but would this not violate the rule that:

squart(a*b) = squart(a)*squart(b)?

Now I know the proof is wrong, anyone can demonstrate it's wrong, but not everyone can provide a proof that shows that it's wrong. If anyone could do that I'd appreciate it.

- Lord Kranar, human Archwizard reg

quote:

---

Originally posted by LordKranar:
6 = squart(36)

---

The problem is in this line. Replacing 6 for a moment with a vector X; X != root (6^2). X = +/- root(6^2). X = <6,-6>. It's a second order solution vector. I'm not going to write a proof on why this is true.

6 = +/- root(36)

Go from there and you're set. 6 = +/- -6. reg

You can easily prove that 6 != -6, however I showed a proof that states that 6 DOES equal -6. This I have no doubt is wrong, but the question is, how can someone show it's wrong since it doesn't seem like I violated any algebraic rules.

Maybe however I did, I don't see anything wrong with the proof above, but if you can show where in that proof I made an error and prove why it's an error, I'd appreciate it.

- Lord Kranar, human Archwizard reg

A mathematician about 200 years ago made the statement that:

A^n + B^n can never = C^n if n > 2

and said he would explain why this is but "there is not enough room on the piece of paper to do so." Then some guy about 10 years ago figured it it was in fact true. He dedicated his entire life to it and figured it out. Anyone know what I'm talking about or am I just babbling?

Johnny reg

Since squart(36) = +/- 6, and your proof only assumed it = 6, that's why it works.

Sonic reg

a^n + b^n + z^n = 0 has no solution for all values of n > 2 when n is an element of the natural numbers.

The theorem was proven I think 2 or 3 years ago.

Back to 6 = -6, the statement that

6 = +/-squart(36) is wrong.

6 = +squart(36) and only +squart(36) not -squart(36) which equals -6.

One has to keep in mind that there exists two types of square roots, the principle root and the negative root. The pricinple root is the radical sign (that weird looking checkmark with a roof) and the negative root is the radical sign with a minus infront of it. When discussing both the principle root and negative root you symbolize that with a radical sign with a +/- infront of it. In my proof however since there is no character for the radical sign, it should be interpreted as the principle root and only the principle root since the value we're dealing with 6 is equal to the principle root of 36 and is NOT equal to the negative root.

- Lord Kranar, human Archwizard

[This message has been edited by LordKranar (edited 11-29-2001).] reg

I just noticed a flaw with this anyways (even though it's irrelevant).

You can't take the square root of a vector. Vectors have their own algebraic rules, in just as much as you can't divide a vector, you can't multiply one nor can you take the root of a vector.

Therefor the statement that X = +/-squart(6^2) is false when X is a vector.

A vector is NOT a number. It's not a negative or a positive, you can't add a number to a vector, or subtract a number with a vector, or multiply or do anything with a number since a vector requires both a magnitude (which itself isn't a number but can be quantified) and a direction (which can not be quantified).

- Lord Kranar, human Archwizard

[This message has been edited by LordKranar (edited 11-29-2001).] reg

I think the flaw has to do with order of operations involving exponents when dealing with complex numbers. With real numbers it's probably not important what order you evaluate the exponents, but this may not be the case with a complex number.

- Lord Kranar, human Archwizard reg

quote:

---

Originally posted by LordKranar:
Someone please PROVE that the following is infact incorrect.

A formal proof will be most appreciated.

6 = squart(36)
squart(36) = squart(4*9)
squart(4*9) = squart(-4*-9)
squart(-4*-9) = squart(-4) * squart(-9)
squart(-4) * squart(-9) = 2i * 3i
2i * 3i = 2*3*i^2
2*3*i^2 = 6*i^2
6*i^2 = -6
therefor 6 = -6

- Lord Kranar, human Archwizard

---

6 = squart(36) - Correct

squart(36) = squart(4*9) - Correct

squart(4*9) = squart(-4*-9) - Incorrect

squart(-4 * -9) really equals 2 * 3 * i^2 = -6
Always remove the imaginary before performing a calculation.

------------------
2*3*i^2 = 6*i^2
6*i^2 = -6
therefor 6 = -6

----------------

The above is false also. What you have really come up with is -6 = -6 not 6 = -6.

i^2 = -1 not 1.

reg

I mean, heck, while I'm no engineer or nuclear physicist, I can't think of a situation where they'd need to use all of these weird types of math. Anything I can think of, it all comes down to basic math.

Need a lil more of this and less of that? Okay. Too hot for this and need to cool it down? Okay. Two things conflicting? Okay. Oy! I just can't find a situation where something like this would be applied..unless you were trying to write your own type of math. Of course, taking those courses would probably help me understand more, lol.

Just curious is all,

- Pudgee reg

Alot of times engineers use imaginary numbers to solve their equations. Its used in quantum mechanics also. Its big with the fractal crowd too, or so im told (though i dont get into that).

'lood

reg

Well complex numbers have a lot of use in more advanced algebra and geometry. I'm not too into calculus, but that's not to say calculus isn't interesting or not important. I do however find algebra and geometry very useful when programing, infact it's math I use on a daily basis. Matrices, vectors, lines and planes, complex numbers, all are an essential part of programming anything from 3d engines, to simple 2d engines. You'll find that any trig ratio does actually involve complex numbers, infact so much about math involved complex numbers although you probably wouldn't have guessed (cause you got your calculator to do all the dirty work for you!).

But math is interesting for more than simply its practical application. Apart from being an extremely powerful tool, math is a way of thinking, and a realm of its own, infinitely large and with infinitely more theorems and patterns to be discovered. It's kind of neat asking questions like what's 1 to the power of infinite, or taking the tangent of a complex number, or just knowing how to calculate irrational numbers like pi.

Without math, those questions are meaningless. With math, those questions have very unusual and facinating answers.

<< Its big with the fractal crowd too, or so im told (though i dont get into that). >>

It's not that it's really big with the fractal crowd, but it does make an appearance in the Mandelbrot set.

- Lord Kranar, human Archwizard

[This message has been edited by LordKranar (edited 03-12-2002).] reg

First, it helps to define what we mean when we talk about the square root of a complex number. In general, any non-zero complex number z will have two distinct (and opposite) square roots u and v such that u^2 = v^2 = z.

We can, however, make an arbitrary choice between these two square roots, which we might call the principal square root. We could, for example, define the principal square root of a complex number z to have positive real part (when z is not 0 or on the negative real axis), to have a positive imaginary part (when z is on the negative real axis), or to be 0 (when z is 0). That lets us choose between the square roots of z in all cases and gives us a one-to-one function. Only by making such a choice can we talk about a complex square root having a single value, which is what the 6 = -6 proof has done.

Alas, however, our principal square root function is not without problems. It lacks certain properties we take for granted with the square root function we are used to dealing with on the non-negative real numbers. For example, it is discontinuous along the negative real axis.

More to the point, let u and v be complex numbers. For our principal square root function, the purported identity

sqrt(u*v) = sqrt(u)*sqrt(v)

does not hold. Basically, the problem arises because although sqrt(u) and sqrt(v) must have non-negative real components, their product need not do so, while sqrt(u*v) must also have a non-negative real component. And that's the problem with the 6 = -6 proof. It uses a false identity that looks true, but isn't.

Gemexchange

reg

Sometimes things that are just so obvious and in your face... are so very hard to prove.

<< how about you give me six bucks... ... ... >>

Yeah... let's all use money to disprove mathematical theorems involving complex numbers.

- Lord Kranar, human Archwizard

[This message has been edited by LordKranar (edited 03-26-2002).] reg