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BasPilot · Expert Joined: 12 Jun 2005 Forum ID: 5836

Okay, here's a question for you guys...

Because I cannot use a radical, I will use ^-1 to mean square root...

If, -a X -a = a (positive) then why does i X i = -1?

Discuss that one...

PS - for you lowerlings, i is not just some variable as 'a' is in the first equation... i is the imaginery number equal to (-1)^-1 (or the square root of negative one)

The reason i was named is because there is not a possible way that any number squared can equal negative one.

Let's see who can come up with this concept...

epiphanic · Pro Joined: 05 Sep 2005 Forum ID: 6173

I may be WAY off, but if ^2 * ^2 = 2, then it stands that the squaring and square root is itself.

Here's something that blew my mind,

Let a=x

a+x = 2X, subtract 2a from both sides, giving

-a+x = 2x - 2a, factor the 2,

-a+x = 2(-a+x), divide by (-a+x) and you get

1 = 2
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Durin · Posted: Thu Nov 17, 2005 10:37 pm Post subject:

The basic, easy explaination is that i is an imaginary number (not real), so the standard rules of algebra can't always be applied in scenarios involving i.

On the other hand, if i is the square root of negative one, it's only logical that by multiplying it by itself, you are squaring it, and by squaring it, you are performing the inverse of the square root function, and thus can assume that the number inside the square root remains the same.

In other words, if f(x)=x^2 and (f^-1)(x)=x^(1/2), then f[(f^-1)(x)]=x.

*x^2 denotes x squared, (f^-1)(x) denotes the inverse function of f(x), x^(1/2) denotes x raised to the power of one-half (which is the same as the square root of x), f[(f^-1)(x)] is the composition of the function f(x) with its inverse function (f^-1)(x).

I confused myself a little here, but I think what I typed makes sense.
Very Happy

Edit (to avoide double posting)-

epiphanic · Pro Joined: 05 Sep 2005 Forum ID: 6173

Yea Durin! You guys figured it out alot quicker than I did. Good work.
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yagadan05 · Guest Forum ID: -1

Math makes me throw-up. It really does. Mad

Durin · Posted: Fri Nov 18, 2005 12:20 am Post subject:

Lola Lola · Posted: Fri Nov 18, 2005 12:43 am Post subject:

In another thread, I asked for the square root of i and no one responded. i has two square roots and they both lie in the field of complex numbers (all complex numbers have two square roots in the field)

Hint: The field of complex numbers can be represented a a two dimensional field with real numbers being the x axis and imaginary numbers the y axis. The square root of i has to have an absolute value of 1.
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ALT · Posted: Fri Nov 18, 2005 11:47 am Post subject:

Chippy · Posted: Fri Nov 18, 2005 12:30 pm Post subject:

What Durin said is, to the best of my knowledge, basically right. i is an imaginary number, thus basic rules of algebra can't always be applied to it.
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Beaner · Posted: Fri Nov 18, 2005 1:18 pm Post subject:

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Lola Lola · Posted: Fri Nov 18, 2005 2:24 pm Post subject:

Durin said:

BP · Veteran Joined: 14 Jun 2003 Forum ID: 1901

Durin · Posted: Fri Nov 18, 2005 6:17 pm Post subject:

Lola Lola · Posted: Sat Nov 19, 2005 9:05 am Post subject:

The solution to the square root of i can be found at the following website:

http://www.math.toronto.edu/mathnet/questionCorner/rootofi.html
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Whatever Lola wants, Lola gets
And little man, little Lola wants you
Make up your mind to have no regrets
Recline yourself, resign yourself you're through

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