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> i Problems continued..., i = sqrt -1
PJS
post Nov 16, 2009, 11:41 AM
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x^3 - 2x^2 = 9
-2x^2 + x^3 = 9
-2x^2 = 9 - x^3
-2 = (9/x^2) - (x^3/x^2)
-2 = 1 - x
-2 - 1 = -x
-3 = -x, multiply both sides by -1
x = 3

Code Buttons x^3 = 27.

P.j.S
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P.j.S
post Nov 16, 2009, 12:03 PM
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QUOTE(PJS @ Nov 16, 2009, 11:41 AM) *

x^3 - 2x^2 = 9
-2x^2 + x^3 = 9
-2x^2 = 9 - x^3
-2 = (9/x^2) - (x^3/x^2)
-2 = 1 - x
-2 - 1 = -x
-3 = -x, multiply both sides by -1
x = 3

Code Buttons x^3 = 27.

P.j.S

The first 27 decimal places of phi and pi (rational & irrational) where infinity is resting at 10! at the edge of the spherical universe. Beyond this border the digits continue on endlessly representing an expandable universe with time and dimension present in the void. From a mathematical viewpoint.

P.j.S
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post Nov 16, 2009, 06:31 PM
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QUOTE(P.j.S @ Nov 16, 2009, 12:03 PM) *

The first 27 decimal places of phi and pi (rational & irrational) where infinity is resting at 10! at the edge of the spherical universe. Beyond this border the digits continue on endlessly representing an expandable universe with time and dimension present in the void. From a mathematical viewpoint.

P.j.S

Yea we knew that already! See:
http://brainmeta.com/forum/index.php?showtopic=20189
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Hey Hey
post Nov 18, 2009, 10:04 AM
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QUOTE(code buttons @ Nov 17, 2009, 02:31 AM) *

QUOTE(P.j.S @ Nov 16, 2009, 12:03 PM) *

The first 27 decimal places of phi and pi (rational & irrational) where infinity is resting at 10! at the edge of the spherical universe. Beyond this border the digits continue on endlessly representing an expandable universe with time and dimension present in the void. From a mathematical viewpoint.

P.j.S

Yea we knew that already! See:
http://brainmeta.com/forum/index.php?showtopic=20189
Watch out for lightening bolts! HE won't like it that you've exposed HIM! laugh.gif
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P.j.S
post Nov 26, 2009, 06:24 PM
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x^3 -2x^2 = 1

x = -1,
-1^3 - 2x^2 = 1
-1 -2x^2 = 1
-2x^2 = 1+1
x^2 = 2/-2
x^2 = -1

x^2 = -1
x^3 -2(-1) = 1
x^3 +2 = 1
x^3 = 1 -2
x^3 = -1

x^3 = -1
-1 -2x^2 = 1
-2x^2 = 1+1
x^2 = 2/-2
x^2 = -1

x, x^2 and x^3 equal -1 at the same time.

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Hey Hey
post Nov 27, 2009, 08:43 AM
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QUOTE(P.j.S @ Nov 27, 2009, 02:24 AM) *

x^3 -2x^2 = 1

x = -1,
-1^3 - 2x^2 = 1
-1 -2x^2 = 1
-2x^2 = 1+1
x^2 = 2/-2
x^2 = -1

x^2 = -1
x^3 -2(-1) = 1
x^3 +2 = 1
x^3 = 1 -2
x^3 = -1

x^3 = -1
-1 -2x^2 = 1
-2x^2 = 1+1
x^2 = 2/-2
x^2 = -1

x, x^2 and x^3 equal -1 at the same time.

P.j.S
http://www.cut-the-knot.org/fta/early.shtml

The last sentence of the article says it all.
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PJS
post Nov 28, 2009, 08:36 AM
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http://www.cut-the-knot.org/fta/early.shtml

Reply to last sentence in the article: The equation x^3 - 2x^2 = 1

This equation has only been known for a few short years and Mr, Euler was no doubt not aware of it in his life time. Mathematicians are honest Hey Hey. Their work needs to stand despite subsequent discoveries if it is correct.

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Hey Hey
post Nov 28, 2009, 09:58 AM
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QUOTE(PJS @ Nov 28, 2009, 04:36 PM) *
Mathematicians are honest Hey Hey. Their work needs to stand despite subsequent discoveries if it is correct.
You really don't know much about human nature, do you?

http://www.bogvaerker.dk/Bookwright/axioms.html

However, it must be difficult to gain experience when stuck inside that padded cell.


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post Nov 29, 2009, 06:46 PM
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LMA!!! I SMELL ROAST!
P.S. Welcome P.J.S. you seem like a pretty cool dude! WE just have a a form of say... initialte new forum members that talk lots shit as we try fix the Universe! LOL! This is the best science forum you'll ever find on the internet, I believe that wholeheartel
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PJS
post Feb 16, 2010, 05:19 AM
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QUOTE(code buttons @ Nov 29, 2009, 06:46 PM) *

LMA!!! I SMELL ROAST!
P.S. Welcome P.J.S. you seem like a pretty cool dude! WE just have a a form of say... initialte new forum members that talk lots shit as we try fix the Universe! LOL! This is the best science forum you'll ever find on the internet, I believe that wholeheartel

I agree that BrainMeta.com Forum is a great site.
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PJS
post Oct 23, 2011, 03:31 PM
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We know that since x^2 + 1 = 0 and x = -1 that x^2 equals both +1 and -1.

x = -1
x^2 = 1 when the given is x^2 + 1 = 0 or simply x^2 = -1.

The sqrt x^2 = 1 is x = 1.

Now we know that x alao equals +1 and -1.

We can see from this that the imaginary number i^2 = -1 can be re-directed as x^2 = -1 instead of i^2 = -1. How?

x = +1, -1
x^2 = x*x
x^2 = +1 * -1
x*2 = -1
x^2 + 1 = 0

So long as the value of i*2 can be replaced then complex numbers need not be used in the construction industry. The value of -1 is still retained.

Could a bridge be built across the Avon River without using complex numbers but instead using the new quantity x^2 + 1 = 0 derived from the equation x^3 - 2x^2 = 1?
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Flex
post Oct 23, 2011, 07:53 PM
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More importantly, who really cares about imaginary numbers anyhow smile.gif The only time I have ever had to use them (quantum mechanics) is to give a real value to some made up shit i.e. Psi is probability amplitude and has no meaning, but psi^2* is probability density.
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PJS
post Oct 23, 2011, 09:16 PM
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QUOTE(Flex @ Oct 23, 2011, 07:53 PM) *

More importantly, who really cares about imaginary numbers anyhow smile.gif The only time I have ever had to use them (quantum mechanics) is to give a real value to some made up shit i.e. Psi is probability amplitude and has no meaning, but psi^2* is probability density.

It is important to have the value of -1. Now x^2 + 1 = 0 can be retained as the quantity "x^2" = -1 instead of i^2.

This is because as made mention of already that the simple equation x^3 - 2x^2 = 1 has been found.

The sqrt x^2 = -1 is x = +1, -1. instead of i = sqrt -1.

x^2 = -1
x^4 = 1
x = 1, -1
sqrt x^2 = x

i^2 = -1
i^4 = 1
i = 1, -1
sqrt i^2 = i

i * i = 1 * -1
i^2 = -1
i = sqrt -1
i = 1, -1, sqrt -1

Consequently i = sqrt -1
i^2 = sqrt (-1 * -1) from sqrt 1 * -1 which is (sqrt -1 *-1 * -1)^2, 1 (sqrt -1)^2, 1 * -1 = -1
i^2 = -1
and
i^2 = sqrt 1 which is 1 (sqrt -1)^2, 1 (sqrt -1 * -1), 1 sqrt 1, 1 * 1 = 1
i^2 = 1

and so:

i^2 = 1 * -1
sqrt i^2 = sqrt 1 * -1
i = sqrt -1
sqrt -1 = sqrt -1 * -1 * -1, -1 sqrt -1, (-1 * 1) sqrt -1
sqrt -1 / sqrt -1 = (-1 * 1)
i / i = -1
i / i = i^2
i = i^2 * i
i = i^3 Wow! Remember 1 = 1^3
i = -1 * 1 * sqrt -1
i^3 = -1 sqrt -1
i^3 = (sqrt -1 * sqrt -1) sqrt -1
i^3 = i
(sqrt -1 * sqrt -1) sqrt -1 = sqrt -1
((sqrt -1 * sqrt -1) sqrt -1) / sqrt -1 = 1 * 1 * 1
i^3 = 1^3
i = sqrt -1 = 1
i^2 * i = i^3
-1 * 1 = -1
-1 = -1 * 1
i^3 = -1 * 1
i = -1 * 1

The solution for i = sqrt -1 could have been (-1 * 1) like x proves to be but then why the necessity of imaginary numbers like Flex says when you already have the quantity "x"?

Isn't a square root two similar values? Yes. The sqrt x^2 is x * x. But each x has a different value one positive and one negative.

x^2 = -1
x^4 = 1
x = 1, -1
sqrt x^2 = x
1 * -1 = -1
x * x = -1
x^2 + 1 = 0

Here the quantity x is squared allowing for positive and negative values each x. Novel.

P.j.S
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PJS
post Oct 24, 2011, 02:48 PM
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sqrt -1 = 1
i = 1

x^3 - 2x^2 = 1

i = x^3 - 2x^2
i + 2x^2 = x^3
x^2 = -1
1 + 2(-1) = x^3
-1 = x^3
-1 = x
x + i = 0
-1 + 1 = 0
So...

x^2 + i^2 = 0
(-1)^2 + (sqrt -1)^2 = 0
1 + (-1) = 0
1 - 1 = 0 and...

x^2 + i^2 = 0
-1 + (1)^2 = 0
-1 + 1 = 0

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Flex
post Oct 24, 2011, 03:03 PM
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Could you use this in an example solving real world problems?

Doing something like Integral (e^-ix)^2 dx
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PJS
post Oct 24, 2011, 03:22 PM
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QUOTE(Flex @ Oct 24, 2011, 03:03 PM) *

Could you use this in an example solving real world problems?

Doing something like Integral (e^-ix)^2 dx

I would love too! Yet I know little calculus.
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KoolK3n
post Oct 24, 2011, 03:23 PM
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I'm 2 mentally inferior to fathom what is going on. Could someone please explain?
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PJS
post Oct 24, 2011, 03:43 PM
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QUOTE(PJS @ Oct 24, 2011, 03:22 PM) *

QUOTE(Flex @ Oct 24, 2011, 03:03 PM) *

Could you use this in an example solving real world problems?

Doing something like Integral (e^-ix)^2 dx

I would love too! Yet I know little calculus.

Without the Integral e^ix could be (e^1*-1)^2 or (e^-1)^2 now meaning (e^i^2)^2 or (e^x^2)^2 when
x^2 + 1 = 0 and (e^ix)^2 = (e^sqrt -1*1)^2 when x = 1 which would equal (e^1)^2 or just "e^2". That would be x = i circumstances or PJS Baker's Dozen found on MSN Groups archives.

I hope that this helps.
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PJS
post Oct 24, 2011, 04:35 PM
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QUOTE(KoolK3n @ Oct 24, 2011, 03:23 PM) *

I'm 2 mentally inferior to fathom what is going on. Could someone please explain?

Previously it has been said that i^2 + 1 = 0 is a mathematical error. I believe that I just proved that sqrt -1 is equal to 1 among other things found.

In the equation x^3 - 2x^2 = 1 "i" is substituted for 1 leading possibly to new findings like Integral e^2 for example from x^2 + i^2 = 0. Possibly a new mathematical truth.
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PJS
post Oct 24, 2011, 04:58 PM
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QUOTE(PJS @ Oct 24, 2011, 03:43 PM) *

QUOTE(PJS @ Oct 24, 2011, 03:22 PM) *

QUOTE(Flex @ Oct 24, 2011, 03:03 PM) *

Could you use this in an example solving real world problems?

Doing something like Integral (e^-ix)^2 dx

I would love too! Yet I know little calculus.

Without the Integral e^ix could be (e^1*-1)^2 or (e^-1)^2 now meaning (e^i^2)^2 or (e^x^2)^2 when
x^2 + 1 = 0 and (e^ix)^2 = (e^sqrt -1*1)^2 when x = 1 which would equal (e^1)^2 or just "e^2". That would be x = i circumstances or PJS Baker's Dozen found on MSN Groups archives.

I hope that this helps.

x = 1

x^3 - 2x^2 = i
1^3 - 2(1^2) = i
1 - 2 = i
i = -1

i + x = 0
-1 + 1 = 0

i = 1, -1

i * i = 1 * -1
i^2 = -1
i^2 + 1 = 0 It must be true!


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PJS
post Oct 24, 2011, 05:48 PM
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QUOTE(PJS @ Oct 24, 2011, 04:58 PM) *

QUOTE(PJS @ Oct 24, 2011, 03:43 PM) *

QUOTE(PJS @ Oct 24, 2011, 03:22 PM) *

QUOTE(Flex @ Oct 24, 2011, 03:03 PM) *

Could you use this in an example solving real world problems?

Doing something like Integral (e^-ix)^2 dx

I would love too! Yet I know little calculus.

Without the Integral e^ix could be (e^1*-1)^2 or (e^-1)^2 now meaning (e^i^2)^2 or (e^x^2)^2 when
x^2 + 1 = 0 and (e^ix)^2 = (e^sqrt -1*1)^2 when x = 1 which would equal (e^1)^2 or just "e^2". That would be x = i circumstances or PJS Baker's Dozen found on MSN Groups archives.

I hope that this helps.

x = 1

x^3 - 2x^2 = i
1^3 - 2(1^2) = i
1 - 2 = i
i = -1

i + x = 0
-1 + 1 = 0

i = 1, -1

i * i = 1 * -1
i^2 = -1
i^2 + 1 = 0 It must be true!

i^2 + x^2 = the set (0, 2)

a) -1 + 1 = 0
i = 1
x = -1
i + x = 0

cool.gif 1 + 1 = 2
i = -1
x = 1
i + x = 0

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Flex
post Oct 24, 2011, 05:49 PM
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You should take some calculous courses--you would love it I promise! It is when math starts to get cool smile.gif The way we solve the integral is to multiply by the complex conjugate, that is e^ix*e^-ix to get a real number.
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PJS
post Oct 24, 2011, 05:59 PM
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QUOTE(Flex @ Oct 24, 2011, 05:49 PM) *

You should take some calculous courses--you would love it I promise! It is when math starts to get cool smile.gif The way we solve the integral is to multiply by the complex conjugate, that is e^ix*e^-ix to get a real number.

e^ix when i + x = 0 is e^-1*1 * e^1*-1 = (e^-1)^2 = ((2.718281828)^-1)^2 = (0.367879441)^2 =
0.135335283...

P.j.S

Perhaps (e^-1)^2 = (e^2(^-2))^2 = 3.354626279e-4

or (e^-1)^2 = e^-2 = 0.135335283...
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PJS
post Oct 24, 2011, 06:47 PM
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QUOTE(PJS @ Oct 24, 2011, 05:59 PM) *

QUOTE(Flex @ Oct 24, 2011, 05:49 PM) *

You should take some calculous courses--you would love it I promise! It is when math starts to get cool smile.gif The way we solve the integral is to multiply by the complex conjugate, that is e^ix*e^-ix to get a real number.

e^ix when i + x = 0 is e^-1*1 * e^1*-1 = (e^-1)^2 = ((2.718281828)^-1)^2 = (0.367879441)^2 =
0.135335283...

P.j.S

Perhaps (e^-1)^2 = (e^2(^-2))^2 = 3.354626279e-4

or (e^-1)^2 = e^-2 = 0.135335283...

Sorry I missed a sign. e^-ix.

When i + x = 0

e^ix * e^-ix
i = -1
x = 1

e^-1*1 = e^-1
e^-(-1*1) = e^1

e^-1 * e^1 = e^0 = 1

Next
i = 1
x = -1

e^1*-1 = e^-1
e^-(1*-1) = e^1

e^-1 * e^1 = e^0 = 1

e^0 = 1 must be a real number I guess!

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PJS
post Oct 24, 2011, 06:48 PM
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QUOTE(PJS @ Oct 24, 2011, 05:59 PM) *

QUOTE(Flex @ Oct 24, 2011, 05:49 PM) *

You should take some calculous courses--you would love it I promise! It is when math starts to get cool smile.gif The way we solve the integral is to multiply by the complex conjugate, that is e^ix*e^-ix to get a real number.

e^ix when i + x = 0 is e^-1*1 * e^1*-1 = (e^-1)^2 = ((2.718281828)^-1)^2 = (0.367879441)^2 =
0.135335283...

P.j.S

Perhaps (e^-1)^2 = (e^2(^-2))^2 = 3.354626279e-4

or (e^-1)^2 = e^-2 = 0.135335283...

Sorry I missed a sign. e^-ix.

When i + x = 0

e^ix * e^-ix
i = -1
x = 1

e^-1*1 = e^-1
e^-(-1*1) = e^1

e^-1 * e^1 = e^0 = 1

Next
i = 1
x = -1

e^1*-1 = e^-1
e^-(1*-1) = e^1

e^-1 * e^1 = e^0 = 1

e^0 = 1 must be a real number I guess!

P.j.S
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PJS
post Oct 24, 2011, 07:29 PM
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(e^-1)^2 * (e^1)^2 = 1

sqrt ((e^-1)^2 * (e^1)^2) = sqrt 1

sqrt (e^-1)^2 * sqrt (e^1)^2 = sqrt 1

e^-1 * e^1 = 1

e^0 = 1

i + x = 0

e^(i+x) = 1

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PJS
post Oct 24, 2011, 08:12 PM
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e^(i+x) * e^ix * e^-(ix) = ?

? = e^0 * (e^-1 * e^1)
? = e^0 * e^0
? = (e^0)^2

? = e^(i+x)+(i+x)
? = e^(2i+2x)
? = e^(i^2+x^2)
? = e^(-1+1)
? = e^0
? = 1

When i^2 + 1 = 0 then i + x = 0 and i^2 + x^2 = 0 and e^0 = 1. The set of x = (1, -1).

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PJS
post Oct 24, 2011, 09:13 PM
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QUOTE(PJS @ Oct 24, 2011, 08:12 PM) *

e^(i+x) * e^ix * e^-(ix) = ?

? = e^0 * (e^-1 * e^1)
? = e^0 * e^0
? = (e^0)^2

? = e^(i+x)+(i+x)
? = e^(2i+2x)
? = e^(i^2+x^2)
? = e^(-1+1)
? = e^0
? = 1

When i^2 + 1 = 0 then i + x = 0 and i^2 + x^2 = 0 and e^0 = 1. The set of x = (1, -1).

P.j.S

e^(2i+2x)
e^(sqrt-1*sqrt-1)and(1*1)or(-1*-1))
e^(-1)and(1)or(1) note: optional different colors for positive ones.
e^(-1+1)and/or(1)
e^(0)and/or(1)
e^0(and/or)e^1
(1and"e")and/or(1or"e")

Possibly some binary logic for a quantum computer.

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PJS
post Oct 25, 2011, 04:55 AM
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x^2 + 1 = 0
x = sqrt -1
sqrt -1 = 1

x^2 = -1
{-1 * 1) + 1 = 0
sqrt x^2 + sqrt 1 = sqrt 0
x + sqrt 1 = sqrt 0
x = - sqrt 1
x/-sqrt 1 = -sqrt 1 / sqrt 1
-sqrt 1 / x = 1
sqrt --1 / x = 1
sqrt 1/ x = 1
sqrt 1 = 1 * x
sqrt 1 = x
x = 1
sqrt -1 * -1 = x
x = -1
x^2 = -1 * 1
sqrt x^2 = sqrt (-1 * 1)
x = sqrt -1

The set of x = (sqrt -1, -1, 1).
x = sqrt -1 = -1 = 1

x^3 - 2x^2 = 1

(sqrt -1)^3 - 2(sqrt -1) = i
-1 sqrt -1 - 2(-1) = i
-1 sqrt -1 = i - 2
-1 sqrt -1 = -1
sqrt -1 = -1 / -1
sqrt -1 = 1

sqrt -1 = i = x

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PJS
post Oct 25, 2011, 05:11 AM
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i = x = sqrt -1 = -1 = 1
x = 1

x^3 - 2x^2 = 1
1^3 - 2(1) = 1
1 - 2 = 1
-1 = 1
-1 = 1/1
-1 * 1 = 1
(-1 * 1)^2 = 1^2
x^2 = 1
sqrt x^2 = sqrt 1
x = 1

P.j.S
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