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> i Problems continued..., i = sqrt -1
PJS
post Oct 25, 2011, 05:32 AM
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QUOTE(PJS @ Oct 25, 2011, 04:55 AM) *

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

P.j.S

sqrt --1 / x = 1
x = sqrt --1

1)
x = sqrt -(1 * -1)
x = sqrt -1 * 1
x = sqrt -1

2)
x = sqrt -(-1 * -1 * -1)
x = sqrt 1 * 1 * 1
x = sqrt 1
x = 1

x = sqrt --1 = sqrt -1 = 1

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PJS
post Oct 25, 2011, 05:48 AM
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sqrt - = ?

sqrt - / sqrt - = 1

sqrt - / sqrt - = i/x
sqrt -x / sqrt -i = 1

sqrt -x = sqrt -i

sqrt -x + sqrt -i = 0

My use of Imaginary Numbers.

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PJS
post Oct 25, 2011, 06:05 AM
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QUOTE(PJS @ Oct 25, 2011, 05:48 AM) *

sqrt - = ?

sqrt - / sqrt - = 1

sqrt - / sqrt - = i/x
sqrt -x / sqrt -i = 1

sqrt -x = sqrt -i

sqrt -x + sqrt -i = 0

My use of Imaginary Numbers.

P.j.S

sqrt -x + sqrt -i = 0

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

-1 + -i = (i + 1) * (i + 1) In order x = 1 then i = -1.

i = -1

-1 + --1 = (-1 + 1) * (-1 + 1)
-1 + --1 = 0 * 0
-1 + --1 = 0
-1 = -1
-1 + 1 = 0
i + x = 0

Perhaps the proper use of mathematical language without short cuts. Everything seems to be in order with the use of Imaginary Numbers.

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PJS
post Oct 25, 2011, 08:09 AM
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-1 + --1 = 0 * 0

(-1 + --1) / 0 = 0
(-1 + 1) / 0 = 0
0 / 0 = 0

-1 + --1 = 0 * 0
-1 = 0 * 0 + (-1)
-1 / 0 = 0 * + (-1)
-1 / 0 = 0 * ( + (-1))
-1 / 0 = 0 * (-1)
-1 = 0 * (-1) * 0
-1 / (-1) = 0 * 0
1 = 0 * 0

1 = 0 * 0
1 / 0 = 0

1 = 0 * 0
1 * 1 = 0 * 0
(1 * x) + 1 = 0 * 0
+ 1 = (0 * 0) / (1 * x)
+ 1 = (0 * 0)/ (1 * x)
x = 0
+ 1 = (0 * 0) / (1 * 0)
+ 1 = 0 / 0
? = (0 / 0) - 1
? = 0 - 1 Zero minus 1 = negative 1
? = -1
+ 1 = 0 / 0 Plus 1 = 0 + 1 = positive 1
+1 = 0 / 0
1 = 0 / 0

1 = 0 / 0
? = (0 / 0)-1
? = (0 / 0)-1
? = - 1
- 1 = (0 / 0)-1
- 1 / -1 = 0
- 1 - -1 = 0 minus 1 minus positive 1 = 0
0 = 0

1 = (0 * 0) and (0 / 0)
+2 = 1 and 1
? = (1 and 1) - (+2)
? = 0
0 = 2 - (+2)
0 = 2 + -2 note: 2 + negative 2
0 = 0 "and" = "+"

(0 * 0) / 1 = ?
? = 1
-1 + ((0 * 0) / 1))
-1 + (1 / 1)
-1 + 1 = 0
? = 0
-0 + (0 * 0) / 1
-0 + (1 / 1)
-0 + 1 = +1,
(0 * 0) / 1 = the set (0, +1)

0 / 1 = ?, given above that 0 / 0 = 0 and 0 * 0 = 1
(0 / 0) / (0 * 0) = ?
(0 / 0) = ? * (0 * 0)
(0 / 0) / ? = (0 * 0)
? = 0
-0 + ((0 / 0) / (0 * 0)) = x
-0 + ( 0 / 1) = x
(0 / 1) = x + 0
0 = (x + 0) * 1
0 = x + 0
0 = 0 + 0
x = 0
? = 0
x = ?
(0 / 0) / (0 * 0) = x
0 / 1 = 0

(0 / 0) / (0 * 0) = ?
? = 1, given the above that 0 / 0 = 1 and 0 * 0 = 1
-1 + ((0 / 0) / (0 * 0)) = x
-1 + (1 / 1) = x
-1 + 1 = x
0 = x
0 / 1 = 0

In the set "?" = (0, 1) both solutions described for 0 / 1 equal 0. The Perfect Zero.

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PJS
post Oct 25, 2011, 08:40 AM
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1 / 0 = ?

(0 / 0) / (0 * 0) = ?
? = 0
-0 + ((0 / 0) / (0 * 0)) = x
(0 / 0) / (0 * 0) = x + 0
(0 / 0) = (x + 0) * (0 * 0)
? = x, as proven above
x = 0
1 = (0 + 0) * (0 * 0)
1 = (2(0)) * (0)^2
1 = 2(0)^2
1 = (2 * 0)^2
1 = (0)^2
1 = 0^(1+1)
1 = 0 * 0
1 = 1

1 / 0 = 0. The equation balances. The Perfect Zero makes an appearance again.

For example 10/0 = 0 but it is razed to 0. This means 10/0 = 0 over time. In time 10/0 will be 0. But if the process is stopped then one is left at the value of the quantity x which is 10 --> x -->/= 0. 10 razed to 0 is 10/0 = 0.

10/0 = 3 is 10 razed to a value of x = 3.
10 / 0 = 0 is 10 brought to nothing. x = 0.

eg.
106/0 = 32
-32 + (106 / 0) = x
-32 = x - (106 / 0)
-32 - x = ( - (106 / 0))
(-32 - x) * 0 = - (106), the multiplying stance is removed from both sides.
-32 - x + 0 = -106
- x + 0 = -106 + 32
- x + 0 = -74
- x = -74 - 0
x = 74

x = 74 lost or shared.

106 / 0 = 0
- 0 + (106 / 0) = x
- 0 - x = - (106 / 0)
(- 0 - x) * 0 = - (106)
- 0 - x + 0 = - 106
0 = - 106 + x
x = 106

x = 106 lost or completely shared.
106/0 = 106, 106 / 0 = 0 The spacing is critical.

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PJS
post Oct 25, 2011, 01:17 PM
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10/0 = 0^1 --> (x/1/0 = 0^1) --> 0/0 = 1 whole razed to 0.
0^1 = 0

10/3 = 0^1 + x
- 0^1 + (10/3) = x
- 0^1 = x - (10/3)
- 0^1 - x = - (10/3)
0^1 + x = (10/3)
0^1 = 0^n
(0^1 + x^1) * 3^1 = (10)^1
(3 * 0^1) + (3x^1) = (10)^1
0^1 + 3x^1 = 10^1
-10 + 3x^1 = -0^1
-10/3 = -x
-x + x^1 = -0^1
0^1 = -0^1

Remainder "R" = -0^1 - 0^1
R = -2(0^1)
R = ((-2 * 0^1) * (-2 * 0^1))
R = (0 * 0)
R = 1
10/3 = x + R + 0
10 = (3x) + R + 0
10^n / 10^n = (3x/10^n) +(1/10^n) + 0/10^n
1 = 0.9n + (+1/10^n) + 0.0^n
1 = 0.9n + (0.1^n) + 0.0^n
n = 6
1 = 0.999999 + R + 0.0^n
1 = 0.999999 + 0.000001 + 0.000000

1 does not equal 0.9... repeating.

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PJS
post Oct 25, 2011, 02:14 PM
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1 = 0.9n + (0.1^n)
n = infinity
1.0 whole

R = 0.1^(n+1), the zero dimension "0^n" is divided.
1.0 whole + R = 0.1^(n+1), in lesser than zero space "0^n".

1.0 --> (R = 0.1^n) 1:0^n:1 (R = 0.1^(n+1) --> -1.1 --> -infinity:0:-infinity.1
compressed penetrating energy for dividing 0/0 = 1.0 whole or in other words visible matter, in the i^2 dimension in lesser than zero "0^n" space. Consequently -infinity.1 divides 0 which produces an exhaust void to keep balance in the universe.

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PJS
post Oct 25, 2011, 02:59 PM
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0 / 0 = 1
(0 / 0) / 0^0 = 1 / 0^0
((0 / 0) / 0^0) * 0^0 = 1
1 / 0^0 * 0^0 = 1
1 / (0^0)^2 = 1
1 / 0^(0+0) = 1
1 = 1 * 0^(0+0)
1 / 1 = 0^0
1 = 0^0

0 = The Perfect Zero
0^0 = 1
0^1 = 0.0
0^2 = 0.00 = money
0^6 = 0.000000
0^6 * 0^1 = 0^(6+1) = 0.000000 + 0.0 = 0^7
0^7 = 0.0000000
0^10 = 0.0000000000
0^n = the zero dimension
0^(n+1) = the zero dimension divided. Enter the i^2 dimension "-1.0 whole" in lesser than zero space "i^2" from "0^n" the zero dimension .

Lately it seems like the quantum computer is being refined since i = sqrt -1 = 1 has been discovered making for me that i^2 + 1 = 0 is not necessarily a mathematical lie if my recent work in this discussion is correct.

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PJS
post Oct 26, 2011, 06:48 AM
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QUOTE(PJS @ Oct 25, 2011, 05:48 AM) *

sqrt - = ?

sqrt - / sqrt - = 1

sqrt - / sqrt - = i/x
sqrt -x / sqrt -i = 1

sqrt -x = sqrt -i

sqrt -x + sqrt -i = 0

My use of Imaginary Numbers.

P.j.S

Perhaps a sign was missed.

sqrt -x = sqrt -i
sqrt -x - sqrt -i = 0

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PJS
post Oct 26, 2011, 07:09 AM
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QUOTE(PJS @ Oct 26, 2011, 06:48 AM) *

QUOTE(PJS @ Oct 25, 2011, 05:48 AM) *

sqrt - = ?

sqrt - / sqrt - = 1

sqrt - / sqrt - = i/x
sqrt -x / sqrt -i = 1

sqrt -x = sqrt -i

sqrt -x + sqrt -i = 0

My use of Imaginary Numbers.

P.j.S

Perhaps a sign was missed.

sqrt -x = sqrt -i
sqrt -x - sqrt -i = 0

P.j.S

Perhaps sqrt -x + sqrt -i = |2|.
Possibly forming the set (0, |2|).

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PJS
post Oct 26, 2011, 09:55 AM
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:
QUOTE(PJS @ Oct 25, 2011, 02:14 PM) *

1 = 0.9n + (0.1^n)
n = infinity
1.0 whole

R = 0.1^(n+1), the zero dimension "0^n" is divided.
1.0 whole + R = 0.1^(n+1), in lesser than zero space "0^n".

1.0 --> (R = 0.1^n) 1:0^n:1 (R = 0.1^(n+1) --> -1.1 --> -infinity:0:-infinity.1
compressed penetrating energy for dividing 0/0 = 1.0 whole or in other words visible matter, in the i^2 dimension in lesser than zero "0^n" space. Consequently -infinity.1 divides 0 which produces an exhaust void to keep balance in the universe.

P.j.S

1 = 0.9n + (0.1^n) + 0^n + 0^(n+1) = :0^n:1n. 1 whole compressed penetrating dark energy.

1 = 0.9n + (0.1^n) + 0^n + 0^(n+1) = :0^n:1n --> -1.1 --> -infinity.1 --> :0/0:1n = :x: --> :x^2: --> :x^3:1n.

The zero dimension (:x:) is penetrated. Enter (:x^2:) the dimension (:i^2:). The third dimension of :phi: (:x^3:) is penetrated as well (:0/0:1n). What is developing as visible matter cannot stay very long at a time in :x^3:. It is expelled as an exhaust void by means of
darklight (z^3).

(1.0 --> 1:0^n:1n --> -1.0) to balance the universe with. It is proposed that (:0/0:1n.:.) cannot divide :0/0: into 2 separate pieces of nothing. The True Void.

It seems that the math for The Theory of Space 0/0 Dynamics is improving.:.100?.

P.j.S

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P.j.S
post Oct 26, 2011, 12:28 PM
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QUOTE(PJS @ Oct 23, 2011, 03:31 PM) *

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?

God willing perhaps. Who is God.?. See this link for a current discussion on this important topic:

What God is is a Title? Then who is God? http://brainmeta.com/forum/index.php?showtopic=20953

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post Oct 26, 2011, 12:57 PM
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Here is a simple puzzle for math lovers of all ages.

x^2 + x^2 = 0 when x or x^2 does not equal 0. It is true. Can you prove this?

hint: x = 1

Show your work please.

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P.j.S
post Oct 26, 2011, 07:44 PM
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QUOTE(P.j.S @ Oct 26, 2011, 12:57 PM) *

Here is a simple puzzle for math lovers of all ages.

x^2 + x^2 = 0 when x or x^2 does not equal 0. It is true. Can you prove this?

hint: x = 1

Show your work please.

P.j.S

Answer:

x^2 + x^2 = 0 with x = 1

1 + 1 = 0
x^2 = 1 note: substituting one x^2 value
1 + x^2 = 0
x^2 = -1 Substituting the same x^2 value
1 + -1 = 0
x^2 + x^2 = 0

So x = sqrt -1
x + x = 2
2 = 1 + sqrt -1
2 - 1 = sqrt -1
sqrt -1 = 1
x + x^2 = 2
1 + 1 = 0
x = -1
1 + (-1) = 0
x = x^2 = (1, -1)

x^2 + x^2 = 2 when x = (1, -1)
2x^2 = 2
(2 * 1)^2 = 4
sqrt 2x^2 = 4
2x = 2
x = 2/2
x = 1

S0 4 = (2 * 1) + (2 * 1) AND (2 * 1) * (2 * 1) = 2x^2
4 = 2x^2
0 = 2x^2 - 4
sqrt 0 = sqrt (2x^2 - 4)
sqrt 0 = sqrt (4 = 2x^2)
sqrt 0 = Sqrt (4 = 2x * 2x)
sqrt 0 = sqrt (4 = 4x)
sqrt 0 = sqrt (4/4 = x)
sqrt 0 = sqrt (1 = x)

sqrt 0 = sqrt (2x^2 - 4)
sqrt 0 = sqrt (((2 * 1)^2) - 4)
sqrt 0 = sqrt ((2)^2) - 4
sqrt 0 = sqrt 4 - 4
sqrt 0 = sqrt 0
0 = 0
sqrt (1 = x)
sqrt 1 = 1
x = 1

x^2 + x^2 = (0, 2)

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post Oct 27, 2011, 10:33 AM
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Quantum Computer Binary Logic:

x = (1, -1)
x^2 = (1, -1)

i = (1, -1)
i^2 = (1, -1)

i + x = ?
? = (0, 2)

? = 0:
a. "i/x"
1/1 = 1 + 1 = 2
-1/1 = -1 + 1 = 0
-1/-1 = -1 + (-1) = -2
1/-1 = 1 + (-1) = 0

? = 2 + 0 + (-2) + 0 = 3(0) = 3*0 = 0 = y

b. invert "1/x"
1/y = 1 / (2(0) */+- 0^2) = 0 */+- 1
1). y = 2(0) * 0^2 = (0+0) * (0*0) = 2(0) * (1, 0) = a. 2(0) * 1 = 2(0) b. 2(0) * 0 = 2(0)^2 = 0 * 1 = 0
2). y = 2(0) / 0^2 = (0+0) / (0*0) = 2(0) / 1 = 2(0) = 0
3). y = 2(0) + 0^2 = (0+0) + (0*0) = 2(0) + 1 = 1.(2(0)) = 1.0 = 1
4). y = 2(0) - 0^2 = (0+0) - (0*0) = 2(0) - 1 = -1.(2(0) = -1.0 = -1

y = 0 = 1/0 = 0 = Z, 1/Z = 0 "razing"
y = 1 = 1/1 = 1
y = -1 = 1/-1 = -1

y = i^2 + x = 0
y = x + x^2 = 1
y = x / i^2 = -1

eg. x=7 Set (13, 56, -7) = 52 + 48 = 100%. Business Partnership considerations.
a. y = i^2 + 7 = -1 + 7 = 6
1/y = 1/6 = 0.1666n = Z
x + y = 7 + 1/Z = 1/0.1666n = 7 + 6 = 13
y*Z = 1

b. 7 + 49 = 56

c. 7/-1 = -7

eg. x=-2 Set (-5, 2, 2)
a. -1 + (-2) = -3
1/y = 1/-3 = -0.3n = Z
x + y = -2 + -3 = -5
1/Z = y

b. -2 + 4 = 2

c. -2/-1 = 2

x axis = -5
y axis = 2
z axis = 2

Does this make any sense at all?

Perhaps the path of a line of darklight at 1c^2 speed from the origin "the sun". 0 to x=-5 to y=2 to z=2 to x=-5 and back to 0 for example.

If 0 --> x=-5 takes 5 seconds how long does it take the darklight to make the circuit at 1c^2 speed or "(186234)^2 miles/second"?

From the School of P.j.S Original Mathematics & Related Sciences

Would you like to join the School of P.j.S Original Mathematics & Related Sciences found on the internet here: http://creativepi.multiply.com/

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post Oct 27, 2011, 03:56 PM
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Dark Energy Geometric 1 Compression Sequence:

i^2 + x = 6 and (2.4 * 2.5) = 6

x = 7

-1 + 7 = 6 * (2.4 * 2.5) = 6 = 6^2 = 36: 3+6=9
6 / 6 = 1
6 + 6 = 12: 1+2=3
6 - 6 = 0

Sum: 49 = 7^2 * x = :7^3:+13=34:3+4=7: ((x=4)*(y=7^7)*(z^2)) / d:= (1080 / ::7^3:: = 3P)!?!:.
7^2::+13=62:6+2=8::1/9=Z=0.1n::9-8=1:+Z = 1.1n.:!;4.(0.1^(n+1):!:!:4. Compress to the phi ":x^3:" dimension!. Perhaps...:4.=10!?!::.

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post Oct 29, 2011, 12:26 PM
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(x + x^2) / (x + 1) = x
x = 6
6 + 36 = 42 / 7 = 6

x = 1
1 + 1 = 2 / 2 = 1

(x + x^2) * ( x + 1) = ?
x^2 + x^3 + x + x^2
x^3 + 2(x^2) + x = ?
x = -1
x^3 + 2x^2 = -x
x^3 + 2x^2 = 1
-1 + 2 = 1
1 = 1
-1 + 1 = 0
? = 0

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

(x + x^2) - (x + 1) = ?
x = -1
0 - 0 = 0
? = 0
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post Oct 31, 2011, 08:49 AM
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x^3 + 2(x^2) + x = ?

? = 0
x = i^2 = -1

a.
(i^2)^3 + 2(i^2)^2 + i^2 = ?
-1 + 2(1) + (-1) = ?
1 + (-1) = 0 true

? = 0
x = x^2 = -1

b.
x^3 + 2(x^2)^2 + x = ?
-1 + 2(1) + (-1) = ?
1 + (-1) = 0
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post Oct 31, 2011, 09:02 AM
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QUOTE(PJS @ Oct 31, 2011, 08:49 AM) *

x^3 + 2(x^2) + x = ?

? = 0
x = i^2 = -1

a.
(i^2)^3 + 2(i^2)^2 + i^2 = ?
-1 + 2(1) + (-1) = ?
1 + (-1) = 0 true

? = 0
x = x^2 = -1

b.
x^3 + 2(x^2)^2 + x = ?
-1 + 2(1) + (-1) = ?
1 + (-1) = 0

x^2 = -1

x^3 + 2(x^2) + x = 0
x^3 + 2(x^2) = -x
(x^3 + 2(x^2)) / -x = -x / -x
-x^2 + 2(-x) = 1
-(-1) + 2(-x) = 1
2(-x) = 0
-x + (-x) = 0
+ (-x) = x
-x = x
-x/-x = x/-x
-x/-x = -x/x
x = 1
(-x / -x) + (-x / 1) = ?
1 + -x = 0
1 - 1 = 0

x^3 + 2(x^2) = -x
x = 1
1 + 2(-1) = -1
-1 = -1
1 - 1 = 0

P.j.S
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PJS
post Oct 31, 2011, 01:34 PM
Post #80


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QUOTE(PJS @ Oct 31, 2011, 09:02 AM) *

QUOTE(PJS @ Oct 31, 2011, 08:49 AM) *

x^3 + 2(x^2) + x = ?

? = 0
x = i^2 = -1

a.
(i^2)^3 + 2(i^2)^2 + i^2 = ?
-1 + 2(1) + (-1) = ?
1 + (-1) = 0 true

? = 0
x = x^2 = -1

b.
x^3 + 2(x^2)^2 + x = ?
-1 + 2(1) + (-1) = ?
1 + (-1) = 0

x^2 = -1

x^3 + 2(x^2) + x = 0
x^3 + 2(x^2) = -x
(x^3 + 2(x^2)) / -x = -x / -x
-x^2 + 2(-x) = 1
-(-1) + 2(-x) = 1
2(-x) = 0
-x + (-x) = 0
+ (-x) = x
-x = x
-x/-x = x/-x
-x/-x = -x/x
x = 1
(-x / -x) + (-x / 1) = ?
1 + -x = 0
1 - 1 = 0

x^3 + 2(x^2) = -x
x = 1
1 + 2(-1) = -1
-1 = -1
1 - 1 = 0

P.j.S

x^3 + (2x)^2 + x = ?

x = x^2 = -1

x^3 + 4 + (-1) = ?
x^3 + 3 = ?
x^3 = -3 = ?

cube root x^3
x = -1
? = -1.44224957

cube
x^3 + ?^3 = -4
sqrt (x^3 + ?^3) = sqrt -4
sqrt (x^3 + ?^3) = sqrt 4x

x^3 - ?^3 = 2
sqrt (x^3 - ?^3) = sqrt 2

sqrt 2 + sqrt 4x = sqrt (x^3 + ?^3) + sqrt (x^3 - ?^3)
2 + 4x = 2x^3
4x = 2x^3 - 2
x = (2x^3 - 2) / 4
x = (1x^3 - 1) / 2
x = -2 / 2
x = -1
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P.j.S
post Oct 31, 2011, 04:53 PM
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x^2 = i^2

i^2 / x^2 = 1
sqrt i^2 / sqrt x^2 = sqrt 1
sqrt i^2 / x = 1
sqrt i^2 = 1 * x
sqrt -1 = 1 * x
sqrt -1 / x = 1
-1 / x^2 = 1
(-1 / x^2)^2 = 1^2
1 / x^4 = 1
x = 1
x = sqrt x^2
x = sqrt -1
x = 1
i = sqrt -1
i = x
i = 1

P.j.S
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P JayS
post May 16, 2012, 10:49 AM
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2x^2 + 3y + z = 1

y = 1/3

2x^2 + z = 1 - 1
2x^2 + z = 0
2x^2 = 0 - z
x^2 = -z / 2
z = 2
x^2 = -2 / 2
x^2 = -1

3y = x^3
3 * 1/3 = x^3
1 = x^3
x = 1

x^2 = -1
x = sqrt -1 = i
x^3 = i^3 = 3y

2x^2 + i^3 + z = 1
z = 2
2x^2 + i^3 = 1 - 2
x^2 = -1
-2 + i^3 = -1
i^3 = -1 + 2
i^3 = 1
i = 1

x = sqrt -1 = 1
x^3 = 3y = i^3
x^2 = -1

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

i = sqrt -1 = 1
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P JayS
post Jul 07, 2012, 05:35 AM
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y = 1/3

1.0 - 3y = 1/10^n which is compressed dark energy. This point can be infinitely small. But the place where penetrating energy enters the zero dimension is 1/10^(n+1).

For example:
1.0 - 0.99999 = 0.00001:0: compressed dark energy next to the zero dimension.

Next: 0.00000:0:1 or 1/10^(5+1) not 1/10^6 for penetrating energy. Now dark energy can decompress within the point and accumulate mass in the i^2 or -1 dimension.

Time is a circumscribed square housing the zero dimension and the much smaller point which expells mass into lesser than zero space or the i^2 dimension. The smaller point is averaged phi space. When averaged phi space expands to reach the square of time then infinity rests and the project is complete. When average phi reaches the square of time housing the universe as opposed the space for a galaxy for example, then the capacity for the universe is full.

The sphere of time can maintain many universes with a radius in time of 0.5 second making the diameter of the sphere of time as 1.0 second. Expansion from infinity resting which can contain irrational numbers will always be a diameter of 1.0 second but for relativity the space will go up the numberline from 10! to 10!+1 etc...

The overall distance will only be 1 span of 1 second in reality. Expansion of the sphere of time requires faster orders of Dark Light. This allows the diameter of time to be referred to as the infinity second. So the distance can be physically longer for the radius of time while the diameter remains only 1.0 second in time to travel for that order of dark light.
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P JayS
post Jan 06, 2013, 10:31 PM
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x = -1
x = 1
x * x = i^2

-1 * 1 = x^2 = i^2 = -1 This is when -1 and 1 are quantities of x.

-1 * x = -1
x = 1

1 * x = -1
x = -1

x * x = x^2 = -1
x = -1
x = 1
x^2 = i^2
x^2 / i^2 = 1
x^2 * i^2 = 1
x^2 = -1
i^2 = -1

P.j.S
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P JayS
post Jan 06, 2013, 11:06 PM
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QUOTE(P JayS @ Jan 06, 2013, 10:31 PM) *

x = -1
x = 1
x * x = i^2

-1 * 1 = x^2 = i^2 = -1 This is when -1 and 1 are quantities of x.

-1 * x = -1
x = 1

1 * x = -1
x = -1

x * x = x^2 = -1
x = -1
x = 1
x^2 = i^2
x^2 / i^2 = 1
x^2 * i^2 = 1
x^2 = -1
i^2 = -1

P.j.S

Now 1 + 1 = 2 or (x^2 / i^2) + 1 = 2 or (x^2 * i^2) + 1 = 2 and 2^3 = 8
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P JayS
post Jan 06, 2013, 11:13 PM
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QUOTE(P JayS @ Jan 06, 2013, 11:06 PM) *

QUOTE(P JayS @ Jan 06, 2013, 10:31 PM) *

x = -1
x = 1
x * x = i^2

-1 * 1 = x^2 = i^2 = -1 This is when -1 and 1 are quantities of x.

-1 * x = -1
x = 1

1 * x = -1
x = -1

x * x = x^2 = -1
x = -1
x = 1
x^2 = i^2
x^2 / i^2 = 1
x^2 * i^2 = 1
x^2 = -1
i^2 = -1

P.j.S

Now 1 + 1 = 2 or (x^2 / i^2) + 1 = 2 or (x^2 * i^2) + 1 = 2 and 2^3 = 8

1 + (x^2 / i^2) + (x^2 * i^2) = 3 so called PJS Trinity in The Room of Objectivity with averaged phi conditions.

2^3 = |8| The Happy Spirit. 8^3 for x, y and z dimensions = 512

1) 512 = (sqrt 5 + 1) / 2 = phi
2) 512 = (sqrt 5 - 1) / 2 = 1/phi
3) (sqrt 5 * sqrt 5) = 5, (1 * -1) = -1, (2 * 2) = 4. (5 - 1) / 4 = 1.
+1 - 1 = |0|
phi - 1/phi = |1|

1 and 0 happily married in the room of objectivity. |1| = male energy and |0| = female energy.
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P JayS
post Jan 07, 2013, 06:37 AM
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x^2 = -1
x = sqrt -1
x * i = -1
sqrt (x * i) = sqrt -1
sqrt (sqrt -1 * sqrt -1) = sqrt -1
sqrt ( sqrt (-1 * -1)) = sqrt -1
sqrt (sqrt 1) = sqrt -1
sqrt 1 = sqrt -1
1 = sqrt -1

or

sqrt (sqrt -1 * sqrt -1) = sqrt -1
sqrt (sqrt (-1 * -1)) = sqrt -1
sqrt -1 = sqrt -1
x = i

P.j.S
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P JayS
post Jan 07, 2013, 10:02 AM
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QUOTE(P JayS @ Jan 07, 2013, 06:37 AM) *

x^2 = -1
x = sqrt -1
x * i = -1
sqrt (x * i) = sqrt -1
sqrt (sqrt -1 * sqrt -1) = sqrt -1
sqrt ( sqrt (-1 * -1)) = sqrt -1
sqrt (sqrt 1) = sqrt -1
sqrt 1 = sqrt -1
1 = sqrt -1

or

sqrt (sqrt -1 * sqrt -1) = sqrt -1
sqrt (sqrt (-1 * -1)) = sqrt -1
sqrt -1 = sqrt -1
x = i

P.j.S

1 * 1 = 1
1 * 1 = sqrt -1

1 / 1 = 1
1 / 1 = sqrt -1

1 = sqrt -1

P.j.S
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