i Problems continued..., i = sqrt 1 
i Problems continued..., i = sqrt 1 
PJS 
Nov 16, 2009, 11:41 AM
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#31

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
P.j.S 
Nov 16, 2009, 12:03 PM
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#32

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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 
code buttons 
Nov 16, 2009, 06:31 PM
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#33

Supreme God Group: Basic Member Posts: 2451 Joined: Oct 05, 2005 Member No.: 4556 
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 
Hey Hey 
Nov 18, 2009, 10:04 AM
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#34

Supreme God Group: Basic Member Posts: 7767 Joined: Dec 31, 2003 Member No.: 845 
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 
P.j.S 
Nov 26, 2009, 06:24 PM
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#35

Overlord Group: Basic Member Posts: 358 Joined: Jun 12, 2009 Member No.: 32189 
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 
Hey Hey 
Nov 27, 2009, 08:43 AM
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#36

Supreme God Group: Basic Member Posts: 7767 Joined: Dec 31, 2003 Member No.: 845 
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 The last sentence of the article says it all. 
PJS 
Nov 28, 2009, 08:36 AM
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#37

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
http://www.cuttheknot.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. P.j.S 
Hey Hey 
Nov 28, 2009, 09:58 AM
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#38

Supreme God Group: Basic Member Posts: 7767 Joined: Dec 31, 2003 Member No.: 845 
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. 
code buttons 
Nov 29, 2009, 06:46 PM
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#39

Supreme God Group: Basic Member Posts: 2451 Joined: Oct 05, 2005 Member No.: 4556 
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 
PJS 
Feb 16, 2010, 05:19 AM
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#40

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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. 
PJS 
Oct 23, 2011, 03:31 PM
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#41

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 redirected 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? 
Flex 
Oct 23, 2011, 07:53 PM
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#42

God Group: Basic Member Posts: 1954 Joined: Oct 17, 2006 From: Bay area CA Member No.: 5877 
More importantly, who really cares about imaginary numbers anyhow 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.

PJS 
Oct 23, 2011, 09:16 PM
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#43

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
More importantly, who really cares about imaginary numbers anyhow 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 
PJS 
Oct 24, 2011, 02:48 PM
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#44

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 P.j.S 
Flex 
Oct 24, 2011, 03:03 PM
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#45

God Group: Basic Member Posts: 1954 Joined: Oct 17, 2006 From: Bay area CA Member No.: 5877 
Could you use this in an example solving real world problems?
Doing something like Integral (e^ix)^2 dx 
PJS 
Oct 24, 2011, 03:22 PM
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#46

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 

KoolK3n 
Oct 24, 2011, 03:23 PM
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#47

Overlord Group: Basic Member Posts: 455 Joined: Aug 20, 2011 From: Minnesota Member No.: 33523 
I'm 2 mentally inferior to fathom what is going on. Could someone please explain?

PJS 
Oct 24, 2011, 03:43 PM
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#48

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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. 
PJS 
Oct 24, 2011, 04:35 PM
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#49

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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. 
PJS 
Oct 24, 2011, 04:58 PM
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#50

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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! 
PJS 
Oct 24, 2011, 05:48 PM
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#51

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 1 + 1 = 2 i = 1 x = 1 i + x = 0 P.j.S 
Flex 
Oct 24, 2011, 05:49 PM
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#52

God Group: Basic Member Posts: 1954 Joined: Oct 17, 2006 From: Bay area CA Member No.: 5877 
You should take some calculous coursesyou would love it I promise! It is when math starts to get cool The way we solve the integral is to multiply by the complex conjugate, that is e^ix*e^ix to get a real number.

PJS 
Oct 24, 2011, 05:59 PM
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#53

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
You should take some calculous coursesyou would love it I promise! It is when math starts to get cool 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 = 3.354626279e4 or (e^1)^2 = e^2 = 0.135335283... 
PJS 
Oct 24, 2011, 06:47 PM
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#54

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
You should take some calculous coursesyou would love it I promise! It is when math starts to get cool 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 = 3.354626279e4 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 
PJS 
Oct 24, 2011, 06:48 PM
Post
#55

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
You should take some calculous coursesyou would love it I promise! It is when math starts to get cool 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 = 3.354626279e4 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 
PJS 
Oct 24, 2011, 07:29 PM
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#56

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
(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 P.j.S 
PJS 
Oct 24, 2011, 08:12 PM
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#57

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
PJS 
Oct 24, 2011, 09:13 PM
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#58

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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^(sqrt1*sqrt1)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. P.j.S 
PJS 
Oct 25, 2011, 04:55 AM
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#59

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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 
PJS 
Oct 25, 2011, 05:11 AM
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#60

Awakening Group: Basic Member Posts: 216 Joined: Aug 17, 2006 Member No.: 5507 
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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