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Solving for 0... It is NOT impossible! What does this mean?

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posted on Nov, 27 2010 @ 09:29 PM
Here is the equation:
i101.photobucket.com...

The equation is deemed impossible, because one side has to be divided by zero. However, what happens if we plug in q for anytime we get 0?

Maybe i just found something simple like radioactive decay? I really dont know what this means.

I am not a scientist. But hopefully some genius folks out there can make some use of this find and give us some new technology.

www.abovetopsecret.com...

Please ignore the q = -1/4 part. That is incorrect.

the equation in question is:

q = q + 1/2

edit on 27-11-2010 by demonseed because: (no reason given)

posted on Nov, 27 2010 @ 09:31 PM
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edit on 27-11-2010 by nolabel because: (no reason given)

posted on Nov, 27 2010 @ 09:31 PM

Originally posted by demonseed
Here is the equation:
i101.photobucket.com...

The equation is deemed impossible, because one side has to be divided by zero. However, by plugging in a variable for 0, i was able to figure out a substitute for 0 and solve the equation.

Maybe i just found something simple like radioactive decay? I really dont know what this means.

I am not a scientist. But hopefully some genius folks out there can make some use of this find and give us some new technology.

www.abovetopsecret.com...

Special thanks to user ParkerCramer for figuring out q=q + 1/2. Most of us, myself included, thought it was unsolvable.

If anybody would please replicate these results for other equations maybe we could figure out a reasonable substitute for 0 in all cases.

If it was deemed impossible, how did you figure it out? Did you use 'x' as a substitute?
What happens if you divide it by 1/2?
edit on 27-11-2010 by nolabel because: (no reason given)

posted on Nov, 27 2010 @ 09:36 PM

q=q+(1/2) --> q-q=1/2 --> 0=1/2 --> 0*2=1 --> ?

edit on 11/27/2010 by abecedarian because: (no reason given)

posted on Nov, 27 2010 @ 09:38 PM
double post

edit on 27-11-2010 by demonseed because: (no reason given)

posted on Nov, 27 2010 @ 09:38 PM

Originally posted by abecedarian

q=q+(1/2) --> q-q=1/2 --> 0=1/2 --> 0/2=1 --> 0/0=1

edit on 11/27/2010 by abecedarian because: (no reason given)

no....

posted on Nov, 27 2010 @ 09:39 PM

notices I messed up too and edited while you were posting.

posted on Nov, 27 2010 @ 09:43 PM

so now you have q=q+(1/2), where q=-(1/4)
thus -(1/4)=-(1/4)+(1/2) ---> which is not true as -(1/4)+(1/2)= (1/4); not -(1/4)

q=0 q=-(1/4)
edit on 11/27/2010 by abecedarian because: (no reason given)

posted on Nov, 27 2010 @ 09:54 PM

Originally posted by abecedarian

so now you have q=q+(1/2), where q=-(1/4)
thus -(1/4)=-(1/4)+(1/2) ---> which is not true as -(1/4)+(1/2)= (1/4); not -(1/4)

q=0 q=-(1/4)
edit on 11/27/2010 by abecedarian because: (no reason given)

oh wow oops...

now i feel really dumb lol.

Well i guess its just q= q + 1/2 lol. Back to square 1.

posted on Nov, 27 2010 @ 10:06 PM

Isnt that what it is... it looks familiar....

posted on Nov, 28 2010 @ 04:44 AM

Maybe I'm just looking at the wrong part, but right at the top of the page, you've got x/(x+4) - 4/(x-4) = (x2 + 16)/(x+4)(x-4), wouldn't

x/(x+4)-4/(x-4) = (x2 – 8x -16)/(x2 +4x – 4x - 16)

be more correct? and then solving for 0, x = 0

I'm probably just missing the point... LONG time since I did any maths other than statistics, and I can't for the life of me see where q came in in the first place.

posted on Dec, 2 2010 @ 10:26 PM
Where are you trying to get with this?

The equation in question is not a well-formed equation, it's gibberish.

It's just as useless as trying to solve 1 = 2. There is no answer, becuase it's not a solvable equation.

It has no essence, no meaning, no practical application; it does not represent anything.

The only way you could possibly solve such an equation is by inventing new math; math where numbers don't have a meaning. Math which is not math, but gibberish.

I'm having trouble understanding what it is that you're trying to do here. There are a million different ways to write undefined mathematical expressions. Just because you can put it into an expression it doesn't mean it's something real which can or should be solved.
edit on 2-12-2010 by Deran because: (no reason given)

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