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Find the flaw with this proof: sqrt(-1)<0


Best Answer dark_magician_92, 05 April 2013 - 07:37 AM

Object: to prove that  i < 0  ( that is, sqrt(-1) < 0  )
 

 ( .5 + sqrt(3/4)*i )^3 = (-1)^3

            which means that .5 + sqrt(3/4)*i = -1

So then      1 + sqrt(3)*i = -2

             sqrt(3)*i = -1

             i = -1/sqrt(3)

Therefore i is a negative number.  QED.
 

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#1 BMAD

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Posted 05 April 2013 - 02:59 AM

Object: to prove that  i < 0  ( that is, sqrt(-1) < 0  )
 

 ( .5 + sqrt(3/4)*i )^3 = (-1)^3

            which means that .5 + sqrt(3/4)*i = -1

So then      1 + sqrt(3)*i = -2

             sqrt(3)*i = -1

             i = -1/sqrt(3)

Therefore i is a negative number.  QED.
 


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#2 dark_magician_92

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Posted 05 April 2013 - 07:37 AM   Best Answer

Object: to prove that  i < 0  ( that is, sqrt(-1) < 0  )
 

 ( .5 + sqrt(3/4)*i )^3 = (-1)^3

            which means that .5 + sqrt(3/4)*i = -1

So then      1 + sqrt(3)*i = -2

             sqrt(3)*i = -1

             i = -1/sqrt(3)

Therefore i is a negative number.  QED.
 

Spoiler for


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#3 rbn491

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Posted 05 April 2013 - 12:35 PM

Obvious error in the proof is the part "which means that .5 + sqrt(3/4)*i = -1".. Generally when x^3=y^3, you can never directly say x is always equal to y.. It only leads to the conclusion that (x^3-y^3)=0, which on factorisation leads to (x-y)*(x^2+xy+y^2)=0.. So any further solution after that particular statement is not accepted in algebra..


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