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A flawed proof? (2) (Posted on 2012-09-27) Difficulty: 2 of 5
-1/1 = 1/-1
⇒ √(-1/1) = √(1/-1) (taking the square root of both sides)
⇒ √(-1)/√1 = √1/√(-1)
⇒ i/1 = 1/i
⇒ i/2 = 1/(2i)
⇒ i/2 + 3/(2i)= 1/(2i) + 3/(2i)
⇒ i(i/2 + 3/(2i))= i(1/(2i) + 3/(2i))
⇒ i2/2 + (3i)/(2i)= i/(2i) + (3i)/(2i)
⇒ (-1)/2 + 3/2 = 1/2 + 3/2
⇒ 1=2

Can you spot the flaw in the above proof?

No Solution Yet Submitted by K Sengupta    
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Hints/Tips EXTRACT ROOTS with caution Comment 3 of 3 |

Even  for real numbers a=b does not imply sqrt(a)=sqrt(b),

we should cautiously write sqrt(a)= ±sqrt(b) and explore both cases,

Therefore the 2nd  line should be amended , inserting ± on one of the sides,


  Posted by Ady TZIDON on 2012-09-27 16:47:27
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