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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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Solution You can't do that | Comment 1 of 3
√(a/b) =√(a)/√(b)  does not apply if a<0  or b<0 for the very reason shown.

The flaw is in the 3rd line as can be seen in the 4th line
i/1 = i but 1/i = -i

Edited on September 27, 2012, 2:05 pm
  Posted by Jer on 2012-09-27 10:48:59

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