We have $i^2 = -1$ where $i = \sqrt{-1}$. Now consider $$\sqrt{-1}\cdot \sqrt{-1} = \sqrt{(-1)\cdot(-1)} = \sqrt1 = 1$$ Which proves $-1 = 1$. Is there anything wrong with the above manipulation?
Asked
Active
Viewed 99 times
-1
-
$\surd a \surd b = \surd ab$ is valid only if $a,b\geq 0$ – Wang Kah Lun Jan 27 '16 at 12:39
-
1No flaw, you've just shown in a few steps that mathematics is nonsense. – Thomas Andrews Jan 27 '16 at 12:39
-
I like your comment Thomas.... :P – x2da Jan 27 '16 at 12:41
-
1Since $-1$ is obviously not equal to $1$, the question isn't "Is there a flaw", but rather "Where is the flaw" – 5xum Jan 27 '16 at 12:42
-
@ThomasAndrews Nice comment, but you failed to take Poe's law into consideration... – skyking Jan 27 '16 at 12:57
2 Answers
1
You can not use the radical properties like you have done in a passage. The product of the square root of $-1$ is not the square root of the product!
BLAZE
- 8,458
Giovanni Siclari
- 183