# 1 = -1 ???Watch

#1
Ok, follow this reasoning step by step... seems to make sense if it wasnt that it breaks just about every axiom of mathematics

1 = √1 = √(-1)(-1) = (√(-1))(√(-1)) = (i) (i) = iÂ² = -1

0
13 years ago
#2
i think the √(-1)(-1) is invalid, as you can't route negative numbers, but i may be totally wrong
0
13 years ago
#3
(Original post by Eru)
i think the √(-1)(-1) is invalid, as you can't route negative numbers, but i may be totally wrong
You can root negative numbers (post alevel anyway)
0
13 years ago
#4
the imaginary number!!! (complex numbers)
0
13 years ago
#5
(Original post by nicolo)
Ok, follow this reasoning step by step... seems to make sense if it wasnt that it breaks just about every axiom of mathematics

1 = √1 = √(-1)(-1) = (√(-1))(√(-1)) = (i) (i) = iÂ² = -1

The error is that is only true if and are real.
0
13 years ago
#6
ohhh so cool
0
#7
Oooooooooh

haha i knew there had to be something wrong with it

well thanx
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13 years ago
#8
(Original post by Gaz031)
The error is that is only true if and are real.
and are real. I think you mean positive.
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13 years ago
#9
(Original post by Neapolitan)
and are real. I think you mean positive.
Yes, thank you.
0
13 years ago
#10
(Original post by Gaz031)
The error is that is only true if and are real.
I thought it was that √1 = Â±1, too.
0
13 years ago
#11
(Original post by Gaz031)
The error is that is only true if and are real.
I thought it was that √1 = Â±1, too.
0
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