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llduncanll
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#1
Report Thread starter 15 years ago
#1
√(-1) = √(-1)
√(-1/1) = √(1/-1)
√(-1)/√(1) = √(1)/√(-1)
√(-1)√(-1) = √(1)√(1)
-1 = 1
Call me thick but i really dunno whats wrong with it..? Damn maths to hell :mad:
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JamesF
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#2
Report 15 years ago
#2
Square root isnt a valid function, √1 = + or -1
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nollaig
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#3
Report 15 years ago
#3
Square root isnt a valid function, √1 = + or -1
I disagree. √x means the positive square root of x only.

This is often confused with the correct statements : x^2 = 4 => x = +2 or -2 .

llduncanll:

Your third line, √(-1)/√(1) = √(1)/√(-1), is not a valid statement: i = 1/i is not true.

Why does it not follow logically from the second statement?

The distributive law for square roots only holds if a,b are positive real numbers i.e. √(a/b) = √a/√b only if a,b are positive real numbers
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Willla2
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#4
Report 15 years ago
#4
can you prove the distributive law is only true for positive real numbers?
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jpowell
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#5
Report 15 years ago
#5
Simple. Assume it is true for negative real numbers, see that leads to a contradiction. QED.
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Willla2
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#6
Report 15 years ago
#6
gosh that seems such a copout
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fishpaste
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#7
Report 15 years ago
#7
How do you prove it holds for positive real numbers though?
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Willla2
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#8
Report 15 years ago
#8
we want to show root(a/b) = root(a)/root(b)

a/b = ab^-1

root(ab^-1) = a^.5b^-.5 = a^.5/b^.5

is that a proof?
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llduncanll
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#9
Report Thread starter 15 years ago
#9
oh yeah, good point guys i was just browsing around last night and all of a sudden it jumped out at me n slapped me round the face a few times... cheers for clearing it up!
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Squishy
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#10
Report 15 years ago
#10
Yup, the flaw in my sig is the jump from the second to the third line. Division by complex numbers is not the same as division by real numbers...didn't mean it to be confusing, it's just one of those warnings about generalising the properties of real numbers to all complex numbers.
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