You're all idiots. SQRT(a*b)=SQRT(a)*SQRT(b) only holds for positive real a and b. Square roots are much more complicated for negative and complex numbers.
Name:
Anonymous2006-08-08 16:05
>>11
You're an idiot if you think square roots are complicated, even if they do involve negative numbers.
What the hell are you talking about. SQRT(-4) = SQRT(4)*SQRT(-1) = 2i. >>4 knows what he's talking about.
Name:
Anonymous2006-08-08 16:14
>>13
No, >>4 is completely wrong and so are you. SQRT literally means "THE POSITIVE SQUARE ROOT OF". It is NOT multivalued. x^2=1 has two real solutions for x, but x=sqrt(1) has ONLY the positive solution x=1.
Look at this part of the theorem:
SQRT(-1*-1)=1
theorem: SQRT(a*b)=SQRT(a)*SQRT(b)
Thus SQRT(-1)*SQRT(-1)=1
But SQRT(-1)^2=i^2=-1. See where it goes wrong now? When you break up a square root with negative signs, you're off by a complex factor. The SQRT(-4) example you just showed is a special case of taking negative and complex roots.
Name:
Anonymous2006-08-08 16:20
>>12
And no, breaking up a negative root into its positive and imaginary part is not complicated, but breaking a positive root into two negative roots (as in the OP's fallacy) IS complicated, because it arises as a special case of complex square roots, and taking roots of complex numbers is extremely complicated because you need to take a complex log, multiply it by 1/2 and take the complex exponential. Take a course on complex analysis to see how it's done.
Name:
Anonymous2006-08-08 16:24
>>15
Again, you are an idiot for thinking it is complicated.
Name:
Anonymous2006-08-08 16:34
>>16
For fuck sakes, I'm just trying to give Anonymous a little credit. Have you looked around this forum lately? Between all the dickcheese and 0.999...=1 debates, I'd say yes, complex roots are slightly outside of the average Anon's skillset. Get off my back about fucking semantics already.
Name:
Anonymous2006-08-09 2:51
Yes it is complicated, it involves complex numbers.
HAY GUYS DID YOU SEE MY PUN
Name:
Anonymous2006-08-09 10:12
>>17
We don't talk about Anonymous, he's usually a middle-school/high-school student who thinks he knows more than PhD candidates in any given subject.
Name:
Anonymous2006-08-10 4:21
11 is right, that theorem: SQRT(a*b)=SQRT(a)*sQRT(b) only holds for a and b being positive real no's.
Wrong. It's basic fucking knowledge. You can't create an equation that proves that one number is another. It's just wrong.
Name:
Anonymous2006-08-10 23:39
>>20
The problem is "ax" is the Futurama pronounciation of the word "ask". It would still be written out as "ask".
Name:
Anonymous2006-08-11 2:02
>>23
Why on earth did you bother typing this response? The OP clearly asked WHY this "proof" was incorrect, and your respond by telling him it's wrong because it can't be right, and go on to insult him for not knowing that it was wrong despite that his original post clearly indicates he realizes that.
first of all, the operation of radicals aren't closed.. so your theorem isn't well defined.. second.. you yourself probably know the proof is false, why bother??
SQRT(a*b)=SQRT(a)*SQRT(b) only applies when a and b are whole numbers. A whole number is an non-negative and non-zero integer.
Name:
Anonymous2006-08-15 12:36
>>31
Wow, you need to get out of middle school and learn about the numbers between integers before you post here. SQRT(a*b)=SQRT(a)*SQRT(b) applies for all positive, real numbers. Why the fuck would it only apply to integers?
Name:
Anonymous2008-09-08 6:45
stop bumping this shit
Name:
Anonymous2008-09-08 15:15
The square root of -1 is an imaginary number. Meaning it does not exist in the set of real number. Meaning it is not 1.