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In the complex number system, why can't 1+1 = 0 ? 
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#1
Feb1414, 01:17 AM

P: 70

in the following i will demonstrate a 'proof' that 1+1=0
1+1=√1 +1 =√(1)(1) +1 = (√(1))(√(1)) +1 = (i)(i) +1 = i^{2} +1 = 1 + 1 = 0 I know I'm not the first to come up with this 'proof', and i have been told that the problem lies with splitting the radical between the 2nd and 3rd lines of working. But in the complex number system there is no problem with going from, say, √(4) = √(1)(4) = 2i so why is there a problem with the above working, isn't it the same line of thought? thanks, Michael. 


#2
Feb1414, 03:27 AM

Sci Advisor
PF Gold
P: 1,354

There's a FAQ on the subject: http://www.physicsforums.com/showthread.php?t=637214



#4
Feb1414, 09:25 AM

P: 70

In the complex number system, why can't 1+1 = 0 ?



#5
Feb1414, 09:48 AM

PF Gold
P: 6,335




#6
Feb1814, 07:45 PM

P: 19




#7
Feb1814, 09:33 PM

P: 255

say: ##i^{i} = e^{\frac{\pi}{2}}## 


#8
Feb1814, 10:12 PM

HW Helper
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#9
Feb1914, 02:19 AM

P: 70

okay, why are we allowed to go from ((4)*(1))^{0.5} to (4)^{0.5}*(1)^{0.5} (which is equal to 2i) but we aren't allowed to go from ((1)(1))^{0.5} to (1)^{0.5}*(1)^{0.5}?



#10
Feb1914, 02:24 AM

P: 70

and on another note, in my most recent post, isn't the first example a negative number and the second a positive (ie (4)(1)= ve and (1)(1)= +ve)



#13
Feb1914, 09:46 AM

P: 223

It's not possible. If 1+1 = 0, then +1 would have to be equal to 1, but it can't be equal to 1 and +1 at the same time. Same way you could try and prove for any "A" that A+A = 0



#14
Feb1914, 10:55 AM

P: 471

In other words, ##\sqrt{4}=i\sqrt{(4)}=i\sqrt{4}=2i## because somebody a long time ago decided that's what those symbols mean, and everybody else went along with it. It only looks like we're using ##\sqrt{4}=\sqrt{1\cdot 4}=\sqrt{1}\cdot\sqrt{4}=i\cdot2## because either (a) you were clever enough to notice that might be the case based on your experience using rules that you learned in precalc or (b) you had a teacher (or tutor or helper on the interwebs) that told you a little white lie to convince you that the answer they gave was correct without realizing (or caring) that it might later cause serious strife for you and other teachers (or tutors or helpers on the interwebs) or (c) you had a teacher (or tutor or helper on the interwebs) who was incompetent tell you this is what was going on without understanding that it really wasn't. 


#15
Feb1914, 11:03 AM

P: 70

a+a=√(a^2 )+a =√((a)(a) )+a =√(a) √(a)+a =√((1)(a) ) √((1)(a) )+a =(√(1))(√a)(√(1))(√a)+a =(i)(√a)(i)(√a)+a =i^2 a+a =a+a =0 


#16
Feb1914, 11:21 AM

P: 70




#17
Feb1914, 01:07 PM

Sci Advisor
Thanks
P: 3,746

That's a definition, and we can use that definition to show that if ##a## and ##b## are both nonnegative reals, then ##\sqrt{ab} = \sqrt{a}\sqrt{b}##. However, we can also use this definition to show that, in general, this equality does not hold for negative numbers. 


#18
Feb1914, 01:09 PM

P: 471

If ##a## is a negative real number, the ##\sqrt{a}=i\sqrt{a}##. There is nothing in between the left and right hand side that explains why it is true. There is not proof of this identity. It is not a matter of mathematics, but a matter of mathematical notation. It is simply true by definition of the notation being used. I'm not "taking √(i)^{2} out of the radical". The ##i## essentially materializes out of thin air, just like the ##\times## does in the definition of the exponential notation ##x^2=x\times x##. 


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