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

  1. Feb 14, 2014 #1
    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
    = i2 +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. jcsd
  3. Feb 14, 2014 #2

    DrClaude

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    Staff: Mentor

    There's a FAQ on the subject: https://www.physicsforums.com/showthread.php?t=637214 [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Feb 14, 2014 #3

    phinds

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    Gold Member

    Doesn't simple common sense tell you this is nonsense?
     
  5. Feb 14, 2014 #4
    yes, which is why i asked the question. it just seemed mathematically sound to me so i wanted to see why we disregard it. But it seems i have a lot to catch up on with complex exponentiation!
     
  6. Feb 14, 2014 #5

    phinds

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    Gold Member

    Ah ... yeah, I see from rereading your question that you aren't quite saying that you believe it, just that you are puzzled by it.
     
  7. Feb 18, 2014 #6
    The problem is that for complex numbers, √(a+bi) has two complex roots. With real numbers, one is positive and the other is negative. We can assign the positive one to equal the positive square root. However, with complex numbers it isn't so simple. A complex root is neither negative nor positive, sense it doesn't make any sense to say a complex number is positive. So, it doesn't make sense to have a "positive" square root. This is where the issue arises.
     
  8. Feb 18, 2014 #7
    Here's the explanation:

    The fun thing is you can go 'crazy' with the rule if you know it.

    say:

    ##i^{-i} = e^{\frac{\pi}{2}}##
     
  9. Feb 18, 2014 #8

    Curious3141

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    Homework Helper

    Umm, that actually is valid. One of the values of ##i^{-i}## is ##e^{\frac{\pi}{2}}##.
     
  10. Feb 19, 2014 #9
    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?
     
  11. Feb 19, 2014 #10
    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)
     
  12. Feb 19, 2014 #11
    Yes it is valid, I don't think I imply otherwise.
     
  13. Feb 19, 2014 #12

    Curious3141

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    Oh, OK. Sorry that I misunderstood your post.
     
  14. Feb 19, 2014 #13
    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
     
  15. Feb 19, 2014 #14
    Alright, here's the rub. This isn't actually a "valid" operation. It just looks like what's going on. The notation convention used by every mathematician and math educator that I know is that ##\sqrt{a}## denotes the non-negative solution of ##x^2=a## if ##a## is a non-negative real number and ##\sqrt{a}## denotes the root ##i\sqrt{-a}## of ##x^2=a## when ##a## is a negative real number (so we've already agreed on what ##\sqrt{-a}## means), where ##i## denotes either (a) some made-up thing with the property that ##i^2=-1## or (b) the principle complex root (we singled one of them out and said "you're it") of ##x^2+1=0## depending on where you are in your math studies. There are no rules of exponents or multiplicative identities of roots or anything of that sort going on. It's all just definitions of notation; these are the symbols that we use to talk about these abstract mathematical concepts.

    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.
     
  16. Feb 19, 2014 #15
    if the same exponential laws applied to negative numbers as they do positive.

    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
     
  17. Feb 19, 2014 #16
    Isn't that pretty much equivalent to what your doing when you say √(-4) = i√(-(-4)) because your just taking √(i)2 out of the radical, which is the same as √(-1). so really your doing the exact same thing as me?
     
  18. Feb 19, 2014 #17

    Nugatory

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    Staff: Mentor

    No, he is not doing the same thing. He is saying that we use the sequence of symbols "##\sqrt{a}##" as a convenient shorthand for "the non-negative solution of ##x^2=a## if ##a## is a non-negative real number; and the root ##i\sqrt{-a}## of ##x^2=a## when ##a## is a negative real number, where ##i## denotes some made-up thing with the property that ##i^2=-1##"

    That's a definition, and we can use that definition to show that if ##a## and ##b## are both non-negative 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.
     
  19. Feb 19, 2014 #18
    No it's not. Read again what I wrote regarding the definition of the meaning of the collection of symbols ##\sqrt{a}## when ##a## is a negative real number:

    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##.
     
  20. Feb 19, 2014 #19
    okay i think i understand what you said, sort of.

    Are you saying that 'the square root' symbol is used to refer to only the principle (positive) root: ie √(a2) = |a| where a can be +ve or -ve but the radical implies the absolute (positive) value of it?

    what i was getting most confused about was that we can go from ##√(-4) = √(-1)(4) = √(-1)√(4) ## but we cant go from ##√16 = √(-4)(-4) = √(-4)√(-4)## I realise now that its because the property ##√ab = √a√b## holds only if a and b are positive or one of them is negative. It does not hold if BOTH a and b are negative as this leads to inconsistencies.

    But what if we weren't to classify these as inconsistencies and hence didn't ban the property √ab = √a√b for a negative a and negative b. could this mean that there would then be solutions to the system of equations (for example): (sorry this may seem off track, but this is why i started considering my original question on this thread)
    a - 2b + 3c = -2
    -a + b - 2c = 3
    2a -b + 3c = 1
    putting into a co efficient matrix:
    [tex]
    \begin{pmatrix}
    1 & -2 & 3 & -2\\
    -1 & 1 & -2 & 3\\
    2 & -1 & 3 & 1
    \end{pmatrix}
    [/tex]
    putting into row echelon form:
    [tex]
    \begin{pmatrix}
    1 & -2 & 3 & -2\\
    0 & -1 & 1 & 1\\
    0 & 0 & 0 & 8
    \end{pmatrix}
    [/tex]
    the last row says 0=8 which is not possible and leads this system to have no solutions.
    what if we said that 0=8 is an identity that is necessary for there to be a solution to this system of equations and provable by (im now assuming that ##√ab = √a√b## for negative a and negative b is not treated as an inconsistency):
    0 = -4 + 4
    0 = 4i2 + 4
    0 = (2i)(2i) + 4
    0 = √(4)(-1)√(4)(-1) + 4
    0 = √(-4)(-4) + 4
    0 = 4 + 4
    0 = 8
    similarly it can be shown the 0=-8 which yields -8=0=8
    now, going back to the matrix in row echelon form, treating 'c' as a free variable the solution space is ##{(-4-c,c-1,c)}##
    lets say c = 2, then a=-6 and b=1 and plugging these values into the three equations we get:
    a - 2b +3c = -2 <-------- this is true for these values
    -a + b - 2c = 3 <-------- this is true for these values
    2a - b + 3c = 1
    2(-6) - 1 + 3(2) = 1
    -7 = 1 <----- since we showed that -8=0=8 this is true also (under the assumptions that i stated)

    so what im getting at is; if √ab = √a√b for negative a and b was not treated as an inconsistency and was allowed then could this mean that certain systems of equations that as it stands in modern mathematics have 'No solution' could actually have a solution?
     
  21. Feb 19, 2014 #20
    ehhhh scratch all that i just realised that if -8=0=8 then by the same logic every single number is equal to every other number, which is definitely an inconsistency!!
     
  22. Feb 20, 2014 #21
    It makes me wonder, this same problem appears again and again, so I try to put my 2
    cents in. I think that the problem lies already at the OP's question: "why can't 1+1=0?"

    To me this kind of question means that the OP already knows that the result of his
    calculation will be 0, because he wrote it that way. Had he instead asked: "what is 1+1= ?",
    his problem might be avoided. It would mean that he had to admit that he does know the
    answer. To me these kind of problems have always been very hard to understand, why
    in mathematics we are many times calculating although the answer is known? Perhaps
    we should use mathematics and calculate only those things whose answers are not known.
     
  23. Feb 20, 2014 #22
    if you read my post #19 in this thread, i explicitly state that i was exploring a system of equations
    (i then go on to give this system and tell you the problem) which yielded no solutions and this is what spurred me to investigate and ask the original question. so in fact i did not know the answer before i asked the question. and in any case i was wrong, so i didn't know the answer at any point.
     
  24. Feb 20, 2014 #23
    Ok, I did not know what you knew, that's why I asked. In your post #15, you arrive
    at result a+a= -a+a. Lets not take the last step yet (the last step is that a+a=0).
    Instead, lets assume that we meet some aliens who have no knowledge how to calculate with our rules. So that they don't know for example what are a+a and -a+a equal to. We give them two equations and ask them to calculate an answer, the question
    is : What is a+a= ? What is a-a= ?
    The alien might think that a+a=a-a, is this a correct conclusion? And even if it is, does
    it automatically mean that ?=0, that you can introduce the number 0 and does it mean
    anything to our alien. I don't know if I am very good at explaining, I hope you understand
    at least something.
     
  25. Feb 20, 2014 #24
    we have already established that i was wrong, and i accept that. i don't see what the problem is? and an alien wouldn't know what the operations +, - or even = meant so we would have to teach it our symbols and arithmetic before it could start to answer the question.
     
  26. Feb 20, 2014 #25
    and sorry i missed the part where you asked me what i knew.
     
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