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1+1=0 help please

  1. Dec 22, 2008 #1
    i'm kind of new to this website and i was just wondering if anyone could tell me if my math is wrong here.

    for the purposes of this thread sqrt means square root

    my logic for 1+1=0 is this:









    if i did anything that is against math rules can someone please tell me. this is really throwing me off because now i dont know if anything i know is correct

    also if this is correct then 4=5 and all my math teachers say that isnt true
  2. jcsd
  3. Dec 22, 2008 #2
    sqrt(a*b)=sqrt(a)sqrt(b) only works for [tex]a,b\geq 0[/tex].
  4. Dec 22, 2008 #3


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    You can't do what I highlighted in bold. For one you are essentially equating +1 to be -1 and that clearly is not true. You can't do that: every number has the property of uniqueness from another as long as the two numbers are not the same value.

    Basically every statement that follows on in every line must yield an equivalent for proof or whatever to hold true and in this case it doesn't hold true.

    The way to state this more explicitly is that we state a class of numbers that have specific algebraic properties. Because SQR(1) != SQR(-1) we specify explicit allowable algebraic properties for each class of numbers (ie the integers, rationals, reals, and complex). By sticking to the premise of keeping "uniqueness" within our number system and by obeying the specific rules of our number classes with their algebraic properties (this is actually deep in algebra and I know i'm not communicating this quite correctly) then we arrive at no contradictions such as this.

    Group theory is a good way of understanding modern algebra. In groups we have operations which act upon group elements and we have the idea of a null element, inverse elements and closure.

    If you think about the case of a null element as applied to an operation like addition, you can see where this idea of "uniqueness" comes from. Basically if we did not have uniqueness in our number system everything would fall apart because nothing would be deterministic.
  5. Dec 22, 2008 #4
    Most of these invalid proofs you find are based on algebraic rules which have restrictions on them. Most people remember the rule, but forget about the restriction.

    Another common "proof":

    Let a = b
    a^2 = a*b
    a^2 - b^2 = a*b - b^2
    (a-b)(a+b) = (a-b)b [factor both sides]
    a+b = b [divide both sides by a-b]
    2b = b [turn the a back into a b]
    2 = 1

    The problem with this one is b = a, so b - a = 0. In the division step, you divide by a-b, so you're really dividing by zero. The rule for division explicitly prohibits doing this, but most people don't recognize it. But whether or not its obvious, you have to do a check at that step to see the divisor isn't zero or your statement may not hold!
  6. Dec 22, 2008 #5


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    Well in Z[2] it's correct.
  7. Dec 22, 2008 #6
    I'm not sure any of the respondents has really answered the question. The main problem here is that sqrt(x) has 2 values, plus or minus sqrt(x). When writing sqrt(x) there is a usual convention that takes the positive root as the value, but this convention has to be adopted for an agreeable reason, for which there is no basis here.

    Actually, the "sqrt(-1)*sqrt(-1)" has 4 possibilities (2 roots, each of which can be plus or minus). Of those 4, two give i^2 and the other two give -i^2. So, as you see, proper selection of the sign of the sqrt will give you two different, valid identities (-1 = i^2, or 1 = -i^2) as well as two wrong ones, if signs are chosen improperly.
  8. Dec 22, 2008 #7
    This is incorrect.

    Equal things are equal. If sqrt(-1) = i, then sqrt(-1) * sqrt(-1) = i * i = -1. If sqrt(-1) = -i, then sqrt(-1) * sqrt(-1) = -i * -i = -1. In neither case does sqrt(-1) * sqrt(-1) = +1.

    You are free to choose whatever definition you like. But once you pick one, you have to stick with it. You can't choose sqrt(-1) = i in the first half of the equation and sqrt(-1) = -i in the second half. If you did, you'd have to conclude i = -i. It's the same function applied to the same argument, so the result (regardless of whether you choose +i or -i) is the same both times.
  9. Dec 22, 2008 #8
    what is Z[2]?
  10. Dec 22, 2008 #9
  11. Dec 22, 2008 #10
  12. Dec 22, 2008 #11
    Can't agree with that - the problem is precisely that sqrt is not a function, it has two values (for a non-zero argument). And each occurrence of a root has these two, independent choices.

    Notice that this is not the same concept as defining which of the square roots of unity will be the imaginary unit. The latter is a definition and, once chosen, you stick to it consistently. This does not change the former: that sqrt(-1) (or of any other non-zero argument) has two values.
  13. Dec 22, 2008 #12
    Sqrt is darn well a function. It is a function defined on the non-negative reals satisfying the equation sqrt(x) = y iff y^2 = x and x >= 0. It is single-valued. It is a partial inverse to the squaring function square(x) = x^2. If you need the negative root, use a minus sign and take -sqrt(x)!

    Similarly, sqrt(-1), by the usual convention, is defined as i. If you need the negative root, you use -sqrt(-1) = -i.

    What I was illustrating is that regardless of whether you use "sqrt" to mean the traditional square root or the negative square root, it doesn't matter.
  14. Dec 22, 2008 #13
    I presume we are not talking about sqrt() in a programming language; I understand that the OP used 'sqrt(1)' and 'sqrt(-1)' just to avoid the hassle of writing [itex]\sqrt 1[/itex] and [itex]\sqrt -1[/itex] in LaTex. As a math concept, y = sqrt(x) is the equation of a parabola with an horizontal axis, and thus for every every positive x there are two corresponding y's - so it can't be a function.
  15. Dec 22, 2008 #14


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    No it isn't! In mathematics the "square root", "sqrt", "square root function", "principle square root" all mean the same thing namely

    [tex]f : \mathbb{R}^* \rightarrow \mathbb{R}^*[/tex]

    [tex]f : x \mapsto \sqrt{x}[/tex]

    Here [itex]\mathbb{R}^* = \mathbb{R}^+\setminus\mathbb{R}^-[/itex].
  16. Dec 22, 2008 #15
    If you wish to define a 'principal square root', you are free to do so, but this was not the sense in which 'sqrt' was used in the original post.

    Actually, from the very definition given in post #12,
    it is clear that both 2 and -2 are valid values for y when x=4.
  17. Dec 22, 2008 #16


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    I do wish you would quote the whole post and not just the bits that suit you:
  18. Dec 22, 2008 #17
    The short quote is the part I would agree on - why restricting domain and codomain to the positive reals, when the original post is using complex numbers?

    (Edit: Actually, I quoted too much - I should have removed the x >= 0 bit as well.)
  19. Dec 22, 2008 #18


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    The square root is restricted to the non-negative reals because we cannot define the square root for complex numbers.

    Although I hesitate to reference wikipedia, it's the best I can do at short notice: http://en.wikipedia.org/wiki/Square_root#Square_roots_of_negative_and_complex_numbers
  20. Dec 22, 2008 #19
    If you mean you cannot define it as a function, I can't agree more. But you certainly can define roots of numbers in the complex numbers (see 'roots of unity' for an example). My claim is that the OP fails to realize sqrt(x) is more than one number.
  21. Dec 22, 2008 #20


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    Actually [itex]\sqrt{x}[/itex] is by definition only one number.

    No it isn't...[itex]x=y^2[/itex] is a parabola....[itex]y=\sqrt{x}[/itex] is only half of that parabola.

    See here for a definition of square root.

  22. Dec 22, 2008 #21
    I actually made a mistake in my definition. I meant to say y >= 0, not x >= 0. So sqrt(x) = y iff x = y^2 and y >= 0.
  23. Dec 28, 2008 #22

    If you work within the real Nos system then your mistake is in the line where you substitute :

    ......sqrt[(-1)*(-1)] by sqrt(-1)*sqrt(-1)............

    Here you are not allowed to do so by the theorem in real Nos that says that :

    ........sqrt(ab)= sqrt(a)*sqrt(b) only if a>=0 and b>=0.........,but -1<0.

    However if you work in the complex Nossystem then:

    ........sqrt(1) = 1 or -1..................................

    SO one has to consider two cases:

    ..........sqrt(1)= 1 or sqrt(1) =-1........................................

    Now what you actually did , you considered only the case sqrt(1) = -1 and so you ended up with the result 2=0.

    But in doing so you violated one of the basic laws in logiccalled proof by cases or disjunction elimination.

    It is the same law one has to consider to prove the well known theorem in algebra :

    .............lxl<y <====> -y<x<y.............................

    in proving that theorem we have to consider x>=0 or x<0 and not only one of the two cases,providing one follows this way of proof.

    SO in considering both cases i.e sqrt(1) =1 or sqrt(1) = -1 =sqrt(-1)*sqrt(-1)=i*i=i^2= -1

    you will end with : 2=2 or 2=0 which is true.

    If you have any doubts i can write a detailed proof for you.

    Finally the law of logic i mentioned you can find ,apart from other books of logic in:

    .............schaum's outline series....................

    .................THEORY AND PROBLEMS OF...........

    ...............LOGIC...................... page 50
  24. Dec 28, 2008 #23

    Actualy in your definition you have to mention both:

    x>=0 ,y>=0 and the definition is : [x>=0, sqrt(x) =y]====>[ y^2=x , y>=0]

    we do not need double implication,a single one will do
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