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-1 = 1 ?

  1. May 30, 2010 #1
    -1 = 1 ?!?!

    [tex]-1 = -1[/tex]

    [tex]\Rightarrow \frac{-1}{1}=\frac{1}{-1}[/tex]

    [tex]\Rightarrow \sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}[/tex]

    [tex]\Rightarrow \frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}[/tex]

    [tex]\Rightarrow \frac{i}{1}=\frac{1}{i}[/tex]

    [tex]\Rightarrow i^2=1[/tex]

    [tex]\Rightarrow -1 = 1[/tex]

    As you can see, -1 clearly equals 1.
     
  2. jcsd
  3. May 30, 2010 #2

    Borek

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  4. May 30, 2010 #3

    mathman

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    Re: -1 = 1 ?!?!

    √1 = ±1
    √-1 = ±i
     
  5. May 31, 2010 #4
    Re: -1 = 1 ?!?!

    Why not go the whole hog and say [itex]\sqrt{x}=\mathbb{C}[/itex]? Similarly [itex]{x+y=x-y=xy=x/y=x^y=\mathbb{C}}[/itex]. You'll be guaranteed a correct result every time!
     
  6. May 31, 2010 #5

    mathman

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    Re: -1 = 1 ?!?!







    ???????????????????????????????????????????????????????????????????
     
  7. Jun 1, 2010 #6
    Re: -1 = 1 ?!?!

    I don't disagree with either of the statements in your original post if the symbols "[itex]\sqrt{ }[/itex]" and "[itex]\pm[/itex]" have (what I understand to be) their standard meanings.

    E.g. with these meanings:

    (a) [itex]\sqrt{1}=1[/itex]
    (b) [itex]\sqrt{1}=\pm{1}[/itex] means [itex](\sqrt{1}=1)\vee(\sqrt{1}=-1)[/itex].

    The first of the disjuncts in (b) would therefore be true and the second false, rendering the disjunction true.

    Similarly [itex]\sqrt{-1}=i[/itex], so [itex]\sqrt{-1}=\pm{i}[/itex] is true.

    By the same token, [itex](\sqrt{1}=1)\vee(\sqrt{1}=-1)\vee(\sqrt{1}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i)[/itex] is true, as indeed is:

    (c) [itex](\exists x\in\mathbb{C})\sqrt{1}=x[/itex]

    By including your original post in this thread, I assumed you were suggesting that whenever square root symbols appear in an equation the equation can be replaced with a disjunction similar to the one in (b), leaving permanently open the the question of which equality actually holds. This is obviously simpler than working out the answer.

    What I am suggesting is that you could make things simpler still by always replacing an equation:

    [itex]\mathcal{E}_1(\sqrt{a_1},\sqrt{a_2},...,\sqrt{a_m})=\mathcal{E}_2(\sqrt{b_1},\sqrt{b_2},...,\sqrt{b_n})[/itex]

    by:

    [itex](\exists x_1,x_2\dots x_m,y_1,y_2,\dots y_n\in\mathbb{C})
    \mathcal{E}_1(x_1,x_2,\dots ,x_m)=[/itex][itex]\mathcal{E}_2(y_1,y_2,\dots ,y_n)[/itex]

    and leaving permanently open the question of for what values of [itex]x_1,\dots,x_m,y_1,\dots,y_n[/itex] the equation holds.

    In fact, if there are a large number of "[itex]\sqrt{}[/itex]" symbols in an expression, an ambiguity with [itex]2^{\aleph_0}[/itex] possible meanings is probably not significantly less definite intuitively than an ambiguity with [itex]2^{m+n}[/itex] possible meanings.

    Indeed this seems like such a good idea that you could extend it to all arithmetical operations so, for example, replacing any equation:

    [itex]\mathcal{E}_1+\mathcal{E}_2=\mathcal{E}_3+\mathcal{E}_4[/itex]

    by:

    [itex](\exists x,y\in\mathbb{C})x=y[/itex]

    and leaving permanently open the question of what [itex]x[/itex] and [itex]y[/itex] actually are.

    Then you can get to the pub quicker.
     
  8. Jun 1, 2010 #7

    mathman

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    Re: -1 = 1 ?!?!

    I was just trying to keep things simple, to show that the flaw in original "proof" was the failure to take into account the sign ambiguity for square root.
     
  9. Jun 1, 2010 #8
    Re: -1 = 1 ?!?!

    Just keep them as sets from the start, sqrt(1)={1,-1} by using this notation the problem with the OP is gone. Also saying that sqrt(1)=1 is wrong unless you use the real version of the square root which only allows for positive real numbers, but as long as we use the complex one we must think of square roots as sets since it is an inverse to the square operation.
     
  10. Jun 2, 2010 #9
    Re: -1 = 1 ?!?!

    With the standard definition of square root there is no sign ambiguity. OP's mistake was on the fourth line, not the third, where he's erroneously assumed that
    [tex]\sqrt{a/b}=\sqrt{a}/\sqrt{b}[/tex]
    is an identity.

    For real [itex]a[/itex] and [itex]b\neq 0[/itex] the rule is
    [tex]\sqrt{a/b}=\sqrt{a}/\sqrt{b}\text{ unless }a>0\text{ and }b<0\text{, when }\sqrt{a/b}=-\sqrt{a}/\sqrt{b}[/tex].
     
    Last edited: Jun 2, 2010
  11. Jun 2, 2010 #10
    Re: -1 = 1 ?!?!

    Not according to Apostol, Mathematical Analysis, Addison Wesley, among many others.

    The square function has no inverse function. A function [itex]\sqrt{}:x\mapsto \{y|y^2=x\}[/itex] is not even the inverse relation, which would be a relation in [itex]\mathbb{C}[/itex] not between [itex]\mathbb{C}[/itex] and [itex]2^\mathbb{C}[/itex].

    Unless you also extend the definitions of arithmetical operations to include operations between sets the problem with the OP would be gone on line four because both [itex]{\{i,-i\}/\{1,-1\}}[/itex] and [itex]{\{1,-1\}/\{i,-i\}}[/itex] would be undefined and the "proof" could progress no further.

    Because root signs can appear in any arithmetic operation in conjuction with other root signs or numbers you would need to extend the definition of every operation to cover the cases where one or more operands, but not necessarily all, are sets. But this is likely to create more problems than it solves. (Not that I agree that changing the definitions to give ambiguous results solves anything.) For example if you were to propose the following for addition and multiplication

    [tex]\text{For }n\in\mathbb{C}\text{ and }A,B\subset \mathbb{C}[/tex]
    [tex]A+B=_{df}\{a+b|a\in A\text{ and }b\in B\}[/tex]
    [tex]nA=_{df}\{na|a\in A\}[/tex]

    then it would follow for any [itex]a\neq 0\in \mathbb{C}[/itex] that

    [tex]2\sqrt{a}\neq \sqrt{a}+\sqrt{a}[/itex]

    because letting [itex]b[/itex] denote one of the square roots of [itex]a[/itex], the set on the left is [itex]{\{-2b,2b\}}[/itex], whereas that on the right is [itex]{\{-2b,0,2b\}}.[/itex]

    More awkward would be deciding how to define limits of expressions involving square roots. E.g. Viete's formula

    [tex]\pi=\lim_{n\rightarrow\infty}2^n\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt{2}}}}}}\text{ (n root signs in total)}[/itex]

    would most likely end up looking a bit sick. Here the [itex]n^{th}[/itex] convergent would be a set with [itex]2^n[/itex] members if the following definition be assumed.

    [tex]\text{For }A\subset\mathbb{C}\text{, }\sqrt{A}=_{df}\{\sqrt{a}|a\in A\}[/tex].

    Potential problems include

    (a) The triangle inequality in [itex]\mathbb{R}^n[/itex] would have to be paraphrased along the lines, "at least one of the four possible sums of two sides of a triangle is greater than either of the third", which has lost something in pithiness.

    (b) In general equations with more square root signs on one side than the other would be insoluble. This includes cases where one side contains no roots, e.g. [itex]\sqrt{x}=2[/itex] would be insoluble because the left hand side is a set containing two numbers, while the right hand side is just the number [itex]2[/itex].

    (c) You'd need to find an alternative to the usual way of defining [itex]x^y[/itex] for positive [itex]x[/itex] and real irrational [itex]y[/itex]

    (d) No expression containing square roots would be continuous, differentiable or integrable (either as an indefinite or a definite integral).

    What definitions would you propose for arithmetical operations using the sets that you propose?

    How would you define limit processes for expressions and sequences of expressions involving square roots?

    Would your proposals solve all the potential problems?
     
    Last edited: Jun 2, 2010
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