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Simple manipulation of x^2 - 1 = 0, getting x=0 as a solution.

  1. Apr 11, 2013 #1
    This isn't homework or anything, just something I tried and it doesn't seem to be consistent.

    I randomly just multiplied x^2 - 1 = 0 by x instead of factoring and solving:
    x^3 - x = 0. Then I factored out an x and got x(x^2 - 1) = 0.
    Then I divided by x^2 - 1 and got x = 0 which isn't a solution?

    I know I can just divide by x in x(x+1)(x-1) = 0 but why can't I divide by the other two factors?
    what am I doing that's algebraically illegal?
     
  2. jcsd
  3. Apr 11, 2013 #2

    micromass

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    What you did is correct, but problem is that your operations are irreversible.

    What do I mean with that? When manipulating equations, you have almost always have an ##\Rightarrow##. This means that the left-hand side implies the right hand side.

    For example, ##x=y~~\Rightarrow ~~ x+3 = y+3##. Indeed: if ##x=y##, then I can add ##3## at both sides.
    Other examples
    [tex]x=y~~ \Rightarrow ~~ x^2 = y^2[/tex]
    [tex]x=y ~~ \Rightarrow ~~ 0x=0y[/tex]

    Some operations are reversible. Which means that the right-hand side also implies the left-hand side. This is a very special property. Students often think that this is always true, but that is not the case. A reversible property can be written as ##\Leftrightarrow##.
    For example ##x=y~~ \Leftrightarrow ~~x+3 = y+3##. So adding ##3## to both sides is reversible. Which means: if I ever encounter the situation x+3=y+3, then I can always conclude that x=y.

    However,
    [tex]x=y~~ \Leftrightarrow ~~ x^2 = y^2[/tex]
    is not true. For example, ##1^2 = (-1)^2##, but we can not deduce that ##1=-1##.

    [tex]x=y ~~ \Leftrightarrow ~~ 0x=0y[/tex]
    is also not true. We have that ##0*1 = 0*2##, but we can not deduce that ##1=2##.

    Now, when solving an equation, you try to do this by manipulating your equation. The problem is however that not all manipulations are reversible.
    An easy case is the following: solve 2x + 4 = 0
    We can do the following:

    \begin{eqnarray*}
    2 x + 4 = 0
    & \Leftrightarrow & 2x = -4\\
    & \Leftrightarrow & x = -2
    \end{eqnarray*}

    What we proved now is actually two things. We have proved the normal direction: IF 2x+4=0 THEN x=-2. We have also proven the backward direction: IF x=-2 THEN 2x+4 = 0.

    However, this backward direction is very often missing when solving equations. For example, let's solve ##\sqrt{x^2+x+3} = x##

    \begin{eqnarray*}
    \sqrt{x^2+x+3} = x
    & \Rightarrow & x^2 + x+ 3 = x^2
    & \Leftrightarrow & x = -3
    \end{eqnarray*}

    In the second step, we have just squared both sides of the equation. But we know that this is not reversible!!
    Indeed, what we have proven now is that IF ##\sqrt{x^2 + x + 3}=x##, then ##x=-3##. But we can make no statement about the converse because the implications are not reversible. Indeed, we see that ##x=-3## is not a solution of the equation.

    Now, in your example, you do the following
    [tex] x^2 - 1 = 0~~\Rightarrow~~ x(x^2 - 1)=0[/tex]

    This is certainly valid. But it is not reversible. So if we are given ##x(x^2 -1)=0##, then we cannot deduce that ##x^2 - 1 = 0##.

    So in your example, you have proven that IF ##x^2 - 1 =0##, then either ##x=0## or ##x=1## or ##x=-1##. This is certainly true. But the converse is not true.
     
  4. Apr 11, 2013 #3

    Mark44

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    A shorter explanation involves the solution set for each step.

    Equation Solution set
    1. x2 - 1 = 0 {1, -1}
    2. x(x2 - 1) = 0 {0, 1, -1}
    3. x = 0 {0}

    Going from 1 to 2, you increased the solution set.
    Going from 2 to 3, you decreased the solution set.

    As micromass explained, your steps are not reversible, meaning that the equations are not equivalent to each other. It's only when equations are equivalent that they have the same solution set.
     
  5. Apr 11, 2013 #4
    You are dividing by zero.

    In the first line you define [itex]x^2-1 =0[/itex]

    Thus when you divide [itex]x(x^2-1) [/itex] by [itex](x^2-1) [/itex] you are dividing by zero.
     
  6. Apr 12, 2013 #5

    Mark44

    Staff: Mentor

    Not necessarily.
    Only if x happens to be either 1 or -1. For all other values of x, the division is valid.
     
  7. Apr 12, 2013 #6
    Yes necessarily!

    The starting equation is [itex]x^2-1 =0[/itex]
    Thus we know that [itex]x^2-1 =0[/itex]. So dividing by [itex]x^2-1 [/itex] is equivalent to dividing by zero.

    Always!

    Again from the original equation [itex]x^2-1 =0[/itex] we know that x has to be 1 or -1. So the division is never valid.
     
  8. Apr 12, 2013 #7

    Mentallic

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    micromass, that was a great post, and I have but one query:

    While it makes sense to say that if we have an equation and the reduction steps that arrive to the solution set are reversible, then the solution set satisfies the original equation, but now that the steps are irreversible, what does it mean to say that

    "IF ##\sqrt{x^2 + x + 3}=x##, then ##x=-3##" ?



    Yes, but all that happens is that we lose its solutions. Since solving [itex]x^2-1=0[/itex] gives us x=1,-1, then dividing by [itex]x^2-1[/itex] means that the equation we're left with has to exclude x=1,-1 from its domain.

    If we instead divided by just x+1, then we're left with x-1=0 and we still have one of the solutions of the original equation left.
     
  9. Apr 12, 2013 #8

    pwsnafu

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    Irrelevant. Even if OP didn't divide ##x(x^2-1)## by ##x^2 - 1##, he would still have ##x = 0## as solution (which isn't a solution in the original equation).
     
  10. Apr 12, 2013 #9

    Mark44

    Staff: Mentor

    Not necessarily.
    Only if x happens to be either 1 or -1. For all other values of x, the division is valid.
    When you say "always", the implication is that you mean "for all x." The expression x2 - 1 is not identically equal to zero. As I pointed out before, this expression is equal to zero only if x = 1 or x = -1. For all other values of x, it is legitimate to divide by x2 - 1.

    In this case, the net effect is that by adding and eliminating factors on the left side, we increase or decrease the solution set.
     
  11. Apr 12, 2013 #10

    pwsnafu

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    Exactly what it says on the tin :tongue:

    If I have ##A \implies B##, then when A is true, so is B.
    What happens when A is false? Well we get no information about B.

    What happens when A is always false? We get no information about B what so ever!
    Nonetheless, we do this in everyday language: "if I had caught the bus to work, I would have been late". This is called a counterafactual.
     
  12. Apr 12, 2013 #11

    micromass

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    It means that the only possible solution to the equation is -3. However, it is not guaranteed that it actually is an solution. We just know that if there is a solution, then it must be -3. However, it may happen that there are no solutions! To check whether -3 is a solution, we need to substitute it in the equation.
     
  13. Apr 12, 2013 #12

    rcgldr

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    x = 0 is one of 3 roots for the cubic equation you created, so there's no conflict.

    For a more general example, you could choose any 3rd root r, and multiply x^2 - 1 = 0 by (x-r), to produce a cubic equation that includes x = r as a root, in addition to x = 1, and x = -1.
     
  14. Apr 14, 2013 #13
    No! I don't mean for all x.

    The starting equation and the title of this post is x^2-1=0. Once you define x^2-1 to be zero it is zero. End of story. If you then try to divide by x^2-1 you're going to have a bad day. And this is why the original poster wrongly gets the x=0 is a root of x^2-1=0.

    Once you define x^2-1 =0 you can infer that x=1 or x=-1.
     
  15. Apr 14, 2013 #14

    pwsnafu

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    No. The division is not the cause of the problem.
    ##x^2 - 1 = 0 \implies x(x^2-1) = 0##
    is valid.
    ##x(x^2-1)=0 \implies x^2-1 = 0##
    is not.
     
  16. Apr 14, 2013 #15
    All [itex]x[/itex] in the set of solutions, not in the set of real numbers. Then, all [itex]x[/itex] means [itex]x=1[/itex] and [itex]x=-1[/itex]. I'm sure that this is what he meant.

    If we are changing the solution set of a polynomial, then we are changing the characteristic property of that polynomial. Hence, we would have a different polynomial.
     
  17. Apr 14, 2013 #16
    By hypothesis ##x^2-1=0##. We don't need that last implies.

    The division is the cause of the problem.

    ##x(x^2-1)=0 \implies x = 0##

    Is invalid. Really, it implies that ##x=0## or ##x=1## or ##x=-1##. But, it doesn't equal zero because that would not satisfy the condition in the hypothesis.
     
    Last edited: Apr 14, 2013
  18. Apr 14, 2013 #17

    pwsnafu

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    But no division is necessary. You still get the same problem.

    We start with
    ##x^2 - 1 = 0##
    ##\implies x(x^2-1) = 0##.
    Now from the fact that ##\mathbb{R}[X]## has no zero divisors, we know
    ##\implies x = 0 \vee x^2-1 =0##
    and now I have the same problem. But I have not done any division at all.

    Re-read micromass's post. In symbols it goes:
    ##x=1 \vee x=-1 \iff x^2-1 = 0 \implies x(x^2-1) = 0 \iff x=1 \vee x=0 \vee x=-1##
    That middle ##\implies## cannot be reversed.

    PS: Khrisstian you seem to not have realised that the hypothesis has swapped is now the cubic, not the quadratic. You can't assume ##x^2 - 1 = 0## is true any more because that is what you are trying to prove now.
     
    Last edited: Apr 14, 2013
  19. Apr 14, 2013 #18
    I dont think there was a hypothesis imposter. The hypothesis was that x^2-1=0. Why would that change?

    Starting with that hypothesis you know that x is 1 or -1. If you multiplied both sides of the equation by anything at all, then the equation would remain true. This is precisely because x is 1 or -1. All other derived solutions must be in the set {1,-1} to be actual solutions for x.
     
  20. Apr 14, 2013 #19
    x has graduated to playing a role in a system of equations once you start multiplying both sides of the first equation. Equation 1 still needs to be true.
     
  21. Apr 14, 2013 #20

    pwsnafu

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    Because we are interested in the opposite direction. That means you read backwards: if you structured the proof vertically you now read bottom up. And you are trying to prove what you assumed at the start.

    You seem to not to understand the problem itself. So I'll break it down. The OP started with an axiom, let's call it Statement A. Statement A says "##x^2-1=0##". He then uses Statement A to prove Statement B. Statement B says "##x(x^2-1) = 0##".

    So we have ##A \implies B##. Up to here no problem. But two things happen. First, he realises that Statement A is equivalent to another Statement A'. A' says ##x=\pm1##.
    Secondly, Statement B is also equivalent...to Statement B', which says (you guessed it!) ##x = 0, \pm 1##.

    And now OP asks us: why does B' contain 0, even though A' does not.

    The reason? A is not equivalent to B. Equivalent means ##A \implies B## and ##B \implies A##. We know the first implication is correct, we already did that one! Nobody is complaining about that one. It's the second implication that causes the problem. OP is unconsciously assuming that A "should be" equivalent to B, because that's what he has previous experience with.

    But in this example ##B \not\Longrightarrow A##. This is why I keep saying the hypothesis swapped. We assume B, and look at whether A can be proved from B. You can't assume what you are trying to prove.

    Edit: PS

    You might ask, why should we care? Well, it's actually the definition of "a solution to the equation..."
    To see this consider
    in symbols we get
    Notice the direction of the arrow. But when you solve the equation you do
    ##2x+4 = -2##
    ##\implies 2x = -6 ##
    ##\implies x = -3##
    The arrows are the wrong way!

    We have not "solved" the problem, actually what we did was show
    ##2x+4 = -2 \implies x=-3##, or in words, "if there is a solution, that solution is -3".
    -3 is not (yet) the solution, it's a candidate to the solution. To complete the problem, "we check our answer":
    ##x=-3 \implies 2x = -6 \implies 2x+4 =-2##
    and our arrows are all in the right direction. The "plug the answer back" is what makes the proof work.
     
    Last edited: Apr 14, 2013
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