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Dumb Proof Question

  1. Sep 29, 2013 #1
    Sorry for the dumb question by my high school teachers never explained this well and I'm an engineering student which means I don't do many proofs...

    Let us say that I have to prove that the following is an identity:

    $$\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }$$

    My high school teachers said that I would have to set the left side of the equation to "LHS" and the right side of the equation to "RHS". Then manipulate the variables "LHS" and "RHS" until LHS=RHS. The rationale given by my high school teachers for the reason why I can't just work on it in the original form is because we don't know yet if LHS=RHS.

    However, what if we make a proposition or we suppose that:

    $$\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }$$

    is true.

    Let's say it is reduced to something like: 0=1. Can we say that by proof of contradiction, the original equation is not an identity?

    Let's say it is reduced to something like: 1=1. Can we say that because the original equation is reduced to an identity, the original equation is an identity?

    Is my choice of words correct?

    So to recap, the below would be accepted as proofs by a university professor, right?

    Question: Prove the following is an identity: ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny } ##
    Answer Case A: Suppose that ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.
    ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }##
    ##{ sin }^{ 2 }x+{ cos }^{ 2 }x={ sin }^{ 2 }y+{ cos }^{ 2 }y##
    1=1
    Therefore, because the original equation is reduced to an identity, ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.

    Answer Case B: Suppose that ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is an identity.
    ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## ... (in this case it would reduce into 1=1, but let's just pretend that doens't happen) 0=1 Therefore, because our original assumption led to a contradiction, our assumption is incorrect, and by proof by contradiction, ##\frac { cosx-siny }{ cosy-sinx } =\frac { cosy+sinx }{ cosx+siny }## is not an identity.
     
    Last edited by a moderator: Sep 30, 2013
  2. jcsd
  3. Sep 29, 2013 #2

    Office_Shredder

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    If it reduces to a contradiction, then you have shown a much stronger result: that there exists no choice of x and y which makes the statement true. all you need to show it isn't an identity is some example of an x and y for which it isn't true.

    As far ad the proof that it is an identity, the key point is that your steps need to be reversible. Then you can start at 1=1 and go backwards to prove the identity.
     
  4. Sep 29, 2013 #3
    No, you can't assume two quantites are equal. If you assume they are equal and get 1=1, you haven't proved anything.

    -3 = 3? Assume they are equal. Square both sides, then divide through by 9, and you get 1 = 1. Does that sound right to you?

    In fact, if you have any x and y, where x and y stand for some complicated expressions, and assume x = y, then all you have to do is divide through by x to get 1 = 1, since y/x = x/x if x and y are equal.
     
  5. Sep 30, 2013 #4

    HallsofIvy

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    You can use what is called "synthetic proof":
    To prove statement "A", begin by assuming A is true, then manipulating the formulas until you arrive at statement "B" which is obviously true (i.e. "sin(x)= sin(x)" or "1= 1")
    only if every step in the manipulation is reversible.

    The real proof, then, would be going back from "B" to "A"- but you don't need to do that explicetely because it is clear that every step was reversible so you could.

    The reason why brocks counterexample fails is that the step "square both sides" is not reversible.
     
  6. Sep 30, 2013 #5
    Why is that? Is it because you lose information?
     
  7. Sep 30, 2013 #6

    Office_Shredder

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    Exactly... if you have x2 = y2, then you cannot conclude that x=y.

    If I tried to reverse the -3 = 3 proof, I would start with

    1=1
    Multiply both side by 9
    9 = 9

    Take the square root of both sides
    -3 = 3.

    But I can't just take an arbitrary square root and know that I will end up with the same number on both sides, so this was not a legitimate step.


    A very strong example of why your steps need to be reversible: Suppose you were asked to prove f(x) = g(x) for two functions f and g.
    Starting with f(x) = g(x):
    Step 1: Multiply both sides by 0:
    0 = 0.

    Now we try to reverse this. Starting with 0=0, I can't divide both sides by 0 to get f(x) = g(x), so I'm obviously stuck and I haven't proved anything.

    In the OP you have to be careful about this kind of effect - you multiply both sides by cos(x)+sin(y), which if it's zero is not a reversible step. Fortunately for your specific identity if it equals zero the identity doesn't hold because you're dividing by zero (and it is implicitly understood that it's not supposed to), so it doesn't affect the short proof, but if you want to be very rigorous about the proof you should make an observation about that in the write-up.
     
  8. Sep 30, 2013 #7
    I see. What are the possible results of a synthetic proof?

    1=1 or something which is obviously true: which would mean that the statement is an identity.
    0=1 or a contradiction: which would mean that there exists no choice of x and y which makes the statement true.

    How about to prove the statement is not an identity?
     
  9. Sep 30, 2013 #8
    Looking back at this, how did you even manage to get -3 on the left side?

    $$9=9\\ \sqrt { 9 } =\sqrt { 9 } \\ 3=3$$
     
  10. Sep 30, 2013 #9

    Office_Shredder

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    This is basically equivalent to what I was saying - squaring/square roots is not a reversible procedure, so you can't use it.

    The third missing result of a synthetic proof is you can never come to a trivial identity or a contradiction. This will be the case if there are some values of your variables which will make the alleged identity true, and some that don't.

    If you just want to prove that something is not an identity, all you have to do is find a single example where it fails. For example, if someone says "Hey, I bet x2 = x3". I reply "Nope", and they say "prove it". All I have to do is:
    Let x = 2.
    x2 = 4, and x3 = 8, so x2 is not equal to x3 when x=2. Therefore x2 = x3 is not an identity.
     
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