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Easy question, unsure if this even counts as calc

  1. Jan 26, 2008 #1
    [SOLVED] Easy question, unsure if this even counts as calc

    1. The problem statement, all variables and given/known data
    14) Show that if a, b, and c are real numbers and a != 0, then there is a unique solution of the equation ax + b = c.

    I'm asking this question here, as it seems TOO easy to be asked in my propositional calculus course.
    The easiest way for me to solve this is to choose numbers a, x, b, and c such that the equation holds, therefore demonstrating that there is a unique solution.

    Is this what they're asking of me?

    Or do I need to find numbers a, b, c such that the equation holds for all values of x?

    I dunno. It just seems too simple...
  2. jcsd
  3. Jan 27, 2008 #2
    Actually, what they're asking is to show that for any real numbers a, b, and c such that a != 0, that exists some value for x in the equation ax + b = c.

    Also, while browsing mathworld, I found their page on propositional calculus: http://mathworld.wolfram.com/PropositionalCalculus.html
  4. Jan 27, 2008 #3
    Thank-you :)

    That's actually just as easy >.>
    Oh well ^^;
  5. Jan 27, 2008 #4

    Ben Niehoff

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    Not quite. They're not asking you to show that the solution exists; they're asking you to show that it is unique. That is, given some a, b, and c (a != 0), show that

    1) There is an x such that ax + b = c, and

    2) There is only one x such that ax + b = c.

    The second one is slightly trickier.

    Note: "Propositional Calculus" is not the same thing as "Differential Calculus" or "Integral Calculus".
  6. Jan 27, 2008 #5
    It doesn't seem too tricky. I did it in 5ish lines?

    Correction, 4:
    ax + b = c
    ax = c – b
    x = (c-b) / a (Note: This is valid, as a != 0.)

    ax + b = c has a unique solution x = (c-b)/a.

    And that's showing my work, too XD
  7. Feb 3, 2008 #6

    Gib Z

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    Well almost - You marked this solved too early, you somewhat missed Ben's point. Yes, rearranging the equation got you one solution.

    But thats like going, [tex]x^2 = 4[/tex], [tex]\sqrt{x^2} = \sqrt{4}[/tex], [tex]x=2[/tex]. That indeed did get you one solution, but x=-2 is another solution.

    You must show your solution is unique. To do that, use proof by contradiction. If there is another solution, it must differ for your first solution by some number, so let that number, whatever it is, be T. From there its as easy as showing that [tex]x= \frac{c-b}{a} + T[/tex] only satisfies ax+b = c if and only if T=0 ie any other proposed solutions must differ from the original one by 0 iee they are the same.
  8. Feb 3, 2008 #7
    Any proof that could REASONABLY be answered by "This is obvious, therefore is true" (Even though such an answer won't get you marks) bugs me :(
    The assignment that this question was for is already handed in, so now I get to kick myself :D
  9. Feb 3, 2008 #8

    Gib Z

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    lol well thats why when I do mathematics, I like to pretend I'm the first person discovering that math and hence I need to find out everything about it myself, not use assumed knowledge. Sure, we know from our familiarity with linear equations that theres only 1 solution, but the first person to have done that might have not! Also, some other person may have assumed that quadratics only have 1 solution by the same line of thought "well When i rearranged it I only got 1 answer", but thats still wrong. Yes, I don't particularly like these kind of proofs either, but get used to it because it can get much much more annoying. I've seen things like "Show ab= ba when a and b are real numbers." It is possible, but it seems pointless doesn't it? Because its so obvious? We just have to live with it =]
  10. Feb 3, 2008 #9


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    Why would you think any proof could be "reasonably" answered that way? What is "obvious" to you may not be obvious to someone else. Any lawyer who went into court and announced that he was not going to present evidence because it was "obvious" that his client was not guilty is going to get his client convicted! He had better explain exactly why it is obvious to him.

    All you are saying is that you are so self-centered, if it's good enough for you it must be good enough for every one.
  11. Feb 3, 2008 #10


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    This is side-tracking a little from the topic, but Goldenwind only meant to say that thinking in such a manner (ie. claiming the proof is obvious) bothered him, not that he was supportive of it.
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