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Math Help - Language and proofs

  1. #1
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    Language and proofs

    I am working on this problem it makes sense to me but there is something inside screaming that it is wrong... here it is.

    Consider the equation x^4y+ay+x=0
    a) Show that the following statement is false. "For all a,x, in real Numbers, there is a unique y such that x^4y+ay+x=0.

    basically i solved for y getting y= -x/(x^4+a). Then explained it as it must be false because x and a are variables spanning R and are conditional to each other so it must be false.

    b was just proving that it was true when the equation had to work for x in all real's

    Help?? or validate.
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  2. #2
    Senior Member Twig's Avatar
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    Well, you found a counter-example, then it cannot be true.

    What if x=1 \mbox{  and  } a = -1 ?
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  3. #3
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    You might be overthinking.... think about this, instead:

    Is your expression for y always defined, or are there values of x and a that cause it to diverge?
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  4. #4
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    Thank you both

    I'm kind of embarrassed to ask this but does unique mean a single y? i thought it meant that for any value of x and a there was a number y which made the equation true.

    Thank you for responding
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  5. #5
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    Quote Originally Posted by j5sawicki View Post
    I'm kind of embarrassed to ask this but does unique mean a single y? i thought it meant that for any value of x and a there was a number y which made the equation true.
    Don't be embarrassed; you're correct in your original thinking. They seem to be asking whether or not it is true that for any x and a there is a corresponding y such that the equation holds. We have shown that the answer is no, since for certain values of x and a, the equation for y is undefined due to a zero denominator.

    There will be many more than just a "single value" for y; the set of y-values should be uncountably infinite.

    Hope that makes it somewhat clearer to you.
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