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Math Help - Natural numbers

  1. #1
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    Natural numbers

    n is a natural unzero number, find x natural so that:

    \sqrt{x-1}<\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}<\sqrt{x}

    I saw that x must be bigger than n, but how I find it?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mexxy View Post
    n is a natural unzero number, find x natural so that:

    \sqrt{x-1}<\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}<\sqrt{x}

    I saw that x must be bigger than n, but how I find it?
    As x increases the difference in [tex]\sqrt{x - 1}[/math[ and [tex]\sqer{x}[/math[ decreases. So x is not likely to be that large. By trial and error (probably the only way to do this one) I get n = 1 and x = 2.

    -Dan
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    "I get n = 1 and x = 2." I don't.
    n=1, x=6;
    n=2, x=18;
    n=3, x=27;
    n=4, x=36;
    n=5, x=45.
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  4. #4
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    Quote Originally Posted by topsquark View Post
    By trial and error (probably the only way to do this one) I get n = 1 and x = 2.

    -Dan
    Are you sure that's correct?
    If x=2 and n=1 is inserted in the original inequality, we get:

    \sqrt{2-1}<\sqrt{1-1}+\sqrt{1}+\sqrt{1+1}<\sqrt{2}

    and I think THAT looks a bit suspicious.
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  5. #5
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    Quote Originally Posted by mexxy View Post
    n is a natural unzero number, find x natural so that:

    \sqrt{x-1}<\sqrt{n-1}+\sqrt{n}+\sqrt{n+1}<\sqrt{x}

    I saw that x must be bigger than n, but how I find it?
    As seen in veileen's post, there's an abundance of solutions to this inequality.
    Therefore, I think that all you can do is simplify this expression a bit by quadrating the whole thing, so that you can get a clearer relationship between n and x.

    If you quadrate the whole inequality, I think that this will be the end result:

    x-1<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x

    but I might have been a bit sloppy while I did it, so I can be a little bit off.
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  6. #6
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    Mexxy said she/ he have to find x... probably something like: x= *stuff, stuff, stuff* >.>
    I mean - that relationship between n and x should be clearer than what you wrote.
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  7. #7
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    Quote Originally Posted by veileen View Post
    Mexxy said she/ he have to find x... probably something like: x= *stuff, stuff, stuff* >.>
    I mean - that relationship between n and x should be clearer than what you wrote.
    Do you mean like something like this:

    3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})+1

    Where the only natural number for a given n, which makes the inequality true, is x.
    I personally don't know how I should go about determining the exact value of x, i.e. x= 'something', when I'm only given this type of inequality, which contains many different solutions.
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  8. #8
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    3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})+1
    3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n}) is not natural ~ "I personally don't know how I should go about determining the exact value of x," - using integer part. (?)
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by scounged View Post
    Are you sure that's correct?
    If x=2 and n=1 is inserted in the original inequality, we get:

    \sqrt{2-1}<\sqrt{1-1}+\sqrt{1}+\sqrt{1+1}<\sqrt{2}

    and I think THAT looks a bit suspicious.
    It was a typo. n = 1 and x = 3. And I see that it was wrong anyway.

    -Dan
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by veileen View Post
    "I get n = 1 and x = 2." I don't.
    n=1, x=6;
    n=2, x=18;
    n=3, x=27;
    n=4, x=36;
    n=5, x=45.
    I'll be darned.

    -Dan
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  11. #11
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    Quote Originally Posted by veileen View Post
    3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})<x<3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n})+1
    3n+2(\sqrt{n^2-n}+\sqrt{n^2-1}+\sqrt{n^2+n}) is not natural
    Does it have to be?

    Quote Originally Posted by veileen View Post
    ~ "I personally don't know how I should go about determining the exact value of x," - using integer part. (?)
    I'm not familiar with integer part. I know that my inequality might not be the best mathematical way to determine x, but it gives a definite solution for x with a given n with the conditions in the original post set.
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  12. #12
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    Does it have to be? - No, that is why I suggested use the integer part.
    3n+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]}\leq 3n+2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})<x

    x<3n+1+2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})\leq3n+1+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]+1

    =3n+2+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]

    [2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]\in \mathbb{N} \Rightarrow x=3n+1+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})] (because it's a natural number between 3n+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})]} and 3n+2+[2(\sqrt{n^{2}-n}+\sqrt{n^{2}-1}+\sqrt{n^{2}+n})])

    * [] - means integer part
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