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Math Help - Prove irrationality!

  1. #1
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    Prove irrationality!

    Prove that the sum is irrational.

    I've planned my approach in 2 ways:

    1. Must show that the sum is not an integer.
    2. Must show it is a zero of a monic polynomial.

    But how to go about it?
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  2. #2
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    Trying to help

    It appears to me that since the first term doesn't cancel the remaining terms, then all you have to do is prove that the first term is irrational.

    Any whole number that's squared which you add one to has to be irrational when you take its square root. I think you can work through the rest.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by wonderboy1953 View Post
    It appears to me that since the first term doesn't cancel the remaining terms, then all you have to do is prove that the first term is irrational.

    Any whole number that's squared which you add one to has to be irrational when you take its square root. I think you can work through the rest.
    Are you really sure thats enough? What about \sqrt{2}+2,1-\sqrt{2} then they don't cancel out but it is nonetheless rational. You need to say a little more if you want to use that argument.
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    Quote Originally Posted by usagi_killer View Post
    Prove that the sum is irrational.
    This is a theorem: If N is a non-square positive integer then \sqrt{N} is irrational.
    What do you know about the sum of irrational numbers?

    Does that help? HOW?
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  5. #5
    Senior Member Sampras's Avatar
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    Quote Originally Posted by usagi_killer View Post
    Prove that the sum is irrational.

    I've planned my approach in 2 ways:

    1. Must show that the sum is not an integer.
    2. Must show it is a zero of a monic polynomial.

    But how to go about it?
    This is equivalent to  \sum_{x= 1001}^{2000} \sqrt{1+x^2} . And  \sqrt{x^2+1} is not a square number which implies it is irrational.

    Since it is monotonic increasing, the sum of the irrationals can't be rational (e.g. we can't have expressions of the form  (2-\sqrt{2}) +(4+\sqrt{2}) ).
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    Quote Originally Posted by Sampras View Post
    This is equivalent to  \sum_{x= 1001}^{2000} \sqrt{1+x^2} . And  \sqrt{x^2+1} is not a square number which implies it is irrational.

    Since it is monotonic increasing, the sum of the irrationals can't be rational (e.g. we can't have expressions of the form  (2-\sqrt{2}) +(4+\sqrt{2}) ).

    Perhaps I'm missing something here, but your very example is an increasing sequence of irrational numbers whose sum is rational! There are others, of course:  \sqrt{2}-1\,,\,\sqrt{2}-\frac{1}{2}\,,\,\sqrt{2}\,,\,7-3\sqrt{2}, etc.

    Tonio
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by tonio View Post
    Perhaps I'm missing something here, but your very example is an increasing sequence of irrational numbers whose sum is rational! There are others, of course:  \sqrt{2}-1\,,\,\sqrt{2}-\frac{1}{2}\,,\,\sqrt{2}\,,\,7-3\sqrt{2}, etc.

    Tonio
    It's not an alternating sequence, so the sum should be irrational.
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  8. #8
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    Okay so had another look at this.

    So to prove the sum is not an integer.



    So,

    Let

    An overestimate for the sum would be

    An underestimate for the sum would be

    But since

    Thus the sum is not an integer.

    Use induction to prove it is a zero of a monic polynomial.

    Base case:

    Let the sum be written as

    Let .

    Thus



    Thus is a zero.

    Now for the inductive hypothesis, assume is true.

    Thus is true.

    Now we wish to prove that is also true.

    We will denote this as .







    Let this be a zero of a monic polynomial

    So using the inductive hypothesis, is a zero , namely

    so

    Now what...?
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  9. #9
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    Quote Originally Posted by usagi_killer View Post
    Okay so had another look at this.

    So to prove the sum is not an integer.



    So,

    Let

    An overestimate for the sum would be

    An underestimate for the sum would be

    But since

    Thus the sum is not an integer.

    Use induction to prove it is a zero of a monic polynomial.

    Base case:

    Let the sum be written as

    Let .

    Thus



    Thus is a zero.

    Now for the inductive hypothesis, assume is true.

    Thus is true.

    Now we wish to prove that is also true.

    We will denote this as .







    Let this be a zero of a monic polynomial

    So using the inductive hypothesis, is a zero , namely

    so

    Now what...?
    if you are familliar with basic concepts of algebraic number theory, you should know this much that the set of integral elements (an elements is called integral if it's a root of a monic polynomial
    with integer coefficient) is a ring and hence it's closed under addition. now each \sqrt{a_i} is integral and thus A=\sum_{i=1}^n \sqrt{a_i} is integral too. it's also a basic fact that \mathbb{Z} is integrally closed, that is an
    integral element which is also a rational number has to be an integer. but, as you almost showed, A is not an integer and so it cannot be rational. Q.E.D.
    Last edited by NonCommAlg; November 18th 2009 at 12:18 PM.
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  10. #10
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    Quote Originally Posted by Sampras View Post
    It's not an alternating sequence, so the sum should be irrational.

    Uuh? But it is not!
    \sqrt{2}-1+\sqrt{2}-\frac{1}{2}+\sqrt{2}+7-3\sqrt{2}=7-\frac{3}{2}=\frac{11}{2}\,\in\mathbb{Q}

    Tonio
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