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    Prove that  \sqrt{17} is irrational.
    Prove that  \log_5 10 is irrational.
    Prove that e is irrational.
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    (Original post by Ano9901whichone)
    Prove that  \sqrt{17} is irrational.
    Prove that  \log_5 10 is irrational.
    Prove that e is irrational.
    Do you need help with this or are you posting it as a challenge?

    If you need help, start with the first one and tell us if you know how to prove that \sqrt{2} is irrational. If you don't then you should start with that.

     \sqrt{17} is similar but slightly harder.
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    (Original post by notnek)
    Do you need help with this or are you posting it as a challenge?

    If you need help, start with the first one and tell us if you know how to prove that \sqrt{2} is irrational. If you don't then you should start with that.

     \sqrt{17} is similar but slightly harder.
    Proving  \sqrt 2 is easy.
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    (Original post by Ano9901whichone)
    Proving  \sqrt 2 is easy.
    Okay. Do you need help with proving \sqrt{17} is irrational?
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    (Original post by notnek)
    Okay. Do you need help with proving \sqrt{17} is irrational?
    Well I think I have it. I would say that assume  \sqrt{17} is rational such that it can be written as  p/q, \ p,q \in \mathbf{Z} .
    Squaring gives  17q^2=p^2 .
    Since  p^2 is clearly divisible by 17 (prime) it follows that p is divisible by 17  \Rightarrow p=17k, \ k \in \mathbf{Z} where gcd =1.
     \Rightarrow 17k=q^2 , by the same logic we deduce that  17|q . This is a contradiction as both p and q cannot be divisible by 17 as they have gcd of 1. So this forces us to conclude that  \sqrt{17} is irrational.
    How's that?
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    (Original post by Ano9901whichone)
    How's that?
    Fine.
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    (Original post by Ano9901whichone)
    Well I think I have it. I would say that assume  \sqrt{17} is rational such that it can be written as  p/q, \ p,q \in \mathbf{Z} .
    Squaring gives  17q^2=p^2 .
    Since  p^2 is clearly divisible by 17 (prime) it follows that p is divisible by 17  \Rightarrow p=17k, \ k \in \mathbf{Z} where gcd =1.
     \Rightarrow 17k=q^2 , by the same logic we deduce that  17|q . This is a contradiction as both p and q cannot be divisible by 17 as they have gcd of 1. So this forces us to conclude that  \sqrt{17} is irrational.
    How's that?
    Good.
    Alternative
    Note from p^2=17q^2 consider p,q in prime factors the LHS has an even number of exponents the RHS has odd number of exponents, hence a contradiction.
    For the log one you do the same and simplify some equations and get a parity argument for p,q>0.

    Note sometimes the first argument does not work as nicely.
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    Oh, and for e is irrational. Look at its series expansion and then multiply both sides by n!, then prove that the remaining terms with factorials in the denominator is bounded in [0,1] and you have a contradiction.
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    (Original post by Zacken)
    Oh, and for e is irrational. Look at its series expansion and then multiply both sides by n!, then prove that the remaining terms with factorials in the denominator is bounded in [0,1] and you have a contradiction.
    Alternatively a much easier way is just to note that e's continued fraction expansion is infinite - this was the first proof of e's irrationality, and was published by Euler.
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    (Original post by HapaxOromenon3)
    Alternatively a much easier way is just to note that e's continued fraction expansion is infinite - this was the first proof of e's irrationality, and was published by Euler.
    Sounds too much like cheating unless you prove that infinite continued fractions => irrationality and that e's continued fraction expansion is infinite.
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    (Original post by Zacken)
    Sounds too much like cheating unless you prove that infinite continued fractions => irrationality and that e's continued fraction expansion is infinite.
    It is not very difficult to compute e's continued fraction expansion, which shows that it is infinite. This paper does it in just over a page: see Proposition 1 and Theorem 1.

    The result about infinite continued fractions is trivial to prove by contrapositive - clearly every rational number's continued fraction must terminate as rational numbers are themselves just fractions.
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    (Original post by HapaxOromenon3)
    It is not very difficult to compute e's continued fraction expansion, which shows that it is infinite. This paper does it in just over a page: see Proposition 1 and Theorem 1.

    The result about infinite continued fractions is trivial to prove by contrapositive - clearly every rational number's continued fraction must terminate as rational numbers are themselves just fractions.
    Nice paper, thanks for the link. The proof via series expansion seems shorter than that, though. I can comfortably do it in 5-8 lines.


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    (Original post by Zacken)
    Nice paper, thanks for the link. The proof via series expansion seems shorter than that, though. I can comfortably do it in 5-8 lines.


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    Not with full rigour you can't. It takes about a page of working - see
    http://www.mathshelper.co.uk/Proof%2...Irrational.pdf
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    (Original post by HapaxOromenon3)
    Not with full rigour you can't. It takes about a page of working - see
    http://www.mathshelper.co.uk/Proof%2...Irrational.pdf
    Lol
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    (Original post by physicsmaths)
    Lol
    At what do you laugh, fool?
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    (Original post by HapaxOromenon3)
    At what do you laugh, fool?
    The document and now you.
    All of that can be shortened to 8 lines as Zacken said.


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    (Original post by physicsmaths)
    The document and now you.
    All of that can be shortened to 8 lines as Zacken said.


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    You have no evidence of that.
 
 
 
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