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    prove irrational

    Prove that sqrt(1 + sqrt(3)) is irrational!
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    Quote Originally Posted by mandy123 View Post
    Prove that sqrt(1 + sqrt(3)) is irrational!
    you can do a proof by contradiction

    assume it is rational. then we can write \sqrt{1 + \sqrt {3}} = \frac ab for a,b \in \mathbb{Z}, b \ne 0.

    we square both sides to obtain: 1 + \sqrt{3} = \frac {a^2}{b^2}

    so \sqrt{3} = \frac {a^2}{b^2} - 1

    now the right side of the equation is a rational number (by our assumption). thus we must have that \sqrt{3} is rational. therefore, we can prove the claim by showing that this is false.

    now proving \sqrt{3} is irrational is also done by contradiction, and usually done the same way \sqrt{2} is proven to be irrational (see here)

    but there is a shorter way.

    consider the equation x^2 - 3 = 0

    by the rational roots theorem, if rational roots exist, they will be one of \pm 1, \mbox{ or } \pm 3. but none of these solve the equation. however, \sqrt{3} solves the equation and is therefore not a rational number.

    so seeing that \sqrt{3} is not rational, we have our contradiction
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    Quote Originally Posted by Jhevon View Post
    [snip]

    now proving \sqrt{3} is irrational is also done by contradiction, and usually done the same way \sqrt{2} is proven to be irrational (see here)

    [snip]
    I'll just toss in my 2 cents worth here and say that proving \sqrt{3} is irrational is subtlely more difficulty than proving \sqrt{2} is irrational ......

    If you're assuming the usual thing, \sqrt{3} = \frac{m}{n} where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

    This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).
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    Quote Originally Posted by mr fantastic View Post
    I'll just toss in my 2 cents worth here and say that proving \sqrt{3} is irrational is subtlely more difficulty than proving \sqrt{2} is irrational ......

    If you're assuming the usual thing, \sqrt{3} = \frac{m}{n} where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

    This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).
    I think this may be overkill, they may already know that the square root of
    any integer that is not a perfect square is irrational (or even n-th root of
    an integer not an n-th power).

    In this form of question I woukd assume they know that \sqrt{3} is irrational.

    RonL
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    Quote Originally Posted by mr fantastic View Post
    I'll just toss in my 2 cents worth here and say that proving \sqrt{3} is irrational is subtlely more difficulty than proving \sqrt{2} is irrational ......

    If you're assuming the usual thing, \sqrt{3} = \frac{m}{n} where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

    This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).
    It is extremely simple.

    Definition: Given a positive integer a and a prime p we define \mbox{ord}_p a = n ( n\geq 0)as the number of times that p divides a, i.e. p^n | a but p^{n+1} \not | a. (Note, this definition is a good definition because eventually powers of p will exceeded a and therefore will not divide it any further thus such an n has to exist).

    Lemma 1: Given two positive integers a \mbox{ and }b with p prime we have that \mbox{ord}_p (ab) = \mbox{ord}_p a + \mbox{ord}_p b.

    Proof: If \mbox{ord}_p a= n then p^n|a but p^n\not | a. If \mbox{ord}_p b = m then p^m | b but p^m \not | b. This means that p^{n+m}| ab but if p^{n+m+1}|ab then p^{n+1} (or p^{m+1}) divides a (or b respectively) which is a contradiction.

    Lemma 2: Given a positive integer a the integer is a power of n if and only if \mbox{ord}_p a is divisible by n for all primes p.

    Proof: If a=b^n then \mbox{ord}_p a = \mbox{ord}_p b^n = n\mbox{ord}_p b. Conversely, if \mbox{ord}_p a is divisible by n then by unique factorization we know that a= \prod_{p \text{ prime}} p^{\text{ord}_p a} and thereofore each exponent is prime and divisible by n hence a is a power of n.

    Theorem: If a\geq 1 is not a power of n then \sqrt[n]{a} is irrational.

    Proof: Proof by contrapositive. Suppose that \sqrt[n]{a} = r/s for positive integers r,s. Then, (r/s)^n = a thus r^n = as^n. Now for any prime p we have \mbox{ord}_p r^n = \mbox{ord}_p as^n thus  n \mbox{ord}_p r = n\mbox{ord}_p s + \mbox{ord}_p a. This shows us that \mbox{ord}_p a is divisible by n for any prime p. Thus, by Lemma 2 we have that a is a power of n.
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