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Math Help - Irrationality and Division

  1. #1
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    Irrationality and Division

    I have two questions in my packet that I'm not completely show how to do. Any help is appreciated.

    Q1) prove that the sum of square root of 2 and square root of 3 is irrational. [hint: prove square root of six is irrational and then use this as a lemma]

    Q2) Suppose a,b,c,d is a set of integers such that a/c and b/c. Show that ab/cd.
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  2. #2
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    #1: Let: n = \sqrt{2} + \sqrt{3} \qquad {\color{red}\star _1}

    Assume n is rational.

    Multiply both sides by the conjugate: n \left(\sqrt{2} - \sqrt{3}\right) = -1 \ \Leftrightarrow \ -\frac{1}{n} = \sqrt{2} - \sqrt{3} \qquad {\color{red}\star _2}

    Add {\color{red}\star _1} and {\color{red}\star _2}: n - \frac{1}{n} = 2\sqrt{2} \ \Leftrightarrow \ \frac{n}{2} - \frac{1}{2n} = \sqrt{2}

    Can you see the contradiction?

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    #2: What's the question?
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  3. #3
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    Quote Originally Posted by TitaniumX View Post
    I have two questions in my packet that I'm not completely show how to do. Any help is appreciated.

    Q2) Suppose a,b,c,d is a set of integers such that a/c and b/c. Show that ab/cd.
    I will assume the question is "If a|c and b|d, then prove that ab|cd" .... where a|b is read as a divides b.

    a|b \Leftrightarrow \exists n, b = an

    c|d \Leftrightarrow \exists m, d = mc

    But this means bd = mnac \Leftrightarrow ac|bd
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  4. #4
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    Rational and irrational numbers

    Hello -

    Quote Originally Posted by o_O View Post
    #1: Let: n = \sqrt{2} + \sqrt{3} \qquad {\color{red}\star _1}

    Assume n is rational.

    Multiply both sides by the conjugate: n \left(\sqrt{2} - \sqrt{3}\right) = -1 \ \Leftrightarrow \ -\frac{1}{n} = \sqrt{2} - \sqrt{3} \qquad {\color{red}\star _2}

    Add {\color{red}\star _1} and {\color{red}\star _2}: n - \frac{1}{n} = 2\sqrt{2} \ \Leftrightarrow \ \frac{n}{2} - \frac{1}{2n} = \sqrt{2}

    Can you see the contradiction?

    ______________________________

    #2: What's the question?
    This is neat. As a matter of interest, it works for any expression of the form \sqrt{a}+\sqrt{b}, where a, b \in \mathbb{N} and \sqrt{a} and \sqrt{b} are individually irrational. Just put \sqrt{a}+\sqrt{b}=n, multiply both sides by \sqrt{a}-\sqrt{b}, and continue as before.

    Also, if you do want to prove that \sqrt{6} is irrational, do it in the same way as proving \sqrt{2} is irrational, as follows:

    Suppose \sqrt{6} is rational, and write it as \frac{p}{q}, where p, q \in \mathbb{N}, and p and q are co-prime. (In other words the fraction \frac{p}{q} is in its lowest terms.)

    Then:

    \frac{p^2}{q^2} = 6

    \implies p^2 = 6q^2

    \implies p^2 has a factor of 2 (and 3 as well, but you don't actually need that)

    \implies p has a factor 2, since 2 is prime

    So write p=2m, say, where m \in \mathbb{N}

    So p^2=4m^2=6q^2

    \implies 3q^2=2m^2

    \implies q^2 has a factor 2

    \implies q has a factor 2

    \implies p and q have a common factor 2.

    Contradiction.

    \implies \sqrt{6} is irrational.

    Grandad
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