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Thread: 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: $\displaystyle n = \sqrt{2} + \sqrt{3} \qquad {\color{red}\star _1}$

    Assume $\displaystyle n$ is rational.

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

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

    Can you see the contradiction?

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    #2: What's the question?
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    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.

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

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

    But this means $\displaystyle bd = mnac \Leftrightarrow ac|bd$
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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: $\displaystyle n = \sqrt{2} + \sqrt{3} \qquad {\color{red}\star _1}$

    Assume $\displaystyle n$ is rational.

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

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

    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 $\displaystyle \sqrt{a}+\sqrt{b}$, where $\displaystyle a, b \in \mathbb{N}$ and $\displaystyle \sqrt{a}$ and $\displaystyle \sqrt{b}$ are individually irrational. Just put $\displaystyle \sqrt{a}+\sqrt{b}=n$, multiply both sides by $\displaystyle \sqrt{a}-\sqrt{b}$, and continue as before.

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

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

    Then:

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

    $\displaystyle \implies p^2 = 6q^2$

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

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

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

    So $\displaystyle p^2=4m^2=6q^2$

    $\displaystyle \implies 3q^2=2m^2$

    $\displaystyle \implies q^2$ has a factor $\displaystyle 2$

    $\displaystyle \implies q$ has a factor $\displaystyle 2$

    $\displaystyle \implies p$ and $\displaystyle q$ have a common factor $\displaystyle 2$.

    Contradiction.

    $\displaystyle \implies \sqrt{6}$ is irrational.

    Grandad
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