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Thread: proof rational + irrational

  1. #1
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    proof rational + irrational

    How do I prove a rational +an irrational number = an irrational number?
    thanks
    Barry
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tinhats View Post
    How do I prove a rational +an irrational number = an irrational number?
    thanks
    Barry
    It is quite simple - let $\displaystyle n \in \mathbb{R} \setminus \mathbb{Q}, p,q,r,s \in \mathbb{Z}$. Then, if $\displaystyle n+p/q=r/s$ we can quite easily manipulate the equality to get $\displaystyle n=x/y, x,y \in \mathbb{Z}$, a contradiction. I shall, however, leave the manipulation up to you.
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  3. #3
    Moo
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    Hello,
    Quote Originally Posted by tinhats View Post
    How do I prove a rational +an irrational number = an irrational number?
    thanks
    Barry
    Here is a similar thing : http://www.mathhelpforum.com/math-he...rrational.html .

    Put $\displaystyle x=1$, that is to say $\displaystyle x'=x''$ and you're done...
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  4. #4
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    here's another example of 2 similar proofs:
    http://www.mathhelpforum.com/math-he...bers-even.html

    Halfway down the page is a proof that an uneven * even number is an even number.
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  5. #5
    Moo
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    Quote Originally Posted by bmp05 View Post
    here's another example of 2 similar proofs:
    http://www.mathhelpforum.com/math-he...bers-even.html

    Halfway down the page is a proof that an uneven * even number is an even number.
    Hmm but yours is dealing with even/uneven, while we're talking about rational/irrational here ?
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  6. #6
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    Yes, well, I feel that any retort would probably be irrational at this stage.
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  7. #7
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Moo View Post
    Hmm but yours is dealing with even/uneven, while we're talking about rational/irrational here ?
    They may not be the same but they are similar.

    In general:

    Let $\displaystyle G$ be a group and $\displaystyle H$ be a subgroup. Then for any $\displaystyle x,y\in G,$ $\displaystyle x\in H$ and $\displaystyle y\notin H$ $\displaystyle \implies$ $\displaystyle xy\notin H.$

    Proof is simple: Suppose $\displaystyle x$ (and therefore $\displaystyle x^{-1})\,\in\,H.$ If $\displaystyle xy\in H,$ then $\displaystyle y=x^{-1}(xy)\in H.$ Therefore by contrapositivity, $\displaystyle y\notin H\ \implies\ xy\notin H.$

    In bmp05’s original example, $\displaystyle G$ is the addibive group of the reals and $\displaystyle H$ is the additive group of the rationals; hence rational + irrational = irrational. In the odd/even example, $\displaystyle G$ is the additive group of the integers and $\displaystyle H$ is the subgroup of the even integers – $\displaystyle \therefore$ even + odd = odd.

    Further examples:

    (i) Even permutation odd permutation = odd permutation. (Take $\displaystyle G$ to be a symmetric group, $\displaystyle H$ to be the corresponding alternating group.)

    (ii) Rotation reflection = reflection. (Take $\displaystyle G$ to be a dihedral group, $\displaystyle H$ to be the subgroup of rotations.)
    Last edited by TheAbstractionist; May 25th 2009 at 06:53 AM.
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