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Proof: there is M s.t for all r>M, 1/2r < 1/100

  1. Jul 28, 2012 #1
    "Prove that there exists an odd integer M such that for all real numbers r larger than
    M, [itex]\frac{1}{2r} < \frac{1}{100}[/itex]"

    How to express this using logical symbols?

    Literally I understand this statement as:
    [itex](\exists M\in [/itex][PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png[itex])[([/itex]M [Broken] is odd[itex])\wedge(\forall r\inℝ)(r>M\wedge\frac{1}{2r} < \frac{1}{100})][/itex]

    But all r in ℝ cannot be greater than some M. There will always be r<=M if the only restriction on r is that it is in ℝ. So this statement can't be proven true. But would it be wrong to interpret the statement as follows:

    [itex](\exists M\in [/itex][PLAIN]http://upload.wikimedia.org/wikipedia/en/math/3/f/3/3f3c78f02a9c53f5460f4bcc2e7dd3cb.png[itex])[([/itex]M [Broken] is odd[itex])\wedge(\forall r\inℝ)(r>M\Rightarrow\frac{1}{2r} < \frac{1}{100})][/itex]

    (the last "and" changed to "imples")....then M=51 would prove the statement true...
     
    Last edited by a moderator: May 6, 2017
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  3. Jul 28, 2012 #2

    Simon Bridge

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    [tex]\forall r \in \mathbb{R} > M \; ; \; \exists M = 2n-1 \; \wedge \; n \in \mathbb{Z} \; : \; \frac{1}{2r} < \frac{1}{100}[/tex]

    but that's not the only restriction on r. You wrote:

    "Prove that there exists an odd integer M such that for all real numbers r larger than
    M
    ..."

    It's asserting that we can always find some odd integer M which makes any real number bigger than M satisfy the relation 1/2r < 1/100.

    My only concern is that this looks a tad on the trivial side. Is the exercise just in writing and interpreting the symbols?
     
    Last edited by a moderator: May 6, 2017
  4. Jul 28, 2012 #3
    Ohh true true...I didn't know you could write something like [tex]\forall r \in \mathbb{R} > M[/tex]...is this a common notation? I mean, would a professor accept it? I am preparing for my transition to advanced math course this fall so this is a concern for me...my book always only uses one restriction in front of the "for all" or "there exists" symbol, so I thought that any additional restriction could be connected with an "and" to the rest of the statement...so to negate [itex]\forall r \in \mathbb{R} > M[/itex], it would just be:
    [itex]\exists r \in \mathbb{R} > M[/itex] , right?

    So the following statement
    "For every positive real number x, there is a positive real number y less
    than x with the property that for all positive real numbers z, yz ≥ z."
    can be expressed as:

    [itex](\forall x\in\mathbb{R^+})(\exists y\in\mathbb{R^+}<x)(\forall z\in\mathbb{R^+})(yz≥z)[/itex]
    right?
    and yea these proofs themselves are meant to be trivial, it's just about using basic proof techniques and knowing how to express statements using logical symbols
     
  5. Jul 29, 2012 #4

    Simon Bridge

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    OK since it is the point of this section, you'd better stick to being pedantic for now. I've never been penalized for that sort of shorthand and, if you skim a bunch of journals you'll find that convention gets abandoned in favor or readability.

    Sadly, what gets emphasized seems to depend on the school.

    What I wanted to show you was that you can make more sense out of things by working backwards.
     
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