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Math Help - Intro to Real Analysis: Question on a completed proof.

  1. #1
    Member ilikedmath's Avatar
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    Exclamation Intro to Real Analysis: Question on a completed proof.

    We had to prove this:
    Let and be sequences of real numbers. If is bounded and = 0,
    then = 0.
    (The fact that is bounded is crucial.)

    I found this:




    My question is where the red arrow is pointing and the red boxed part:
    How do I know how to choose 1/M for ε?
    In this case we want to prove | a_n b_n| < ε
    so I need to choose any ε value right? How do I know to choose 1/M?
    Is there algebra I can do that will lead me to choosing 1/M?
    I think I understand the rest of the proof up to that point where the red arrow is pointing to.

    Any help, suggestions are appreciated. Thanks for your time!
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  2. #2
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    Quote Originally Posted by ilikedmath View Post
    How do I know how to choose 1/M for ε?
    In this case we want to prove | a_n b_n| < ε
    so I need to choose any ε value right? How do I know to choose 1/M?
    Is there algebra I can do that will lead me to choosing 1/M?
    You are to prove that given any \varepsilon >0 it is possible to insure that <br />
\left| {a_n b_n  - 0} \right| < \varepsilon .
    So from the fact that \left( {a_n } \right) \to 0 you use the definition of sequence convergence.
    But in place of just \varepsilon >0 you are free to use \frac{\varepsilon}{M} >0.

    \left| {a_n b_n  - 0} \right| = \left| {a_n b_n } \right| = \left| {a_n } \right|\left| {b_n } \right| \leqslant \left| {a_n } \right|M < \frac{\varepsilon }{M}M = \varepsilon
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  3. #3
    Member ilikedmath's Avatar
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    Quote Originally Posted by Plato View Post
    You are to prove that given any \varepsilon >0 it is possible to insure that <br />
\left| {a_n b_n  - 0} \right| < \varepsilon .
    So from the fact that \left( {a_n } \right) \to 0 you use the definition of sequence convergence.
    But in place of just \varepsilon >0 you are free to use \frac{\varepsilon}{M} >0.

    \left| {a_n b_n  - 0} \right| = \left| {a_n b_n } \right| = \left| {a_n } \right|\left| {b_n } \right| \leqslant \left| {a_n } \right|M < \frac{\varepsilon }{M}M = \varepsilon
    So basically all they did was work with the inequality \varepsilon >0 and divided both sides by M to get the epsilon they wanted? It was that 'simple'?
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    Quote Originally Posted by ilikedmath View Post
    So basically all they did was work with the inequality \varepsilon >0 and divided both sides by M to get the epsilon they wanted? It was that 'simple'?
    NO that is not what happened.
    \frac{\varepsilon}{M} > 0 was used in the definition of sequence convergence as the positive bound.
    There are no "sides" here to be divided.
    It is all a matter of properly using definitions.
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  5. #5
    Member ilikedmath's Avatar
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    Quote Originally Posted by Plato View Post
    It is all a matter of properly using definitions.
    Ok, I see now. Thanks for clarifying.
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