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Math Help - Question on boundedness

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    Question on boundedness

    Let  a\in\mathbb{R} and consider the set S_a=\{a^n:n\in\mathbb{N},n>0\}. Prove carefully that a is not bounded above when a>1.

    My first thoughts are to use a contradiction, by using the definition of supremum, but I can't seem to find one. Could someone help?
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    MHF Contributor Drexel28's Avatar
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    Re: Question on boundedness

    Quote Originally Posted by worc3247 View Post
    Let  a\in\mathbb{R} and consider the set S_a=\{a^n:n\in\mathbb{N},n>0\}. Prove carefully that a is not bounded above when a>1.

    My first thoughts are to use a contradiction, by using the definition of supremum, but I can't seem to find one. Could someone help?
    Yeah, if you want to do it that way. Maybe say something like if it were bounded above it would have a supremum \alpha but by definition of supremum and the fact that a>1 we have that there is some m\in\mathbb{N} for which \displaystyle \frac{\alpha}{a}<a^m and so \alpha< a^{m+1} contradiction.
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    Re: Question on boundedness

    just out of curiosity: how do we know there is such an m? because, to me, this seems like a shell game: we are implcitly assuming that we can make a^m bigger than any real number (in particular, α/a), which is to say, {a^n} is unbounded (circular reasoning).

    i mean, it's clear {a^n} is increasing, since a > 1, but increasing without bound? i don't see it in the proof (that is to say, how exactly do we choose m?).
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    MHF Contributor Drexel28's Avatar
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    Re: Question on boundedness

    Quote Originally Posted by Deveno View Post
    just out of curiosity: how do we know there is such an m? because, to me, this seems like a shell game: we are implcitly assuming that we can make a^m bigger than any real number (in particular, α/a), which is to say, {a^n} is unbounded (circular reasoning).

    i mean, it's clear {a^n} is increasing, since a > 1, but increasing without bound? i don't see it in the proof (that is to say, how exactly do we choose m?).
    Because by definition, since \displaystyle \frac{\alpha}{a}<\alpha it cannot be an upper bound for S_a and thus there must exist some element of S_a which is strictly bigger than \displaystyle \frac{\alpha}{a}--I just generically called that element a^m.
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    Re: Question on boundedness

    thanks. it is important (from a logical standpoint) that one note that α/a < α (this is what uses the condition that a > 1).

    this actually tells us "which" m, it is the first one for which a^m > α/a, that is, the proof becomes constructive.
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