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Math Help - irrational exponents

  1. #1
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    irrational exponents

    hi,

    i am trying to prove that the property x<y implies a^x<a^y for any real numbers x and y. i am having a little trouble using two sequences of rational numbers converging to x and y respectively. using some analysis and the theory of convergent sequences i can show that a^x<=a^y but i cannot show STRICT inequality. could someone help me with this? thanks!
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  2. #2
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    Quote Originally Posted by kkoutsothodoros View Post
    hi,

    i am trying to prove that the property x<y implies a^x<a^y for any real numbers x and y. i am having a little trouble using two sequences of rational numbers converging to x and y respectively. using some analysis and the theory of convergent sequences i can show that a^x<=a^y but i cannot show STRICT inequality. could someone help me with this? thanks!
    I have an easy proof but I just do not know how you defined exponents. The way I define exponents (let a\in \mathbb{R}^+) is by a^x = e^{x\log a}. (The exponential function is defined to be the inverse of the logarithm function, and the logarithm function is defined through an integral). Now if a>1 then \log a >0. Therefore, if x>y \implies x\log a > y\log a which means e^{x\log a} > e^{y\log a} since the regular exponential is an increasing function and so by definition a^x > a^y.
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  3. #3
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    hi, thanks! i am familiar with that definition of exponents and it does make things much easier to deal with.

    the definition i am using for a^x is the limit of a^x_n where x_n is a sequence of rational numbers. i am using the fact that every real number is the limit of a sequence of rational numbers.

    thanks!
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  4. #4
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    Quote Originally Posted by kkoutsothodoros View Post
    the definition i am using for a^x is the limit of a^x_n where x_n is a sequence of rational numbers. i am using the fact that every real number is the limit of a sequence of rational numbers.
    To prove the strict inequality using that (horrible!) definition, let s and t be rational numbers with x<s<t<y. You then know that a^s<a^t, from properties of rational numbers.

    If (x_n),\ (y_n) are sequences of rational numbers with limits x, y respectively, then for all sufficiently large n it will be true that x_n<s and t<y_n. Assuming you can prove that this implies a^x\leqslant a^s and a^t\leqslant a^y, it then follows that a^x\leqslant a^s<a^t\leqslant a^y.
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  5. #5
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    thanks!

    and yes it is a horrible way to define exponents!
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