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    How would you prove 3^1/3 > 2^1/2 ????
    thanks
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    cube both sides....

    3 > 23/2

    3> {23}1/2

    3>81/2

    can you finish it off ?*
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    It doesn't...

    3^1/3 = 2^1/2
    (3^1/3)^6 = (2^1/2)^6
    3^2 = 2^3
    9 = 8

    Alternatively you could just calculate the answers for each side of the equation and see that they're different... I dunno if that helps. :/
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    (Original post by l1002)
    It doesn't...

    3^1/3 = 2^1/2
    (3^1/3)^6 = (2^1/2)^6
    3^2 = 2^3
    9 = 8

    Alternatively you could just calculate the answers for each side of the equation and see that they're different... I dunno if that helps. :/
    Reread the original post : the OP hasn't suggested that they are equal.
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    Bloody hell the imaginary part of that graph
    http://www.wolframalpha.com/input/?i=x%5E(1%2Fx)

    Considering the derivative of x^(1/x) is a way to do the general case for comparing any integers other than x=2, and comparing to that becomes easy once you compare 2 to 4 and show the inequality in the question
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    (Original post by mayjb)
    How would you prove 3^1/3 > 2^1/2 ????
    thanks
    Cube both sides, then square both sides. Get it into integers and it should be true. Directly proven.
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    (Original post by RDKGames)
    Cube both sides, then square both sides. Get it into integers and it should be true. Directly proven.
    With this method you have to note that they're both positive first as well, else implications don't work backwards
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    (Original post by RDKGames)
    Cube both sides, then square both sides. Get it into integers and it should be true. Directly proven.
    That's the wrong direction. You need to prove that 9 > 8 implies the inequality, not the inequality implies 9 > 8. (unless you use iff statements throughout - but you need to be very explicit about that, that is, explicitly justify that the sixth power is strictly increasing over the positive reals).
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    Instead of taking the power of six of both sides, as suggested above, a neat way of doing this without complicated monotonic arguments would be to note:
    \sqrt{2} < {36}/{25},
    {36}/{25} < \sqrt[3]{3},
    both of which can be verified by simple squaring or cubing. Then just order the inequalities, as required.
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    (Original post by Alex:)
    Instead of taking the power of six of both sides, as suggested above, a neat way of doing this without complicated monotonic arguments would be to note:
    \sqrt{2} < {36}/{25},
    {36}/{25} < \sqrt[3]{3},
    both of which can be verified by simple squaring or cubing. Then just order the inequalities, as required.
    still need the monotonic 'note sqrt(2) and 36/25 are both greater than 0 and that x maps one to one onto x^2 for positive reals'
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    (Original post by ValerieKR)
    still need the monotonic 'note sqrt(2) and 36/25 are both greater than 0 and that x maps one to one onto x^2 for positive reals'
    We don't require the monotonic property of the root function at all.

    For instance, the inequality
    \sqrt{2} < 36/25,
    can be verified by simply taking the square of both sides,
    2 < (36/25)^2.

    We're saying that the square root of 2 is the number that when multiplied by itself equals 2. All we actually require is the power law for inequalities
    x < y \Leftrightarrow x^n < y^n
    for n \in \mathbb{N} and x,y positive. And if you want to get slightly deeper into this, then the square root has to exist, because of the completeness of the real numbers.
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    (Original post by Alex:)
    For instance, the inequality
    \sqrt{2} < 36/25,
    can be verified by simply taking the square of both sides,
    2 < (36/25)^2.
    for instance the inequality

    -3>2
    can be verified by simply taking the square of both sides
    9>4

    you've made x, y positive at the end and that accompanied with the power law is just a generalised to x^n version of the monotonic function, you've used a different language to say the same thing
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    (Original post by Alex:)
    Unfortunately, -3 isn't a positive number.
    unfortunately, the power law is a more specific version of the monotonic property, and stating they have to be positive is part of the baggage i was referring to
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    (Original post by ValerieKR)
    unfortunately, the power law is a more specific version of the monotonic property
    Exactly my point. It's a more elementary property than monotonicity of the root function. Positivity hasn't got anything to with monotonicity.
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    (Original post by Alex:)
    Exactly my point. It's a more elementary property than monotonicity of the root function.
    it is less general and therefore less elementary, the monotonic property includes the power law but the power law does not include the monotonic property
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    (Original post by ValerieKR)
    it is less general and therefore less elementary, the monotonic property includes the power law but the power law does not include the monotonic property
    That's not how it works. The power law is a much more elementary tool than monotonicity. Typically, the more general results are less elementary than the specific results.

    Also, the point still stands that you don't need the monotonic property.
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    (Original post by Alex:)
    That's not how it works. The power law is a much more elementary tool than monotonicity. Typically, the more general results are less elementary than the specific results.

    Also, the point still stands that you don't need the monotonic property.
    That point only stands when the power law is more elementary than the monotonic property - which again, I'd dispute.

    But elementary has a lot of meanings and this is now just semantics.
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    (Original post by ValerieKR)
    That point only stands when the power law is more elementary than the monotonic property - which again, I'd dispute.

    But elementary has a lot of meanings and this is now just semantics.
    Fair enough, I can agree with this.
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    Great stuff


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    (Original post by physicsmaths)
    Great stuff


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    thanks for the help!
 
 
 
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