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Math Help - Prove no rational satisfies 2^x = 3

  1. #1
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    Unhappy Prove no rational satisfies 2^x = 3

    Here is the question and i have no idea how to go about it any help would be hugely appreciated.

    Prove no rational satisfies

    2^x = 3

    Many thanks
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    Quote Originally Posted by craig in a post he soft deleted and I hard deleted View Post
    You would firstly take logs of both sides, \log_{2} 2^x=\log_{2} 3.

    As \log_{y} y^x = x, we can say that:

    x=\log_{2} 3

    Put this into your calculator and you get x=1.584962501, not a rational number
    On the contrary, 1.584962501 certainly IS a rational number because it is a terminating decimal! log_2(3),which is NOTequal to 1.585962501 is not rational but I suspect it would be necessary to prove that to get credit for this problem!
    Last edited by mr fantastic; January 8th 2009 at 07:28 PM.
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    Quote Originally Posted by sebjory View Post
    Here is the question and i have no idea how to go about it any help would be hugely appreciated.

    Prove no rational satisfies

    2^x = 3

    Many thanks
    This is in the prealgebra/algebra section, so i am not sure if a proof like this (a proof by contradiction) will be too complicated. if it is, tell me, we can come up with another.

    Assume, to the contrary, that there is a rational number x such that 2^x = 3.

    Then x = \frac ab, where a and b are integers, with b \ne 0. So we have

    2^{\frac ab} = 3

    \Rightarrow 2^a = 3^b

    However, this is absurd, since a and b are integers, this equation contradicts the uniqueness of prime factorization (since the left side is always a product of 2's and the right is always a product of 3's).



    note that a = b = 0 makes the equation make sense... but b cannot be zero!
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    thankyou proof by contradiction is taught over here in the UK.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sebjory View Post
    thankyou proof by contradiction is taught over here in the UK.
    oh, you're from the UK, hehe, I was worried. you have a US flag up

    i suppose you are also familiar with the uniqueness of prime factorization theorem (you might know it under a different name), but it talks about how each integer can be uniquely expressed as a product of primes with integer powers
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    Yes, we're studying this whole semester on these sorts of theorems about rational numbers (the Completeness axiom etx) and with these come a whole host of proofs...

    Do you know any good proof resource where i can learn as many types of proof as possible? (not including Induction since i am fairly happy with my ability to manipulate these sorts of problems)
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    Quote Originally Posted by sebjory View Post
    Yes, we're studying this whole semester on these sorts of theorems about rational numbers (the Completeness axiom etx) and with these come a whole host of proofs...

    Do you know any good proof resource where i can learn as many types of proof as possible? (not including Induction since i am fairly happy with my ability to manipulate these sorts of problems)
    completeness axiom? ...this is prealgebra?!!

    i am sure there are resources online that have what you are looking for. i guess wikipedia is always a good place to start, at least for quick reference. but i have never really searched for any such site, as i have several textbooks that cover such material and am fairly happy with them, so i don't know. sorry
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    Quote Originally Posted by Jhevon View Post
    This is in the prealgebra/algebra section, so i am not sure if a proof like this (a proof by contradiction) will be too complicated. if it is, tell me, we can come up with another.

    Assume, to the contrary, that there is a rational number x such that 2^x = 3.

    Then x = \frac ab, where a and b are integers, with b \ne 0. So we have

    2^{\frac ab} = 3

    \Rightarrow 2^a = 3^b

    However, this is absurd, since a and b are integers, this equation contradicts the uniqueness of prime factorization (since the left side is always a product of 2's and the right is always a product of 3's).



    note that a = b = 0 makes the equation make sense... but b cannot be zero!
    Another argument could be

    2^a is always even and 3^b is always odd.

    Clearly, therefore, 2^a = 3^b where a and b are integers is nonsense, giving the contradiction and completing the proof.
    Last edited by Prove It; January 9th 2009 at 07:48 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Prove It View Post
    Another argument could be

    2^a is always even and 3^b is always odd.

    Clearly, therefore, 2^a = 3^b where a and b are integers is nonsense, giving the contradiction and completing the proof.
    even easier!
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