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irrational- proof help watch

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    prove that if

    2^x = 5 , then x is irrational .


    Need a little help

    thnx
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    Assume that x is rational, so can be written in the form a/b where a/b are coprime integers. Follow that to lead to a contradiction, showing that your assumption was false.
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    (Original post by dbmag9)
    Assume that x is rational, so can be written in the form a/b where a/b are coprime integers. Follow that to lead to a contradiction, showing that your assumption was false.

    x= a/b

    x^2 = a^2/b^2 = 5

    where to go from that
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    (Original post by A.braham)
    x= a/b

    x^2 = a^2/b^2 = 5

    where to go from that
    Is your question x^2 or 2^x?

    For x^2, what you've written is a good start. Then rearrange and think about the factors of the numbers on each side. Notice that the starting assumption involved the numbers being coprime (sharing no factors, so a/b is in its simplest form).
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    sorry it is 2^x

    i'm really stuck and don't know how to start solvin it

    any links ?

    help appreciated
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    Wiki Support Team
    (Original post by A.braham)
    sorry it is 2^x

    i'm really stuck and don't know how to start solvin it

    any links ?

    help appreciated
    What are you expecting links to?

    Suppose that x is rational, say x = a/b, where a and b are positive integers, and that 2^x = 5. We want to use this to get a contradiction. Obviously the two above facts combine to give us 2^(a/b) = 5. Raise both sides to the power of b, notice that the numbers on both sides are positive whole numbers, and look at prime factors.
 
 
 
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Updated: January 31, 2010

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