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Math Help - Prrove common log of 3 is irrational

  1. #1
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    Prrove common log of 3 is irrational

    This is what I've done so far:

    Suppose it is rational, then log3 = a/b where a and b are in lowest terms. Hence 10^(a/b) = 3 which implies 10^a = 3^b which implies 2^a * 5^a = 3^b. This says that 2^a|3^b. Since 2|2^a it follows that 2|3^b which implies 2|3 because 2 is prime.

    So what do I do from here? Can I just say I have a contradiction and thus log3 is irrational?

    EDIT: Here's another one, thanks!

    Prove that if r is a rational number such that r>1 and r /=/ 10^n for any positive integer n, then common log(r) is irrational.

    Suppose logr is rational, then logr = a/b (a,b in lowest terms). This implies 10^(a/b) = r which implies that 10^a = r^b which implies 5^a * 2^a = r^b. This says that 2|r and 5|r, thus the only prime factors of r are 2 and 5. Hence 2^x * 5^y = r which implies 2^xb * 2^yb = r^b. Thus xb=a and yb=a which says x=y. Thus 2^x * 5^x = r, which says 10^x = r, a contradiction. Thus logr is irrational.

    Is this right?
    Last edited by superevilcube; October 29th 2008 at 12:28 PM.
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  2. #2
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    Hello, superevilcube!

    I have a proof . . . I hope it's acceptable.


    Prove that \log_{10}(3) is irrational.

    Assume \log_{10}(3) is rational.

    Then: . \log_{10}(3)  = \frac{a}{b}\;\;\text{ where }a\text{ and }b\text{ are positive integers}

    And we have: . 10^{\frac{a}{b}} \:=\:3

    Raise both sides to the power b\!:\;\;10^a \:=\:3^b


    Since . \begin{Bmatrix}10^a\text{ always ends in 0} \\3^b\text{ ends in 1, 3, 7, or 9}\end{Bmatrix}\quad\text{then: }\:10^a \neq 3^b \quad\hdots\;\text{a contradiction}


    Therefore, \log_{10}(3) is irrational.

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  3. #3
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    Quote Originally Posted by superevilcube View Post
    10^a = 3^b which implies 2^a * 5^a = 3^b.
    This is not true in general. 10^a is not equal to 2^a * 5^a for a>1.
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  4. #4
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    Quote Originally Posted by frankdent1 View Post
    This is not true in general. 10^a is not equal to 2^a * 5^a for a>1.
    How disappointing. To discover that the formula (a \, b)^n = a^n \, b^n is wrong. After all these years of me believing in it. *sigh* Consigned to the same bin of broken dreams as Father Christmas and the Easter Bunny ...... I just can't believe it ...... Either 10^2 \neq 2^2 \, 5^2 or 2 < 1 ..... I feel so ..... foolish
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  5. #5
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    Quote Originally Posted by mr fantastic
    Consigned to the same bin of broken dreams as ... the Easter Bunny
    Hey ... what's that about the Easter Bunny?

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