Results 1 to 3 of 3

Math Help - Proving something irrational

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    17

    Proving something irrational

    This is the question I have:

    Show that if m is a natural number and is not the nth power of an integer, then m^(1/n) is irrational.

    I'm pretty stumped. What I wanted to do was suppose that m^1/n is rational and derive a contradiction that m is equal to some integer to the nth power (contradicting the hypothesis), but I'm not sure how to do that. I tried and soon got stuck.

    Suppose m^1/n is rational, then m^1/n = a/b for some a,b in the integers in lowest terms. Hence m = a^n/b^n which implies that mb^n = a^n. So m|a^n which means mi=a^n , for some integer i. But then this just takes me in a circle... that integer i is just b^n.

    I'm not looking for a full solution, just something that will drive me in the right direction. Thanks for the help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by superevilcube View Post
    This is the question I have:

    Show that if m is a natural number and is not the nth power of an integer, then m^(1/n) is irrational.

    I'm pretty stumped. What I wanted to do was suppose that m^1/n is rational and derive a contradiction that m is equal to some integer to the nth power (contradicting the hypothesis), but I'm not sure how to do that. I tried and soon got stuck.

    Suppose m^1/n is rational, then m^1/n = a/b for some a,b in the integers in lowest terms. Hence m = a^n/b^n which implies that mb^n = a^n. So m|a^n which means mi=a^n , for some integer i. But then this just takes me in a circle... that integer i is just b^n.

    I'm not looking for a full solution, just something that will drive me in the right direction. Thanks for the help.
    Suppose otherwise, that is suppose that m^{1/n} is rational, then there exist integers a , b such that \gcd(a,b)=1 and:

    m^{1/n}=\frac{a}{b}

    or:

    b^n m = a^n

    Now consider the prime factorisation of m and remember that a and b share no prime factors (in particular consider the multiplicity of the prime factors of m ).

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    44
    Quote Originally Posted by superevilcube View Post
    This is the question I have:

    What I wanted to do was suppose that m^1/n is rational
    and derive a contradiction
    Yes U were right.Let's go on.

    Then m^1/n = a/b, a,b is interger, b is not zero ,mean (a/b)^n=m,and m is interger >=0(Nature number), then a/b is interger, but first we said "m is not the nth power of an integer' so ,that is inconsistent.

    proved.


    PS: Think u should say n is positive interger first.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving prime is irrational
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 29th 2010, 09:38 AM
  2. Proving something is irrational
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 28th 2008, 04:21 PM
  3. proving roots irrational
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: October 23rd 2008, 10:50 AM
  4. proving √q is irrational by contradiction
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: October 18th 2008, 06:31 PM
  5. Proving irrational
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: September 25th 2008, 12:12 PM

Search Tags


/mathhelpforum @mathhelpforum