Results 1 to 8 of 8

Math Help - prove that root 6 is irrational

  1. #1
    Junior Member
    Joined
    Jul 2009
    Posts
    68

    prove that root 6 is irrational

    plzz prove that root 6 is irrational.

    i have to show the whole prove not just say that 6 is not a perfect square so root 6 is irrational. i have to prove it wholly.

    thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2010
    Posts
    422
    Suppose \sqrt{2\cdot3} = n/m. Then 2\cdot3\cdot m^2 = n^2. Factor both sides into primes. Then the power of 2 on the RHS is even because it is a square, but the power of 2 on the LHS is odd. This contradicts unique factorization.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    extremely sryy but i can't understand it properly plzz give me detail.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by saha.subham View Post
    plzz prove that root 6 is irrational.

    i have to show the whole prove not just say that 6 is not a perfect square so root 6 is irrational. i have to prove it wholly.

    thanks in advance
    Or note that \sqrt{6} is a solution to x^2-6=0 and the only rational solutions to that are...

    P.S. It follows that \sqrt{2}+\sqrt{3}\notin\mathbb{Q}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    srryy but i can;t understand without details
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,429
    Thanks
    1857
    But I thought you said you had prove it. What do you understand about proving that roots are not rational?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2009
    From
    Melbourne
    Posts
    274
    Thanks
    4
    haha! If you square root anything but a perfect square then the answer will be irrational!

    PROVEN!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,831
    Thanks
    1602
    Quote Originally Posted by jgv115 View Post
    haha! If you square root anything but a perfect square then the answer will be irrational!

    PROVEN!!
    I have a feeling that if the OP is going to say that then that statement has to be proven in the process.


    Assume that \sqrt{6} is rational.

    Then \sqrt{6} = \frac{a}{b}, where a, b are integers, and \frac{a}{b} is in simplest form (since every rational number can be reduced to simplest form).


    Therefore 6 = \frac{a^2}{b^2}

    6b^2 = a^2

    2\cdot 3b^2 = a^2.


    Since a^2 is even, that means a is even. So write it as a = 2c. Then you have

    2\cdot 3b^2 = (2c)^2

    2\cdot 3b^2 = 4c^2

    3b^2 = 2c^2

    b^2 = 2\left(\frac{c^2}{3}\right).


    Remembering that b is an integer, so is b^2. Also, since 2 is a factor, b^2 is even, and so b is even. Write it as b = 2d.


    Remember that \sqrt{6} = \frac{a}{b}

    but a = 2c and b = 2d

    So \sqrt{6} = \frac{2c}{2d}

    \sqrt{6} = \frac{c}{d}.


    But we said that \frac{a}{b} was already in simplest form.

    So we have a contradiction.

    This means that \sqrt{6} can not possibly be rational.

    Therefore \sqrt{6} is irrational.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 11
    Last Post: October 25th 2009, 07:45 PM
  2. Replies: 2
    Last Post: May 26th 2009, 10:09 AM
  3. Replies: 7
    Last Post: January 29th 2009, 04:26 AM
  4. root three irrational
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 13th 2008, 12:23 PM
  5. How to prove square root 2 is irrational?
    Posted in the Math Topics Forum
    Replies: 8
    Last Post: June 24th 2007, 08:40 AM

Search Tags


/mathhelpforum @mathhelpforum