Results 1 to 3 of 3

Math Help - Prove the sqrt(8) is not a rational number

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    3

    Prove the sqrt(8) is not a rational number

    This is the proof I used. Does anyone know if it is correct?

    Assume that the sqrt(8) is a rational number.

    sqrt(8) = n/m Assume no common factors between n and m

    8 = n^2/m^2

    8m^2 = n^2

    So n^2 must be an even number since anything muliplied by 8 is even.
    n must be even as well because even numbers squared are even, odd numbers squared are odd.

    Represent n as 4k
    n^2 = (4k)^2 --> 4^2k^2 --> 16k^2

    8m^2 = 16k^2
    m^2 = 2k^2
    So, m^2 is even
    So, m is even
    Therefore m and n have a common factor, it disproves my assumption
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello Bruins1
    Quote Originally Posted by Bruins1 View Post
    This is the proof I used. Does anyone know if it is correct?

    Assume that the sqrt(8) is a rational number.

    sqrt(8) = n/m Assume no common factors between n and m

    8 = n^2/m^2

    8m^2 = n^2

    So n^2 must be an even number since anything muliplied by 8 is even.
    n must be even as well because even numbers squared are even, odd numbers squared are odd.

    Represent n as 4k
    n^2 = (4k)^2 --> 4^2k^2 --> 16k^2

    8m^2 = 16k^2
    m^2 = 2k^2
    So, m^2 is even
    So, m is even
    Therefore m and n have a common factor, it disproves my assumption
    Perfect!

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by Bruins1 View Post
    This is the proof I used. Does anyone know if it is correct?

    Assume that the sqrt(8) is a rational number.

    sqrt(8) = n/m Assume no common factors between n and m

    8 = n^2/m^2

    8m^2 = n^2

    So n^2 must be an even number since anything muliplied by 8 is even.
    n must be even as well because even numbers squared are even, odd numbers squared are odd.

    Represent n as 4k
    n^2 = (4k)^2 --> 4^2k^2 --> 16k^2

    8m^2 = 16k^2
    m^2 = 2k^2
    So, m^2 is even
    So, m is even
    Therefore m and n have a common factor, it disproves my assumption
    8m^2 is even because 8m^2 = 2(4m^2).The proof is correct
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove that cos is a rational number
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 6th 2011, 12:22 AM
  2. Replies: 4
    Last Post: June 29th 2009, 06:18 PM
  3. Replies: 1
    Last Post: March 23rd 2009, 06:01 PM
  4. Prove no rational number whose square is 98
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 4th 2009, 01:11 PM
  5. prove sqrt(3) + sqrt (5) is an irrational number
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: October 6th 2006, 06:48 PM

Search Tags


/mathhelpforum @mathhelpforum