Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - Irrational

  1. #1
    Junior Member
    Joined
    Nov 2005
    Posts
    43

    Irrational

    Here's an interesting one:

    Prove that \sqrt{2} is irrational by induction
    Follow Math Help Forum on Facebook and Google+

  2. #2
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    Using induction on what?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    dud
    dud is offline
    Newbie
    Joined
    Jan 2006
    From
    Trondheim, Norway
    Posts
    20
    I don't get this, how can you use induction if all there is is 1 constant?
    Induction needs to have a variable n doesn't it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by TD!
    Using induction on what?
    The convergents of the continued fraction expansion of \sqrt2,
    possibly?

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by CaptainBlack
    The convergents of the continued fraction expansion of \sqrt2,
    possibly?

    RonL
    I don't have the time to do this in detail now, but further thought and
    sketches of a construction make me fairly sure it can be done using the
    continued fraction expansion of \sqrt2.

    The key idea is that a rational number has a terminating simple cf expansion,
    but we can show by induction that the simple cf expansion of \sqrt 2 does not
    terminate.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Interesting, I see what you are saying CaptainBlack,
    \sqrt{2}=[1;\=2] But how do you show it does not terminate?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker
    Interesting, I see what you are saying CaptainBlack,
    \sqrt{2}=[1;\=2] But how do you show it does not terminate?
    Take a finite part of the expansion and the non-integer r-term, extend the
    expansion by one term which will be a 2 and the r-term will remain the same.

    Hence by induction the cf expansion does not terminate.

    RonL
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2005
    Posts
    43
    Indeed, that is one way of inducing the proof.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Treadstone 71
    Indeed, that is one way of inducing the proof.
    Method #2:

    Let P_n be the proposition that:

    For all 1\le m \le n there is no k\epsilon \mathbb{Z}_+, such that \sqrt2 = \frac{k}{m}.


    This is clearly true for n=1 as \sqrt2 is not an integer.


    The induction step consists of showing that there is no k\epsilon \mathbb{Z}_+,
    such that \sqrt2 = \frac{k}{n+1}, which together with P_n will prove P_{n+1}.

    This last step I can do, but will not reproduce here.

    RonL
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Treadstone 71
    Indeed, that is one way of inducing the proof.
    Method 3

    I'm not going to do much on this other than suggest it might be made to work.

    Let P_n be something like:

    The decimal expansion of \sqrt2 does not have a peiodic
    tail of any period \le n.

    RonL
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Nov 2005
    Posts
    43
    Method 2 is what I have written down originally. It goes slightly differently than what you wrote.

    \sqrt{2}\neq 1

    Suppose \sqrt{2}\neq \frac{n}{m} for some general n,m\in \mathbb{Z} Details are omitted here, but it can be shown that \sqrt{2}\neq\frac{n+1}{m} and \sqrt{2}\neq\frac{n}{m+1} By induction, it does not equal to any rational number.
    Last edited by Treadstone 71; January 18th 2006 at 11:15 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Nov 2005
    Posts
    43
    Quote Originally Posted by CaptainBlack
    Let P_n be something like:

    The decimal expansion of \sqrt2 does not have a peiodic
    tail of any period \le n.

    RonL
    This is the very fact that you have to prove, because it is equivalent to the definition of an irrational number. You will actually have to compute the decimals of root 2, which comes down to something similar to what you did with the continued fractions.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Treadstone 71
    This is the very fact that you have to prove, because it is equivalent to the definition of an irrational number.
    But aren't they all.

    You will actually have to compute the decimals of root 2, which comes down to something similar to what you did with the continued fractions.
    I was thinking of doing something slightly different. I had no intention of
    computing the decimal expansion of \sqrt 2, but supposing
    that the tail were periodic of period n, then writing the sum of
    the tail in terms of the replicand (if that is the right word) which we don't
    know, and the sum of the geometric series:

    \sum_1^{\infty} \frac{1}{10^{n+1}},

    and show in some way not yet determined that this lead to a contradiction.

    RonL
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Speaking about continued fractions, can someone prove Euler's interesting fraction:
    e=[2;1,2,1,1,4,1,1,6,1,1,8,...]
    and
    \frac{e-1}{e+1}=[0;2,6,10,14...]
    one more,
    \frac{e^2-1}{e^2+1}=[0;1,3,5,7,...]
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Junior Member
    Joined
    Jan 2006
    Posts
    56

    could you please explain your notation i dont undestand them

    Quote Originally Posted by ThePerfectHacker
    Speaking about continued fractions, can someone prove Euler's interesting fraction:
    e=[2;1,2,1,1,4,1,1,6,1,1,8,...]
    and
    \frac{e-1}{e+1}=[0;2,6,10,14...]
    one more,
    \frac{e^2-1}{e^2+1}=[0;1,3,5,7,...]
    I dont understand your notation mister Hacker please could you explain?

    In fact i have been teached a very basic and simple method to proove that any integer that is not a square have an irational square root. I am in doubt that any steps of the tree induction method, not developed and using weird notation for some of them (this is the math puzzle section not the advanced question section), so any steps would not use the same or a more complicated argument that the basic method

    People that dont know it are invited to search in <<square root>>thread !!!
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 19th 2010, 05:04 PM
  2. example of irrational + irrational = rational
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 4th 2010, 03:44 AM
  3. Replies: 1
    Last Post: March 23rd 2010, 12:55 PM
  4. Replies: 2
    Last Post: January 31st 2010, 05:40 AM
  5. Replies: 7
    Last Post: January 29th 2009, 03:26 AM

Search Tags


/mathhelpforum @mathhelpforum