Results 1 to 10 of 10

Math Help - how is sq root of 3 can be proved to be irrational?

  1. #1
    rcs
    rcs is offline
    Senior Member rcs's Avatar
    Joined
    Jul 2010
    From
    iligan city. Philippines
    Posts
    455
    Thanks
    2

    how is sq root of 3 can be proved to be irrational?

    Can anybody help on proving sq. root of 3 is not rational?
    Indirect Method.


    thank you so much.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    May 2011
    Posts
    15

    Re: how is sq root of 3 can be proved to be irrational?

    If sqrt(3) was rational, then by definition it could be written as a/b, where a and and b are both integers. To prove indirectly that sqrt(3) is irrational, we must prove that a and b can not exist.

    Suppose: sqrt(3) = a/b, a and b are integers (it is rational), and a/b is in lowest terms

    1) Squaring both sides yields: 3 = a^2/b^2
    2) Multiplying both sides by b^3 yields: 3b^2 = a^2
    - Suppose b is even. Then b^2 is even and 3b^2 is also even. This means that a^2 and a must also be even. If both a and b are even numbers, then the ratio a/b can be simplified by cancelling a common factor of two, and contradicts our assumption that a/b is in lowest terms. For the sake of this proof, we must then assume that a and b are therefore both odd and that a/b is written in simplest terms.
    3) Since a and b are odd, we can write the equations:
    a = 2x+1
    b = 2y+1

    Where x and y are integers, thus restricting a and b to be any odd integer.
    Substituting these into 3b^2 = a^2, we get:

    3(2y+1)^2 = (2x+1)^2 (simplify)
    3(4y^2+4y+1) = 4x^2+4x+1 (distribute)
    12y^2+12y+3 = 4x^2+4x+1 (subtract 1)
    12y^2+12y+2 = 4x^2+4x (factor)
    2(6y^2+6y+1) = 2(2x^2+2x) (divide by 2)
    6y^2+6y+1 = 2x^2+2x (factor right side)
    6y^2+6y+1 = 2(x^2+1)

    Now, for any choice of "y", the left hand side will be odd. For any choice of "x", the right hand side will be even. Thus, there exist no solutions to this equation, and proof that sqrt(3) is irrational is evident by contradiction.

    Let me know if you have any questions!
    Last edited by rdtedm; July 30th 2011 at 07:08 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2010
    Posts
    193

    Re: how is sq root of 3 can be proved to be irrational?

    That seems to be a lot of work.

    I would start as you did, supposing \frac{a}{b}=\sqrt{3}, with \gcd (a,b)=1. Then, squaring both sides, \frac{a^2}{b^2}=3\Rightarrow a^2=3b^2.

    So 3\mid a^2. Since 3 is prime, this implies 3\mid a, from which we conclude that we actually have 9\mid a^2.

    Hence 9\mid 3b^2 as well. It follows that 3\mid b^2, and finally, 3\mid b.

    Thus, 3\mid \gcd (a,b), or 3\mid 1. This is, of course, a contradiction (well, in \mathbb{Z}).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2011
    Posts
    27

    Re: how is sq root of 3 can be proved to be irrational?

    Quote Originally Posted by topspin1617 View Post
    I would start as you did, supposing \frac{a}{b}=\sqrt{3}, with \gcd (a,b)=1. Then, squaring both sides, \frac{a^2}{b^2}=3\Rightarrow a^2=3b^2.
    Instead of what you do next, here is another way (I think nicer and a bit shorter) :

    a^2 = 3b^2 so b|a^2. Since a\wedge b=1, a^2\wedge b = 1 as well. Euclid's lemma give us so b | 1, that is b=1.

    As well, a^2 = 3b^2 says that a|3b^2 and so a|3 (with Euclid's lemma again). So a\in\{1,3\}. Both gives us a contradiction ( 1=3 or 9=3, which is the same by the way).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2010
    Posts
    193

    Re: how is sq root of 3 can be proved to be irrational?

    Quote Originally Posted by pece View Post
    Instead of what you do next, here is another way (I think nicer and a bit shorter) :

    a^2 = 3b^2 so b|a^2. Since a\wedge b=1, a^2\wedge b = 1 as well. Euclid's lemma give us so b | 1, that is b=1.

    As well, a^2 = 3b^2 says that a|3b^2 and so a|3 (with Euclid's lemma again). So a\in\{1,3\}. Both gives us a contradiction ( 1=3 or 9=3, which is the same by the way).
    I like it, never thought of doing it that way.

    Of course... can't we just say that ANY equation of the form a^2=nb^2, with n\in \mathbb{N}\setminus \{0\} not a square, is a contradiction simply by observing that this must give us some prime occurring to BOTH an odd and even power in the same factorization, contrary to the fact that \mathbb{Z} is a unique factorization domain? Err... contrary to the Fundamental Theorem of Arithmetic, I suppose it is?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jul 2011
    Posts
    27

    Re: how is sq root of 3 can be proved to be irrational?

    Yes, it's a way to do it, clearly.

    What's bothering me it's the use of the factorial structure of the ring \mathbb Z, when our two proofs above seem to only use its euclidian structure...

    (I said 'seem' because one might be have to use the factorial structure for something like 3|a^2 \Rightarrow 3|a [in your proof] or a\wedge b = 1 \Rightarrow a^2 \wedge b = 1 [in mine]. I will check that and then tell if my remark is relevant or not.)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1584
    Awards
    1

    Re: how is sq root of 3 can be proved to be irrational?

    Why not just prove this in general?
    If p is a positive integer which is not a square then \sqrt{p} is irrational.
    The proof by contradiction is easy. If K is the least positive integer having the property that K\sqrt{p}\in\mathbb{Z}^+ then K\sqrt p  - K\left\lfloor {\sqrt p } \right\rfloor is smaller positive integer also having that property.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2010
    Posts
    193

    Re: how is sq root of 3 can be proved to be irrational?

    Quote Originally Posted by pece View Post
    Yes, it's a way to do it, clearly.

    What's bothering me it's the use of the factorial structure of the ring \mathbb Z, when our two proofs above seem to only use its euclidian structure...

    (I said 'seem' because one might be have to use the factorial structure for something like 3|a^2 \Rightarrow 3|a [in your proof] or a\wedge b = 1 \Rightarrow a^2 \wedge b = 1 [in mine]. I will check that and then tell if my remark is relevant or not.)
    Both of those facts do come down to unique factorization. Anyway... isn't this a non-issue, as the Euclidean assumption is a stronger assumption anyway?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: how is sq root of 3 can be proved to be irrational?

    Quote Originally Posted by rcs View Post
    Can anybody help on proving sq. root of 3 is not rational?
    Indirect Method.


    thank you so much.

    Another method:


    Let us denoting by P(x) the polynomial x^2-3,


    P(x)=x^2-3, the only possible solutions of P(x) over \mathbb{Q} are: \pm{1},\pm{3}, and non of them is the solution of P(x).


    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jul 2011
    Posts
    27

    Re: how is sq root of 3 can be proved to be irrational?

    Quote Originally Posted by topspin1617 View Post
    isn't this a non-issue, as the Euclidean assumption is a stronger assumption anyway?
    In fact, I talk too fast without asking me that.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Square root Irrational
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 26th 2010, 02:51 AM
  2. prove that root 6 is irrational
    Posted in the Algebra Forum
    Replies: 7
    Last Post: April 14th 2010, 02:30 AM
  3. root three irrational
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 13th 2008, 11:23 AM
  4. irrational root of polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 13th 2008, 05:49 PM
  5. Proof of square root being irrational
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 14th 2007, 01:07 PM

Search Tags


/mathhelpforum @mathhelpforum