Results 1 to 12 of 12

Math Help - Rational and Irrational numbers

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    60

    Rational and Irrational numbers

    Now, don't think this is just an ordinary question! Here goes:

    a and b are irrational numbers
    a + b is rational
    a x b is rational

    Work out possible values for a and b where they are positive irrational numbers.



    I am sure that I have gotten down all the details, any help will be much appreciated, thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Geometor View Post
    Now, don't think this is just an ordinary question! Here goes:

    a and b are irrational numbers
    a + b is rational
    a x b is rational

    Work out possible values for a and b where they are positive irrational numbers.



    I am sure that I have gotten down all the details, any help will be much appreciated, thank you!

    a = (1 + sqrt(2))
    b = -sqrt(2)
    then a and b are irrational numbers and...

    a + b = (1 + sqrt(2)) + (-sqrt(2)) = 1 which is rational

    a = sqrt(2)
    b = 1/sqrt(2)
    then a and b are both irrational and...

    a*b = sqrt(2)*1/sqrt(2) = 1 which is rational

    i should be using this oppurtunity to try out Latex, i don't know how to use it

    EDIT: o sorry, i forgot a and b must be positive, so the first example is incorrect, i'll come up with another one

    how about
    a = 10 - sqrt(5) + sqrt(2)
    b = sqrt(5) - sqrt(2)

    then a + b = 10 which is rational and both a and b are positive irrational numbers
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2007
    Posts
    60
    Quote Originally Posted by Jhevon View Post
    a = (1 + sqrt(2))
    b = -sqrt(2)
    then a and b are irrational numbers and...

    a + b = (1 + sqrt(2)) + (-sqrt(2)) = 1 which is rational

    a = sqrt(2)
    b = 1/sqrt(2)
    then a and b are both irrational and...

    a*b = sqrt(2)*1/sqrt(2) = 1 which is rational

    i should be using this oppurtunity to try out Latex, i don't know how to use it

    EDIT: o sorry, i forgot a and b must be positive, so the first example is incorrect, i'll come up with another one

    how about
    a = 10 - sqrt(5) + sqrt(2)
    b = sqrt(5) - sqrt(2)

    then a + b = 10 which is rational and both a and b are positive irrational numbers
    Many thanks but I don't think that is correct since
    10 - sqrt(5) + sqrt(2) times sqrt(5) - sqrt(2) is irrational :S
    thanks for trying, lol this is a really complex problem ay?

    is this a trick question and it might be impossible?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Geometor View Post
    Now, don't think this is just an ordinary question! Here goes:

    a and b are irrational numbers
    a + b is rational
    a x b is rational

    Work out possible values for a and b where they are positive irrational numbers.
    a=1-\sqrt{2} \mbox{ and }b=1+\sqrt{2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2007
    Posts
    60
    Quote Originally Posted by ThePerfectHacker View Post
    a=1-\sqrt{2} \mbox{ and }b=1+\sqrt{2}
    erm... the question requires "positive" numbers and I believe that 1- sq. root 2 is negative?

    thanks for helping anyway
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Geometor View Post
    Many thanks but I don't think that is correct since
    10 - sqrt(5) + sqrt(2) times sqrt(5) - sqrt(2) is irrational :S
    thanks for trying, lol this is a really complex problem ay?

    is this a trick question and it might be impossible?
    o, we have to come up with a pair of numbers that satisfy both condtions at the same time? i thought we could use different examples for each.

    EDIT: TPH came up with an example
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Geometor View Post
    erm... the question requires "positive" numbers and I believe that 1- sq. root 2 is negative?

    thanks for helping anyway
    So then change it a little bit,

     a = 2 - \sqrt{2} \mbox{ and } b = 2 + \sqrt{2}

    What is so hard?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Geometor View Post
    Now, don't think this is just an ordinary question! Here goes:

    a and b are irrational numbers
    a + b is rational
    a x b is rational

    Work out possible values for a and b where they are positive irrational numbers.



    I am sure that I have gotten down all the details, any help will be much appreciated, thank you!
    Because their sum is rational a and b are of the form:

    a = A + x,

    b = B - x

    where A and B are rational and x is irrational.

    Then:

    <br />
a\ b = AB + (-A+B)x - x^2 = C<br />

    where C is rational.

    Now we have demanded that A, B and C be rational, but we may as well
    demand that they be integers (the derivation of rational solutions from
    integer solutions is fairly elementary and left to the reader).

    Anyway we have:

    <br />
x^2 + (A-B)x + (C- AB) = 0<br />

    and x is irrational. However:

    <br />
x=\frac{-(A+B) \pm \sqrt{(A-B)^2 - 4(C-AB)}}{2}=\frac{-(A+B) \pm \sqrt{(A+B)^2 - 4C}}{2}<br />

    which is irrational only if q= (A+B)^2 - 4C is not a perfect
    square.

    So here is our algorithm for finding solutions:

    Choose an integer K, and an integer C such that K^2 - 4C is not a perfect square,
    then choose A and B so that A+B=K, then:

    <br />
x=\frac{-(A+B) \pm \sqrt{(A+B)^2 - 4C}}{2}<br />

    is an irrational number such that if:

    a=A+x
    b=B-x

    then a and b are irrational and a+b is an integer as is a \times b

    Of course this does not guarantee that both a and b are positive.

    RonL
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    May 2007
    Posts
    60
    thanks for the help! sorry, can only thank one person a day I believe?
    I find this very confusing though :S
    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 Geometor View Post
    thanks for the help! sorry, can only thank one person a day I believe?
    I find this very confusing though :S
    No you can thank as many as you want.

    RonL
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Geometor View Post
    thanks for the help! sorry, can only thank one person a day I believe?
    how come?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    May 2007
    Posts
    60
    Quote Originally Posted by Jhevon View Post
    how come?
    oh woops, nevermind it's just that I can't thank on the same post so I got mixed up
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Irrational and Rational Numbers
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: September 25th 2011, 02:32 PM
  2. Rational and Irrational numbers
    Posted in the Algebra Forum
    Replies: 9
    Last Post: April 20th 2010, 03:00 PM
  3. Induction on irrational and rational numbers
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: February 2nd 2010, 03:34 AM
  4. Rational and irrational numbers
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 12th 2009, 02:42 AM
  5. Replies: 8
    Last Post: September 15th 2008, 04:33 PM

Search Tags


/mathhelpforum @mathhelpforum