Results 1 to 3 of 3

Thread: More Parsonson - Rational Numbers

  1. #1
    Member
    Joined
    Feb 2011
    Posts
    86
    Thanks
    1

    More Parsonson - Rational Numbers

    Hi all. I'm trying to complete Ex. 11 (photos of the book pages below).
    .would someone be able to give me a hint on how to approach this?

    I have tried replacing (x,1) with (b/a,1) in the given expression to show the answer gives (a.b/a,1) and therefore the expression holds.

    Not sure if this is satisfactory?




    Sent from my iPhone using Tapatalk
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    20,249
    Thanks
    3366

    Re: More Parsonson - Rational Numbers

    When you say "I have tried replacing (x,1) with (b/a,1)" you are completely misunderstanding what is being said!

    The whole point is that a rational number (which might be represented as b/a) can be written as the pair of integers (b, a). You can't replace x by b/a- x must be an integer. The definition of multiplication is given by $\displaystyle (p_1, q_1)\times (p_2, q_2)= (p_1p_2, q_1q_2)$ so that $\displaystyle (a, 1)\times (x, 1)= (ax, 1)$ so saying that $\displaystyle (x, 1)\times (a, 1)= (b, 1)$ means immediately that b= ax.

    But then the definition of "equal" pairs, that $\displaystyle (a_1, b_1)= (a_2, b_2)$ means $\displaystyle a_1b_2= a_2b_1$, (b, a)= (x, 1) is the same as saying that b(1)= b= a(x), exactly what we had.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2011
    Posts
    86
    Thanks
    1

    Re: More Parsonson - Rational Numbers

    Quote Originally Posted by HallsofIvy View Post
    When you say "I have tried replacing (x,1) with (b/a,1)" you are completely misunderstanding what is being said!

    The whole point is that a rational number (which might be represented as b/a) can be written as the pair of integers (b, a). You can't replace x by b/a- x must be an integer. The definition of multiplication is given by $\displaystyle (p_1, q_1)\times (p_2, q_2)= (p_1p_2, q_1q_2)$ so that $\displaystyle (a, 1)\times (x, 1)= (ax, 1)$ so saying that $\displaystyle (x, 1)\times (a, 1)= (b, 1)$ means immediately that b= ax.

    But then the definition of "equal" pairs, that $\displaystyle (a_1, b_1)= (a_2, b_2)$ means $\displaystyle a_1b_2= a_2b_1$, (b, a)= (x, 1) is the same as saying that b(1)= b= a(x), exactly what we had.
    Oh dear. I'm not very good at this. I understand your explanation though. Thanks HallsofIvy.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: Feb 2nd 2019, 10:16 AM
  2. Rational numbers
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Feb 22nd 2014, 06:23 AM
  3. Replies: 2
    Last Post: Oct 4th 2012, 10:15 PM
  4. Proof: All rational numbers are algebraic numbers
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Sep 5th 2010, 10:26 AM
  5. Replies: 8
    Last Post: Sep 15th 2008, 04:33 PM

/mathhelpforum @mathhelpforum