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Rational numbers and Lowest terms proof:

  1. Sep 13, 2010 #1

    silvermane

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    I've been recently reading a book on abstract algebra and number theory, and I stumbled upon a problem that at first glance looked obvious, but I can't seem to figure out how to formally write the proof.

    1.)So, let's say we have 4 integers, r,s,t,u, all greater than or equal to 1. Suppose [tex]\frac{r}{s}[/tex] = [tex]\frac{t}{u}[/tex] where both fractions are in lowest terms. Prove that r=t and s=u.

    For this problem, I was thinking of solving it via contradiction but I can't seem to get there. I was thinking of using the fact that the gcd(r,s) = 1, and if the gcd(t,u)=1 that there would be a contradiction if they were not equal, but I feel like that's not enough. Any tips or hints would be greatly appreciated.

    2.) Now, suppose we have r and s again (lowest terms), and we look at [tex]\frac{r}{s}[/tex]. Prove that an integer N cannot equal [tex]\frac{r}{s}[/tex] unless s = 1.

    I was thinking of saying that we can write r as a product of primes:
    r=[tex]p_{1}[/tex]*[tex]p_{2}[/tex]*...*[tex]p_{k}[/tex]​

    and then writing s as a product of primes, but primes that are all different from r's:
    s=[tex]q_{1}[/tex]*[tex]q_{2}[/tex]*...*[tex]q_{k}[/tex]​

    Obviously through some algebraic manipulation, we see that [tex]\frac{r}{s}[/tex] is also in lowest terms, and thus can't be an integer unless s=1, but I feel like I need more detail in this part of my proof.

    Once again, thank you all in advance for your help and advice :)
     
  2. jcsd
  3. Sep 13, 2010 #2
    You're on the right track.

    Let [tex] r = p_1 * p_2 * ... * p_a. [/tex]

    Let [tex] s = q_1 * q_2 * ... * q_b. [/tex]

    Let [tex] t = p'_1 * p'_2 * ... * p'_c. [/tex]

    Let [tex] u = q'_1 * q'_2 * ... * q'_d. [/tex]

    Cross multiply to get

    [tex] (p_1 * p_2 * ... * p_k)(q'_1 * q'_2 * ... * q'_k) = (p'_1 * p'_2 * ... * p'_k)(q_1 * q_2 * ... * q_k) .[/tex]

    We know that [tex] p_i \neq q_k [/tex] and [tex] p'_i \neq q'_k \ [/tex] for all i and k.

    Can you figure out the rest?
     
  4. Sep 15, 2010 #3

    silvermane

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    Yes, very much so! You've been more than helpful. I thought of doing it that way, but must have been distracted somehow. Thank you for all your help! :)
     
  5. Sep 15, 2010 #4
    For the second one is it not enough to prove by contradiction?

    Like suppose [tex]\frac{r}{s}=N[/tex] then s divides r which leads to a contradiction ,unless s=1, given our choice of s and r namely gcd(r,s)=1.

    Is this what you are doing ?
     
  6. Sep 17, 2010 #5

    silvermane

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    Yes that's exactly what I've done. I just needed to know that I had to break them all into a product of irreducible primes to better understand what was going on. Thank you for your help :)
     
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