1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Rational numbers and Lowest terms proof:

  1. Sep 13, 2010 #1


    User Avatar
    Gold Member

    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:

    and then writing s as a product of primes, but primes that are all different from r's:

    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


    User Avatar
    Gold Member

    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


    User Avatar
    Gold Member

    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 :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook