# Rational numbers and Lowest terms proof:

1. Sep 13, 2010

### silvermane

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 $$\frac{r}{s}$$ = $$\frac{t}{u}$$ 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 $$\frac{r}{s}$$. Prove that an integer N cannot equal $$\frac{r}{s}$$ unless s = 1.

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

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

Obviously through some algebraic manipulation, we see that $$\frac{r}{s}$$ 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. Sep 13, 2010

### Raskolnikov

You're on the right track.

Let $$r = p_1 * p_2 * ... * p_a.$$

Let $$s = q_1 * q_2 * ... * q_b.$$

Let $$t = p'_1 * p'_2 * ... * p'_c.$$

Let $$u = q'_1 * q'_2 * ... * q'_d.$$

Cross multiply to get

$$(p_1 * p_2 * ... * p_k)(q'_1 * q'_2 * ... * q'_k) = (p'_1 * p'_2 * ... * p'_k)(q_1 * q_2 * ... * q_k) .$$

We know that $$p_i \neq q_k$$ and $$p'_i \neq q'_k \$$ for all i and k.

Can you figure out the rest?

3. Sep 15, 2010

### silvermane

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! :)

4. Sep 15, 2010

### ╔(σ_σ)╝

For the second one is it not enough to prove by contradiction?

Like suppose $$\frac{r}{s}=N$$ 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 ?

5. Sep 17, 2010

### silvermane

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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