0

The exercise is

Let $a \geq 1$ and $b \geq 1$ be integers. Show that there exist integers $u \geq 1$ and $v \geq 1$ such that $u \mid a$ and $v \mid b$, $\gcd(u,v)=1$, and $\text{lcm}(u,v)=ab$.

We have $\gcd(u, v) =1$ so $u$ and $v$ are relatively prime, and $u \mid a$ and $v \mid b$ so $u$ and $v$ must be divisors of $a$ and $b$ with prime factors belonging exclusively to $a$ and $b$. From the fact that $\gcd(u,v) \text{lcm}(u,v) = uv$ it follows that $uv=ab$ so the product of two relatively prime numbers must be $ab$. But what if $a$ and $b$ share a factor greater than $1$?

For example, say that $a=3 \cdot 5 \cdot 7$ and $b= 5 \cdot 17 \cdot 19$. If we must find a $u$ and $v$ satisfying the conditions above. We have $uv = ab = 3 \cdot 5^2 \cdot 7 \cdot 17 \cdot 19 = 169575$. So to choose a $u$ and $v$ such that $u \mid a$, $v \mid b$ and $uv=ab$ we can simply let $u=a$ and $v=b$. But $\gcd(u,v) = \gcd(a,b) = 5$. That obviously won't work. Ok, what if we group the fives into one of them so $\gcd(u,v)=1$, let's say $u$? Then $u \nmid a$. And so on.

It seems like it's impossible to choose such a $u$ and $v$ unless $a$ and $b$ are relatively prime. Is this a mistake or am I doing this incorrectly?

iwjueph94rgytbhr
  • 932
  • 5
  • 15
  • 3
    Yeah, probably the theorem is supposed to be $\operatorname{lcm}(u,v)=\operatorname{lcm}(a,b).$ You are correct, it is not true if $\gcd(a,b)\neq 1.$ – Thomas Andrews Aug 06 '23 at 00:27
  • Where did you find this problem? – Brian Moehring Aug 06 '23 at 00:29
  • The formulation of the theorem is fine. There is no need to assume that a and b are relatively prime. – Wlod AA Aug 06 '23 at 00:33
  • 2
    @BrianMoehring exercise 34 of section 1.3 in "Introduction to Abstract Algebra" by W. Keith Nicholson. I've found a lot of errors in this book (I've only been reading it for a few days now), the author has even made some simply arithmetic errors. – iwjueph94rgytbhr Aug 06 '23 at 00:33
  • Yes, I read the exercise in a wrong way. I didn't see what was on the paper/screen; I couldn't help to see the true conclusion $$ u\cdot v = \text{lcm}(a\ b)$$. – Wlod AA Aug 06 '23 at 00:46
  • $ab = \color{#c00}{{\rm lcm}(u,v) = uv}\iff (a/u)(b/v)=1\iff u=a,, v = b,,$ by $(\color{#c00}{u,v)=1},$ so $,(a,b)=1$ – Bill Dubuque Aug 06 '23 at 00:54
  • Almost surely there is a typo and the correct conclusion should be $,{\rm lcm}(a,b) = uv,,$ which is a very common exercise, e.g. see the linked dupe (and its links), both in lcm form and dual gcd form. – Bill Dubuque Aug 06 '23 at 13:27
  • As such you should instead accept wlod's answer (or, better, delete the question since it is based on a typo so has little value). – Bill Dubuque Aug 06 '23 at 13:31

2 Answers2

1

Thomas Andrews' comment is correct, re: it's not true if $\gcd(a,b)\neq 1$. One relatively easy way to see this is, since we're dealing with only positive integers, then $u \mid a \;\to\; u \le a$ and $v \mid b \;\to\; v \le b$, so $uv \le ab$, with equality only if $u=a$ and $v=b$. However, then $\gcd(u,v)=1 \;\to\; \gcd(a,b) = 1$.

Thus, the current problem statement is incorrect. One possible way to fix it is to add the requirement that $\gcd(a,b)=1$. Then we can just choose $u=a$ and $v=b$ since $uv = ab$ in that case.

Another way to adjust the statement is to do as Thomas's comment suggested, i.e., change $\operatorname{lcm}(u,v)=ab$ to $\operatorname{lcm}(u,v)=\operatorname{lcm}(a,b)$ instead (or, since $\operatorname{lcm}(u,v)=uv$, we get the suggestion in Wlod AA's answer). Then with $p_i$, for $1 \le i \le n$, being the $n$ primes dividing $a$ and/or $b$, we get

$$a = \prod_{i=1}^{n}p_i^{q_i}, \;\; b = \prod_{i=1}^{n}p_i^{r_i}$$

We can have $u$ be the product of all $p_i^{q_i}$ where $q_i \ge r_i$, and $v$ be the product of all the $p_i^{r_i}$ where $r_i \gt q_i$. Then $\gcd(u,v)=1$, plus also

$$\operatorname{lcm}(u,v) = uv = \prod_{i=1}^{n}p_i^{\max(q_i,r_i)} = \operatorname{lcm}(a,b)$$

John Omielan
  • 52,653
1

The correct conclusion should be:

$$ u\cdot v\ =\ \text{lcm}(a\,\ b) $$

Thank you.

Wlod AA
  • 2,192