Prove that there exist divisors for a and b with certain conditions

Question

Show that if $a$ and $b$ belong to $\mathbb{Z}_+$ then there are divisors $c$ of $a$ and $d$ of $b$ with $(c,d)=1$ and $cd=\text{lcm}(a,b)$.

My try:

We know that $$\text{gcd}(a,b)\cdot \text{lcm}(a,b)=ab$$ then, $$\text{lcm}(a,b)=\dfrac{ab}{\text{gcd}(a,b)}$$ Let $d=b$ and $c=\frac{a}{\text{gcd}(a,b)}.$

Now $d|b$ and $c|a$ with $\text{lcm}(a,b)=cd$.

Now I am not sure about the following step: $$\text{gcd}(c,d)=\text{gcd}\left(\dfrac{a}{\text{gcd}(a,b)},b\right) =\text{gcd}\left(\dfrac{\text{lcm}(a,b)}{b},b\right)=1$$

Tell me is this correct?

Answer 1

It is not correct.

If $ a = 4, b = 2$, then you have $ d = b = 2, c = a/ \gcd(a,b) = 4/2 = 2$.
This doesn't satisfy $ \gcd(c,d) = 1$.

Now, figure out where your reasoning breaks down, and how to fix it.

Answer 2

Using prime factorization formula for a and b we have: a= $p_1$$^{a1}$ $p_2$$^{a2}$ ... $p_n$$^{an}$ b=$p_1$$^{b1}$ $p_2$$^{b2}$ ... $p_n$$^{bn}$
We know that gcd($p_i$$^s$,$p_k$$^t$)=1 for any i,k $\in$ [1,n] with i $\ne$ k and s,t $\in$ {a1,a2,...,an,b1,b2,...,bn} Now we know that lcm(a,b)= $p_1$$^{max(a1,b1)}$ $p_2$$^{max(a2,b2)}$ ... $p_n$$^{max(an,bn)}$ Let c be the multiplication of factors of lcm(a,b) that are factors of a and let d be the multiplication of factors of lcm(a,b) that are factors of b this implies that c divides a and d divides b Hence cd = lcm(a,b) and since factors of c and factors of d are both of the form $p_i$$^{s}$ Thus gcd(c,d)=1

Answer 3

List the prime factors of $a$ and/or $b$ as $p_1, p_2, .....,p_n$.

Let $a= \prod_{i=1}^n p_i^{k_i}$ for some non-negative integer values of $k_i$ (which may be equal to $0$). And let $b=\prod_{i=1}^n p_i^{j_i}$ for some non-negative integer values of $j_i$.

Then $\text{lcm}(a,b)= \prod_{i=1}^n p_i^{\max(k_i, j_i)}$. We can note that although $k_i$ may be $0$ and $j_i$ may be $0$, they can't be both be. And we can not that if $\max(k_i, j_i)=k_i$ then $p_i^{k_i}|a$. And if $\max(k_i, j_i)=j_i$ then $p_i^{j_i}|b$.

Now if we let $c = \prod_{i=1;k_i\ge j_i} p_i^{k_i} = \prod_{i=1;k_i\ge j_i} p_i^{\max(k_i,j_i)}$ and $d=\prod_{i=1;k_i< j_i} p_i^{j_i} = \prod_{i=1;k_i< j_i} p_i^{\max(k_i,j_i)}$ then:

$c|a$ and $d|b$.

$\gcd(c,d) =\prod_{i=1; k_i\ge j_i\text{ and }k_i < j_i} p_i^{\min(k_i, j_i} = \prod_{i=1; i\in \emptyset}^n p_i^{\min(k_i, j_i} = 1$.

$cd = \prod_{i=1;k_i\ge j_i} p_i^{\max(k_i,j_i)}\prod_{i=1;k_i< j_i} p_i^{\max(k_i,j_i)}=\prod_{i=1}^n p_i^{\max(k_i, j_i)}=\text{lcm}(a,b)$.

....

If that seems too abstract consider $a= 3^{17}5^47^{3}$ and $b=2^{6}3^{8}5^{9}$.

Then $a = 2^03^{17}5^47^3$ and $b=2^{6}3^85^87^0$.

And $\text{lcm}(a,b) = 2^{6}3^{17}5^97^3$.

$k_1=0< j_1=6$ so $2^6$ is a factor of $d$. $k_2=17>j_2=8$ so $3^{17}$ is a factor of $d$. And $k_3=4< j_3=9$ so $5^9$ is a factor of $d$. and $k_4=3>j_4 = 0$ so $7^3$ is a factor of $c$

So $c = 3^{17}7^3$ and $c|a$. And $d=2^65^9$ and $d|b$.

ANd $\gcd(c,d)= 1$ and $cd = 2^63^{17}5^97^3 = \text{lcm}(a,b)$.