$$\gcd(\text {lcm}(x, y), \text {lcm} (x, z)) = \text {lcm}(x,\gcd(y, z))$$ where $x,y,Z$ are three integers.
I came across this property on GCD but was not able to prove this.Can anyone suggest me a method.
Asked
Active
Viewed 123 times
0
-
1Show that any prime power $p^k$ that divides the LHS also divides the RHS, and vice-versa. – saulspatz Aug 29 '20 at 14:23
-
Using only gcd laws yields a more general proof that works in any gcd domain, e.g. use the dual of the proof in my answer in the second link, which boils down to $\ (a,b,c)(ab,bc,ac) = (a,b)(b,c)(a,c)\ \ $ – Bill Dubuque Aug 29 '20 at 15:33
1 Answers
0
Let $x=p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$, $y=p_1^{\beta_1}p_2^{\beta_2}...p_n^{\beta_n}$ and $z=p_1^{\gamma_1}p_2^{\gamma_2}...p_n^{\gamma_n},$ where $\alpha_i,$ $\beta_i$ and $\gamma_i$ are non-negative integers and $p_i$ are different prime numbers.
Thus, we need to prove that: $$\min\{\max\{\alpha_i,\beta_i\},\max\{\alpha_i,\gamma_i\}\}=\max\{\alpha_i,\min\{\beta_i,\gamma_i\}\}.$$ Can you end it now?
Michael Rozenberg
- 203,855
-
-
@cgss Don't agree. $\gcd(4,6)=2=2^{\min{2,1}}3^{\min{0,1}}=2.$` I wait your reaction. – Michael Rozenberg Aug 29 '20 at 15:32
-
1Yes you are right. I just did lcm = max, gcd = min and didn't pay attention to the fact that they were interchanged in the question itself. – cgss Aug 29 '20 at 15:50