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If we have that $[a,b]=k$ meaning $a|k$ and $b|k$. Is there an equation that we can get from this relation? My problem says "find and prove the sufficient condition in which $(a,b)=[a,b]$". I think it is when $a=b$ but I am trying to prove a biconditional statement that $(a,b)=[a,b]$ if and only if $a=b$. I am stuck in the forward direction.

Any input will help.

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  • $\begingroup$ If $a|k$ then it means $k=k_aa$ for some $k_a$, and also $k=k_bb$ for some $k_b$. Also if $(a,b)=n$, then $n|a$ and $n|b$, so $a=n_an$ and $b=n_bn$ for some $n_1,n_b$. Then can you infer something from these decompositions? $\endgroup$
    – Anna SdTC
    Feb 17 '17 at 23:15
  • $\begingroup$ Hint: $(a,b)$ always divides $[a,b]$, and their product is $ab$. Further, $(a,b) \leq a,b \leq [a,b]$. $\endgroup$ Feb 17 '17 at 23:15
  • $\begingroup$ @AnnaSdTC i got something like that Kakba=Kakbb I assume that gcd(a,b)=lcm(a,b) i am trying to prove a bio conditional statment and this is my forward direction. But yeah where can I get from this equation? $\endgroup$
    – Jamie John
    Feb 17 '17 at 23:21
  • $\begingroup$ @астонвіллаолофмэллбэрг do you think that would help in that situation ? $\endgroup$
    – Jamie John
    Feb 17 '17 at 23:22
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The main relation between greatest common divisor and lowest common multiple is $$ (a,b)[a,b]=ab $$ (when $a$ and $b$ are positive integers). Suppose $(a,b)=[a,b]$; then $$ (a,b)^2=ab $$ On the other hand, $(a,b)\le a$, so $(a,b)^2\le a^2$. In particular $$ ab\le a^2 $$ so $b\le a$. Similarly, $a\le b$.

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  • $\begingroup$ Thank you so much! much simpler and way shorter than my way! $\endgroup$
    – Jamie John
    Feb 18 '17 at 0:18
  • $\begingroup$ How do we go from ab ≤ a2 to the next step? I am sorry I know it is an algebra thing but I am confused with it $\endgroup$
    – Jamie John
    Feb 18 '17 at 19:20
  • $\begingroup$ @HananIbraheim Since $(a,b)=[a,b]$ by assumption, we have $ab=(a,b)[a,b]=(a,b)^2\le a^2$. $\endgroup$
    – egreg
    Feb 18 '17 at 20:14
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See, if $a=b$ then it is obvious that $(a,b)= [a,b]$.

The other way, if $(a,b) = [a,b]$, then because $(a,b) \leq a \leq [a,b]$, it follows that $(a,b) = a$ (because above,equality holds in our case).

Similarly, if $(a,b) = [a,b]$, then because $(a,b) \leq b \leq [a,b]$, it follows that $(a,b) = b$.

Finally, $a = (a,b) = b = [a,b]$ is true.

Therefore, $a = b \iff [a,b] = (a,b)$.

EDIT: If $a=b$, then the "least common multiple" of $a$ and $b$ is obviously $a$ (or $b$). (Because both $a$ and $b$ divide $a$, and $a$ is the smallest number with this property).

Similarly, if $a=b$, the "greatest common divisor" of $a$ and $b$ is obviously $a$ (or $b$).(Because $a$ divides both $a$ and $b$, and $a$ is the largest number with this property).

Hence, $(a,b) = a = [a,b]$, nearly by definition of $\operatorname{lcm}$ and $\gcd$.

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  • $\begingroup$ I thought I can use some algebra to get there. I think your way is perfectly fine , I just need to see if my other way is ok otherwise I will use your hints. Thank you so much . It helped alot $\endgroup$
    – Jamie John
    Feb 17 '17 at 23:50
  • $\begingroup$ @HananIbraheim You are welcome! $\endgroup$ Feb 17 '17 at 23:51
  • $\begingroup$ that is what I did , do you think its ok? your way is much simpler tho I claim that (a,b)=[a,b] if and only if a=b Let (a,b)=[a,b]=k Then k|a and k|b so a=kak ,b=kbk [a,b]=k so a|k and b|k so k=aak and k=bbk So a=kaaak we know that a >0 so -ka=ak And b=kbbbk we know that b>0 so -kb=bk Then a =akk , b =-bkk So k =-a/ak , and also k=-b/bk So abk= bak However, bk= k/b and ak= k/a So we substitute we get a(k/b)=b(k/a) So abk=abk And ab=ab so a=b=1 thus a=b $\endgroup$
    – Jamie John
    Feb 17 '17 at 23:53
  • $\begingroup$ @HananIbraheim That's fine, although it wasn't the easiest to read because of no subscripts. I think you have captured the divisibility idea well in the answer. But how is $a=b=1$ in the end? Also, when you get $k_aa_k = 1$, and transpose, you should get $\frac 1{k_a} = a_k$, not $-k_a = a_k$. But you started correctly. $\endgroup$ Feb 17 '17 at 23:56
  • $\begingroup$ yeah I know I am sorry its not easy to read i wish I can upload a word document here. I see , I will go over the algebra again. Thank you so much $\endgroup$
    – Jamie John
    Feb 18 '17 at 0:00

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