Proof involving lcm and biconditional statement.

Question

Suppose $a,b\in\mathbb{Z}$. Then $a = \operatorname{lcm}(a,b)$ if and only if $b\mid a$

Unsure of how to approach this problem.

score 3 · Answer 1 · edited Jul 17 '14 at 23:01

3

$\newcommand{\lcm}{\operatorname{lcm}}$ By definition $\lcm(a,b)=bk$ for some $k\in\Bbb Z$. So if $bk=a$ then $b\mid a$ by definition.

Conversely, $b\mid a$ implies $\lcm(a,b)\le a$ and $a\mid a$ implies $\lcm(a,b)\ge a$.

edited Jul 17 '14 at 23:01

Michael Hardy

1

answered Jul 17 '14 at 22:59

Adam Hughes

37,795
10
61
85

score 1 · Answer 2 · answered Jul 17 '14 at 23:16

1

Hint $\,\ b\mid a \iff a,b\mid a \iff {\rm lcm}(a,b)\mid a\iff {\rm lcm}(a,b) = a$

answered Jul 17 '14 at 23:16

Bill Dubuque

282,220

score 1 · Answer 3 · edited Jan 30 '19 at 17:45

1

As shown in this answer, $$ \operatorname{lcm}(a,b)\gcd(a,b)=ab $$ Thus, we have that $$ \gcd(a,b)=b $$ This means that $b$ is a divisor of $a$; i.e. $b\mid a$.

edited Jan 30 '19 at 17:45

Martin Sleziak

56,060

answered Jul 17 '14 at 23:23

robjohn

353,833

I see that, nowadays, Bezout is deleted. This is much healthier! – Bill Dubuque Jul 17 '14 at 23:44
@BillDubuque: ah, but I do use properties of Euclidean Division in the cited answer; very similar to your Bezout. – robjohn Jul 17 '14 at 23:59

Proof involving lcm and biconditional statement.

3 Answers3