Show that lcm$(a,b)= ab$ if and only if gcd$(a,b)=1$

Not sure how to begin. If gcd$(a,b)=1$ what can I deduce from that?

Solution 1:

Hint $ $ note that $n\mapsto ab/n\,$ bijects the common divisors of $\,a,b\,$ with the common multiples $\le ab.$ Being order-reversing it maps the greatest common divisor to the least common multiple, i.e. $\,\gcd(a,n)\mapsto ab/\gcd(a,b) = {\rm lcm}(a,b).\,$ Thus $\,\gcd(a,b)=1\iff {\rm lcm}(a,b) = ab$.

Remark $\ $ For more on the (involution) duality between gcd and lcm see here and here.

Solution 2:

Outside of the definition of $gcd(a,b)$ and $lcm(a,b)$, this proof only requires you to know that every integer can be expressed as a unique product of primes. I use a contrapositive proof to prove "$\implies$" and a direct proof to prove "$\impliedby$"

By contrapositive, assume $gcd(a,b)=d > 1$. Then $a=dr$ and $b=ds$. So $lcm(a,b) \leq drs < (dr)(ds) = ab$

Now assume $gcd(a,b) = 1$. Writing $a$ and $b$ as a product of primes we have:

$a=p_1p_2...p_n$

$b=q_1q_2...q_m$

Where $p$,$q$ are prime and $p_i \neq q_j$ $\forall i,j$ (or else a and b would have a common divisor greater than 1).

Then, any number divisible by a and b must be of the form:

$k(p_1...p_n)(q_1...q_n)$ for some natural number $k$. Then, the smallest such number that is still divisble by a and b is when k=1. So, $lcm(a,b)=ab$.

Solution 3:

It is known that:

$$gcd(a,b) \cdot lcm(a,b)=a \cdot b $$

• IF $gcd(a,b)=1$: $$ gcd(a,b) \cdot lcm(a,b)=a \cdot b \Rightarrow 1 \cdot lcm(a,b)=a \cdot b \Rightarrow lcm(a,b)=a \cdot b$$
• IF $lcm(a,b)=a \cdot b$: $$gcd(a,b) \cdot lcm(a,b)=a \cdot b \Rightarrow gcd(a,b) \cdot a \cdot b=a \cdot b \Rightarrow gcd(a,b)=1 $$