Question

Show that \(a\)and\(b\)are positive integers, then\(ab = \gcd \left( {a,\,b} \right) \cdot lcm\left( {a,\,b} \right)\). (Hint: Use the prime factorizations of\(a\)and\(b\)also the formula for\(\gcd \left( {a,\,b} \right)\)and\(lcm\left( {a,\,b} \right)\)in terms of this factorization.)

Short answer

It is proved that \(lcm\left( {a,\,b} \right) \cdot \gcd \left( {a,\,b} \right) = ab\)

Step-by-step solution

Greatest common divisor

Let\(a\)and\(b\)be given integers with at least one of them is not zero. The greatest common divisor of\(a\)and\(b\)is denoted by\(\gcd \left( {a,\,b} \right)\)is the positive number\(d\)satisfying,

1. \({d \mathord{\left/
2. {\vphantom {d a}} \right.
3. \kern-\nulldelimiterspace} a}{\rm{ and }}{d \mathord{\left/
4. {\vphantom {d b}} \right.
5. \kern-\nulldelimiterspace} b}\)
6. If\({c \mathord{\left/
7. {\vphantom {c a}} \right.
8. \kern-\nulldelimiterspace} a}{\rm{ and }}{c \mathord{\left/
9. {\vphantom {c b}} \right.
10. \kern-\nulldelimiterspace} b}\)then\(c \le d\).

Proving that \(ab = \gcd \left( {a,\,b} \right) \cdot lcm\left( {a,\,b} \right)\)

Let\(d = \gcd \left( {a,\,b} \right)\)

So by definition of greatest common divisor we have,

\(a = dx\)for some \(x\)

\(b = dy\)for some \(y\)

Here, \(x\) and \(y\) are co-primes because \(\gcd \left( {x,\,y} \right) = 1\).

Now, we know that

\(lcm\left( {a,\,b} \right)\)is divisible by \(a = dx\)

\(lcm\left( {a,\,b} \right)\)is divisible by \(b = dy\).

So, \(lcm\left( {a,\,b} \right) = dxy\)

\(\begin{array}{l}lcm\left( {a,\,b} \right) \cdot \gcd \left( {a,\,b} \right) = dxy \cdot \gcd \left( {a,\,b} \right)\\lcm\left( {a,\,b} \right) \cdot \gcd \left( {a,\,b} \right) = dxy \cdot d\\lcm\left( {a,\,b} \right) \cdot \gcd \left( {a,\,b} \right) = dx \cdot dy\end{array}\)

So, we get \(lcm\left( {a,\,b} \right) \cdot \gcd \left( {a,\,b} \right) = ab\)

Hence proved.