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Show that the product of the least common multiple and the greatest common divisor of two positive integers $a$ and $b$ is $a\times b$.

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$x \text{ and } y$ can be expressed as

• $x = 2^{a_{1}}\cdot 3^{a_{2}}\ldots P^{a_{n}}.$
• $y= 2^{b_{1}}\cdot 3^{b_{2}}\ldots P^{b_{n}}.$

where, $a_{i} \& b_{i} \geq 0$ for $1\leq i \leq n$, and $P$ is a prime number.

$x*y = 2^{a_{1}+b_{1}}\cdot 3^{a_{2}+b_{2}}\ldots P^{a_{n}+b_{n}}$

• $\text{LCM} (x,y) = 2^{\max(a_{1},b_{1})}\cdot 3^{\max(a_{2},b_{2})}\ldots P^{\max(a_{n},b_{n})}$
• $\text{HCF} (x,y) = 2^{\min(a_{1},b_{1})}\cdot 3^{\min(a_{2},b_{2})}\ldots P^{\min(a_{n},b_{n})}$

$\max(a_{i} ,b_{i}) + \min(a_{i},b_{i}) = a_{i} + b_{i}$

So, $\text{LCM}(x,y)\ast \text{HCF}(x,y) = 2^{a_{1}+b_{1}}\cdot 3^{a_{2}+b_{2}}\ldots P^{a_{n}+b_{n}} = x\ast y.$

Hence proved.

by

Let  a ( 120 ) = f1 * f2 * f3 * f4 * f5 ....  ( 2 * 2 * 2 * 3 * 5 ).

b ( 18 )   = F1 * F2 * F3........       ( 2 * 3 * 3 )

LCM of a( 120 )  and b( 18 ) = product of elements of a union b.

= M( {f1, f2, f3, f4, f5 } U { F1, F2, F3 } )    //M- > multiplication.

= M( (f1/F1), f2, f3, (f4/F2), F3, f5 )   ----- (A)      // f1/ F1- > f1 or F1

HCF of a( 120 )  and b( 18 ) = product of elements of a intersection b.

= M( {f1, f2, f3, f4, f5 } $\cap$ { F1, F2, F3 } )

= M( (f1/F1), (f4/F2) )      -----------------------(B)

// Remember if a $\cap$ b= ∅, then HCF = 1.

From equation A and B,

Product of LCM and HCF =  M( (f1/F1), f2, f3, (f4/F2), F3, f5 ) M( (f1/F1), (f4/F2) )

=  M( f1, f2, f3, f4, F3, f5,  F1, F2 )

= Product of a and b.

by

dude, Its a problem of mathematical logic. Is not it?
Is it asking us to write predicate logic ?? Is it mentioned in the question ??

I thought it is asking us to prove it.
I think.(see the topic).

But you have done good thing.
lcm(x,y) : lcm of x and y

gcd(x,y):  gcd of x and y

∀x∀y( x >= 0 ∧ y>=0 ) $\rightarrow$ (LCM(x,y) * GCD(x,y) = x*y )
by
let a=2 and b=4

now lcm of a&b is 4

and hcf of a&b is 2

now if we product we'll get 8 which is equivalent  to a*b

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