GATE1991-15,a

Question

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

Answer 1

$x \text{ and } y$ 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, $a_{i} \text{ & } b_{i} \geq 0$ 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.

Answer 2

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

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

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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 

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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.

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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.

Comments

dude, Its a problem of mathematical logic. Is not it?

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Is it asking us to write predicate logic ?? Is it mentioned in the question ??

I thought it is asking us to prove it.

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I think.(see the topic).

But you have done good thing.

Answer 3

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 )

Answer 4

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