3

Is there any alternate proof for proving that

The product of L.C.M (Least Common Multiple) and G.C.D (Greatest Common Divisor) of any two positive integers = the product of those two integers

?

As of now, there exists a proof which involves use of Number Theory based facts.

I am interested to know if there exists any alternate proof (perhaps in different branch of Mathematics ?!).

You can assume that you want to prove this fact to a person who is aware of basics of mathematics but not NUMBER THEORY

Martin Sleziak
  • 53,687
Karri Chandrasekhar
  • 233

2 Answers2

12

Let us express both numbers in their prime factorisation. Note that by the fundamental theorem of arithmetic, any number can be only represented in one prime factorisation:

$$X = 2^a * 3^b * 5^c * 7^d \cdots (1)$$

$$Y = 2^p * 3^q * 5^r * 7^s \cdots (2)$$

The highest common factor of $X, Y$ is $2^{min(a,p)} * 3^{min(b,q)} * 5^{min(c,r)} \cdots$, which multiplies the smallest powers of $2$, $3$, $5$ and so on.

The lowest common multiple is $2^{max(a,p)} * 3^{max(b,q)} * 5^{max(c,r)} \cdots$, which multiplies the largest powers of $2$, $3$, $5$ and so on.

Since one number cannot have a power which is both the smallest and largest, then $HCF(X,Y) * LCM(X,Y) = 2^a2^p * 3^b3^q * 5^c5^r \cdots$ in some order. This is equal to $(1)$ multiplied by $(2)$, which is equal to $X$ multiplied by $Y$.

Toby Mak
  • 16,827
  • well explained. Thanks. However, here it appears that it boils down to a problem of minimum of maximum values and maximum of minimum values of an 'arranged group/set' of quantities which are always EQUAL. Is it becuase of this that this holds true ? I am sorry, perhaps I am putting what runs through my mind at the time of typing, but it is what I feel as happening. – Karri Chandrasekhar Aug 26 '17 at 08:58
  • You need to mention the term Fundamental theorem of arithmetic in the start – Vidyanshu Mishra Aug 26 '17 at 09:12
  • @KarriChandrasekhar Another way to put this proof (quite vaguely) is that it is just a way of showing that the 'gap' when you have the HCF and the product of two numbers, and vice versa. – Toby Mak Aug 26 '17 at 10:10
  • 1
    Note to future readers: "HCF" means "gcd". – darij grinberg Apr 01 '19 at 18:13
4

Given two numbers, $a,b$, let $d:=\gcd(a,b)$ and then $f:=a/d$ and $g:=b/d$, giving $a=df$ and $b=dg$.

Now we know that the least common multiple, $c$, of $a$ and $b$ has to be divisible by $df$ and by $dg$. We also know that $f$ and $g$ have no common factors (otherwise $d$ would be bigger by that factor). So $c=dfg$ is the smallest number that can fulfill all these requirements.

Then $ab = df\cdot dg = d\cdot dfg = \gcd\cdot {\rm lcm}$

Joffan
  • 39,627