The Mathematical Proof Thread

[Kernel Sohcahtoa]

Hello, everyone.  Thanks for all of the beautiful posts.  I am posting a proposition and my attempted proof of it (it is far from beautiful).  I'd appreciate any tips or incites.  Please do not assume this proof is correct until others have had a chance to pick it apart.  This was exercise 36 (page 130) in Hammack's Book of Proof.  Hammack only gives solutions for the odd exercises, so I'm not sure if this proof is valid.  Also, I was not sure how to use a hide tag, so I apologize for posting the proof out in the open.

Relevant definitions:

Definition of Divisibility: Suppose a and b are integers.  We say that a divides b if b=ac for some integer c.  In this case we also say that a  is a divisor of b, and that b is a multiple of a (Hammack, Book of Proof, pg 90).


Definition of the Least Common Multiple (LCM): The lcm of non-zero integers a and b, denoted lcm(a,b), is the smallest positive integer that is a multiple of both a and b (Hammack, Book of Proof, pg 90).


Proposition: Suppose a and b are natural numbers.  Then a=lcm(a,b) if and only if b divides a

Proof Mini Strategy.

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 We have a bi-conditional statement here (p if and only if q).  Hence, in order to prove that this proposition is true, we'll need to show that p implies q and that q implies p.  In other words, we need to show that if a=lcm(a,b), then b divides a.  In addition, we need to show that if b divides a, then a=lcm(a,b).


Proof.  Suppose a=lcm(a,b).  Then by the definition of the lcm, a is a multiple of a and b, so a=ax and a=by for some integers x and y.  Thus, a=by means that b divides a by the definition of divisibility.

To prove the proposition in the other direction, suppose b divides a.  Then a=bu for some integer u, so a is a  multiple of itself and b.  Thus, a is greater than or equal to lcm(a,b).  On the other hand, lcm(a,b) is a multiple of a.  Thus, lcm(a,b)=aw  for some integer w.  Consequently, by the definition of divisibility, a divides lcm(a,b), which means that a must be less than or equal to lcm(a,b).  Hence, we have demonstrated that a is less than or equal to lcm(a,b) and a is greater than or equal to lcm(a,b), so it follows that  a=lcm(a,b).  Hence, the proof is complete.

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-AK