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If for $a,b,x \in \mathbb{N}$ if we have $$a \mid x, b \mid x$$ prove that $\operatorname{lcm} (a,b) \leq x$

My solution: maximum value of $a,b$ is when they are both $x$, in which case their lcm is also $x$, but if any of them are less than $x$ then their lcm can't get bigger, hence their lcm is always less than $x$. Is this correct?

math_learner
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  • How are you defining the LCM? – lulu Apr 29 '23 at 12:30
  • @lulu lcm is the least common multiple. – math_learner Apr 29 '23 at 12:49
  • Yes, and what is your definition of that? My point is that I can't understand your question. It seems to me to be a literal restatement of the definition. So, perhaps you are using an exotic definition? – lulu Apr 29 '23 at 12:52
  • @math_learner, think calmly. First understand what l.c.m.(a,b) is. This is not even a question. – VizDracViz Apr 29 '23 at 12:56
  • @lulu The least common multiple (LCM) of two numbers is the lowest possible number that can be divisible by both numbers. Now following this definition, is my solution correct? – math_learner Apr 29 '23 at 12:57
  • I don't understand. $x$ is a common multiple. the LCM is the least common multiple, so by definition the LCM is $≤x$. What is there to prove? It's like "Bob is the shortest person in the room. Dave is also in the room, prove that Bob is no taller than Dave." – lulu Apr 29 '23 at 13:03

1 Answers1

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maximum value of $a,b$ is when they are both $x$

This doesn't make sense, because your $a,b, x$ are some fixed integers. If let $c=$max$(a,b)$, then $c$ is also a fixed number, so it is either $c< x, c=x$ or $c> x$. You cannot treat them as to first fix $x$ and then let $a, b$ vary and compare with your previously fixed $x$.

The least common multiple is $L=lcm(a, b)=a_1gb_1$, where $a=ga_1, b=gb_1,$ and $\gcd(a_1, b_1)=1$.

If $a|x$, then we have $x=ka, k\in\mathbb{N}^+$, then it gives $x=kga_1.$ If $b|x$, then we have $gb_1|kga_1\Rightarrow b_1|ka_1$, since $\gcd(a_1, b_1)=1$, it means $b_1|k$, so we have $k=tb_1, t\in \mathbb{N}^+$,

$$x=kga_1=tb_1ga_1=tL\Rightarrow x\ge L$$

MathFail
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