Proving that lcm exists amongst common multiples (Beginner Here, Please give feedback/or rectify on this method of thinking)
I was thinking about proving that a least common multiple exists amongst common multiples (a rather trivial thing)
Lets say we want to investigate about common multiples of b and d, restricting to $gcd(b,d)=1$ (As $bx=dy$ and $(bc)x=(dc)y$ for $c\neq 0$ have same coordinates/locus), and $b,d \in \mathbb{N^{+}}$
$bx-dy=0$
$\dfrac{b}{d}x=y$
Here, We would like to plot only $x,y \in \mathbb{N^{+}}$
Natural Coordinates are guranteed if:
$x=dt$
$y=bt$
Restricting $t \in \mathbb{N^{+}}$
$(x,y) = (dt,bt),\;$ {Parametric Plotting here}
So, $x_{n}=dn\;,y_{n}=bn,\;$ for $n \in \mathbb{N^{+}}$
Therefore,
$c_{n}=(x_{n})b=(dn)b=dnb$
$c_{n}=(y_{n})d=(bn)d=dnb$
$D = \{x_{1},x_{2},..\}\;$; Domain
$R = \{y_{1},y_{2},..\}\;$; Range
$C= \{c_{1},c_{2},..\}\;$;  Set of common multiples
To prove that a least $c_{n}$ exists, I am thinking of first of all proving that if $x_{n}<x_{n+1}$, then $y_{n}<y_{n+1}$ so as to take a $(x_{i},y_{i})$ as the least coordinate in the sense that $x_{i}<x_{j}$ for any $x_{j} \in D \setminus \{x_{i}\}$ and $y_{i}<y_{j}$ for any $y_{j} \in R \setminus \{y_{i}\}$.
Then proving that if $x_{n}<x_{n+1}$ and $y_{n}<y_{n+1}$, then $c_{n}<c_{n+1}$. In that $c_{i}<c_{j}$ for any $c_{j} \in C \setminus \{c_{i}\}$
Hence that $(x_{i},y_{i})$ would represent the least common multiple, say $c_{i}$
Proof:
(1) Positive slope:
if $x_{n}<x_{n+1}, \text{ then } y_{n}<y_{n+1}$
For $n \in \mathbb{N^{+}}$,
$x_{n}=dn$, and corresponding $y_{n}=bn$
$\Delta x =x_{n+1}-x_{n}=d(n+1)-dn=d$
$\Delta y =y_{n+1}-y_{n}=b(n+1)-b(n)=b$
The difference between corresponding $y_{n+1}-y_{n}$ is positive: So $y_{n}<y_{n+1}$
Now, in a certain sense, $(x_{1},y_{1})=(d,b)$ can be taken as to represent the lowest point in the 1st Quadrant,
(2) Ordering of Common Multiples:
Given $(x_{n},y_{n})$ and $(x_{n+1},y_{n+1})$, then $c_{n}<c_{n+1}$
$c_{n}=x_{n}b=y_{n}d$
$c_{n+1}=x_{n+1}b=y_{n+1}d$
$c_{n}=x_{n}b=dnb$
$c_{n+1}=x_{n+1}b=(d(n+1))b=(dn+d)b=dnb+bd$
Clearly $c_{n}<c_{n+1}$
So one can take $c_{1}$ to be the least among all the common multiples.
$c_{1}=x_{1}b=db$
$c_{1}=y_{1}d=bd$
Which also highlights the fact if $gcd(b,d)=1$, then

*

*$lcm(b,d)=bd$.

*Also that coordinates representing lowest point in 1st Quad, are $(d,b)$

*Returning to $(bc)x=(dc)y,c\neq 0$. The lowest point is the same $(d,b)$, So $lcm(bc,dc) = (bc)d = (dc)b$

*Also showing that $lcm(bc,dc)=\dfrac{bc \cdot dc}{gcd(bc,dc)}=\dfrac{bc \cdot dc}{c}=bcd$
In general $a,b \in \mathbb{N^{+}}$ $lcm(a,b)=\dfrac{ab}{gcd(a,b)}$
 A: We usually just use the fact that the natural numbers are well-ordered by their construction. That way, you only need to prove that the set of common multiples of $b$ and $d$ (1) is non-empty, and it (2) is a subset of the natural numbers, and having done that it is a natural conclusion that it contains a unique least element.
The proof of (1) is pretty simple, since $bd$ is a multiple of both $b$ and $d$ and hence is a common multiple, and the proof of (2) comes from defining multiples as being natural numbers.
Of course, proving that this least element is equal to $\frac{bd}{\gcd(b, d)}$ takes a little more effort, although typically in number theory courses the proofs avoid things like coordinates or loci since they're working from a position where the relationship of the natural numbers to the real numbers or the Cartesian plane hasn't been established.
A: This could be made much shorter. Let $a,b\in\mathbb{Z}^+$, and consider the set $V$ of all common multiples of $a$ and $b$. Since $ab\in V$, the set is non-empty. Furthermore, as all common multiples are also positive integers, we have that $V\subseteq\mathbb{Z}^+$. Thus, by well-ordering, $V$ has a smallest element $\operatorname{lcm}(a,b)\in V$. This is enough to prove the existence of the least common multiple.
