Showing that, for $c,d\in\Bbb N$, $c|d$ implies $c\leq d$

Question

I need help solving the following. My idea is to use Euclid's algorithm however I was told that I can simply prove this just with natural numbers.

Prove that for all natural numbers $c$ and $d$, if $c|d$ then $c ≤ d.$

Answer 1

Hint: (broad overview)

Recall what it means for a natural number to divide another - it means one is an integer multiple of the other. Since both are positive, that integer must also be positive (i.e. it is $1$ or $2$ or $3$ or ...).

Consider the ratio of the first two integers and see what you can conclude.

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Solution:

If $c,d \in \Bbb N$ with $c|d$, then there exists $n \in \Bbb Z^+$ such that

$$d = nc$$

Consider the ratio of $d$ and $c$:

$$\frac d c = n$$

Since $n$ is a positive integer,

$$\frac d c = n \geq 1 \implies \frac d c \geq 1 \implies d \geq c$$

concluding the proof.