# Showing that, for $c,d\in\Bbb N$, $c|d$ implies $c\leq d$ [closed]

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 $$c ≤ d.$$

## closed as off-topic by user21820, Xander Henderson, José Carlos Santos, Saad, Parcly TaxelMar 26 at 3:27

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• If $n\in\Bbb N$ then $n\ge1$. – Lord Shark the Unknown Mar 7 at 6:31
• What does $c\mid d$ mean to you? – Arthur Mar 7 at 6:32
• And what does that mean? – Arthur Mar 7 at 6:37
• That's not the definition of the word "divides". What does the phrase "$c$ divides $d$" actually mean? – Arthur Mar 7 at 6:42
• What Arthur is trying to get at, @k.rudin, is the actual definition of what it means for a number to divide another. What is the exact relation between them? There's an equation somewhat associated with this, depending on how it was introduced to you. – Eevee Trainer Mar 7 at 6:46

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.

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.