As the title says, if a number is divisible by a number, is it always divisible by that number's factors?

An example being that $100$ is divisible by $20$, it is also divisible by $10, 5, 4, 2$ as well?

Does this always apply?


Yes. It is indeed true. The proof also follows immediately. Before looking at the proof, lets us understand what it means to say that $x \in \mathbb{Z}$ divides $y \in \mathbb{Z}$.

We say that $x \in \mathbb{Z}$ divides $y \in \mathbb{Z}$, if there exists $n \in \mathbb{Z}$, such that $ y = x \times n$.

For instance, $6$ divides $-30$, since we have $-5 \in \mathbb{Z}$ such that $-30 = 6 \times (-5)$.

Similarly, $27$ divides $108$, since we have $4 \in \mathbb{Z}$ such that $108 = 27 \times 4$.

Now lets prove your claim.

Claim: If $a$ divides $b$ and $b$ divides $c$, then $a$ divides $c$, where $a,b,c \in \mathbb{Z}$.


Since $a$ divides $b$, we have $n_1 \in \mathbb{Z}$ such that $b = a \times n_1$.

Similarly, since $b$ divides $c$, we have $n_2 \in \mathbb{Z}$ such that $c = b \times n_2$.

Making use of the fact that $b = a \times n_1$ in the above equation, we get that $$c = \underbrace{(a \times n_1) \times n_2 = a \times (n_1 \times n_2)}_{\text{By associativity of multiplication}} = a \times n$$ where $n = n_1 \times n_2 \in \mathbb{Z}$.

Hence, $a$ divides $c$.


Yes, suppose $n$ is divisible by $m$, and $m$ is divisible by $k$. This means $n=m\ell$ and $m=kj$, all integers. Then $n=m\ell=(kj)\ell$, so $k\mid n$ by the definition of divisibility.

  • $\begingroup$ (FYI Shawn, that last line reads "k divides n", or "n is divisible by k") $\endgroup$ – The Chaz 2.0 Jun 15 '12 at 1:39

Yes, divisibility is transitive: $\ \ \begin{align}\rm&\color{#90f}{a\:|\:b}\ \ \ \&\ \ \ \color{#0a0}{b\:|\:c}\: \Longrightarrow\ \color{#c00}a\:|\:\color{#0a0}c\\[.2em] {\rm by}\ \ &\color{#90f}{b = a}a',\,\ \color{#0a0}{c = b}b' = \color{#c00}a a'b'\end{align}$

In terms of fractions: $\rm\ a\:|\:b\:|\:c\ \Rightarrow\ \dfrac{b}a,\,\dfrac{c}b\in \mathbb Z\ \Rightarrow \dfrac{b}a\dfrac{c}b = \dfrac{c}a\in\mathbb Z\ \Rightarrow\ a\:|\:c$

i.e. divisibility is transitive because integers are closed under product $\rm\:\mathbb Z \times \mathbb Z \subset \mathbb Z$

Similarly: $\rm\:a\:|\:b,c\:\Rightarrow\:a\:|\:b+c\:$ since integers are closed under sum $\rm\:\mathbb Z + \mathbb Z\subset \mathbb Z,\:$ viz.

$$\qquad {\rm a\:|\:b,c\ \Rightarrow\ \dfrac{b}a,\,\dfrac{c}a\in\mathbb Z\ \Rightarrow\ \dfrac{b}a+\dfrac{c}a = \dfrac{b\!+\!c}a\in\mathbb Z\ \Rightarrow\ a\:|\:b\!+\!c}$$

As such these divisibility properties hold w.r.t. any subring.

  • $\begingroup$ When using fractions we need to handle separately the cases when the denominator $= 0\ \ $ $\endgroup$ – Bill Dubuque Jul 19 '19 at 16:03

Divisibility is a partial order on the positive integers.


Write $a$ as a product of primes. We have:

$$\mu k = b \tag 1$$

where $\mu$ is either a prime factor of $a$ or a product of some of its prime factors, $k$ is the product of the rest of the factors. Now, $(1)$ is the notation of divisibility read as $\mu$ divides $b$.


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