Divisibility is transitive: $\ a\mid b\mid c\,\Rightarrow\ a\mid c$

Question

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?

Answer 1

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 $a,b,c \in \mathbb{Z}$.

Proof:

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 $c$.

Answer 2

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.

Answer 3

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.

Answer 4

Divisibility is a partial order on the positive integers.

Answer 5

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 $b$.

Answer 6

Yes. This always applies in arithmetic and algebra.

Arithmetic:

Let, $a$ is divisible by $\alpha$. According to the condition of divisibility,

$$a=c\alpha\tag{1}$$

$$[\text{where c is any integer}]$$

We have to prove that $a$ is also divisble by any arbitrary factor of $\alpha$, $\beta$.

Now, according to the condition of divisibility, $$\alpha=\beta k\tag{2}$$ $$\text{[where k is any integer]}$$

Now, the stage has been set. Let us see whether $a$ is divisible by $\beta$ or not.

$$\frac{a}{\beta}$$

$$=\frac{ak}{\beta k}$$

$$=\frac{ak}{\alpha}\ [\text{from (2)}]$$

$$=\frac{a}{\alpha}\cdot k$$

$$=\frac{c\alpha}{\alpha}\cdot k\ [\text{from (1)}]$$

$$[\text{The $\alpha$s on the numerator and the denominator cancel out}]$$

$$=ck$$

So, $a$ is divisible by $\beta$.

Algebra

Consider $2x^2+4x-30$. It is divisible by $2x-6$. If you don't believe me, do the long division yourself. Now, the factors of $2x-6$ are $2$ and $x-3$. $2x^2+4x-30$ is independently divisibly by both $2$ and $x-3$.

Algebra 2

Consider $3x^2+20x-63$. It is divisible by $3x-7$. However, one of the factors of $3x-7$, $3$, cannot independently divide $3x^2+20x-63$, while another factor, $x-\frac{7}{3}$, can.