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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?

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4 Answers

up vote 5 down vote accepted

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 $c$.

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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.

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(FYI Shawn, that last line reads "k divides n", or "n is divisible by k") –  The Chaz 2.0 Jun 15 '12 at 1:39
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Yes, divisibility is transitive, i.e. $\rm\:a\:|\:b,\ b\:|\:c\:\Rightarrow\:a\:|\:c,\:$ since $\rm\: b = aa',\ c = bb' = aa'b'.$

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

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.

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

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Divisibility is a partial order on the positive integers.

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