If $b\mid a$ and $b\mid a-c$ then $b\mid c$ proof

Question

I was doing my homework in algebra and I'm not sure if I'm right. Basically, I have to show that $b\mid c$ is true as $b\mid a$ and $b\mid (a-c)$ are true (transitivity).

$ \begin{align*} b\mid a \leftrightarrow a = b\cdot k \\ b\mid a-c \leftrightarrow a-c = b\cdot k \end{align*} $

With the system of equations shown above, I added them up and it resulted on the following equation.

$ \begin{align*} a + (a-c) &= 2b\cdot k \\ 2a - c &= b \cdot 2k && 2a,2b \in \mathbb{Z} \\ \beta - c &= b \cdot \alpha && \text{implying } b\mid\beta - c \\ \end{align*} $

Is it right?

You probably want to differentiate the various $k$... For example, $k_1,k_2,k_3,\dots$ — abiessu, Aug 10 '21 at 00:22
There is really only one scenario where $a=bk$ and also $a-c=bk$, where $c=0$... — abiessu, Aug 10 '21 at 00:23
That's one thing I saw, but I couldn't go further than that. — ranu, Aug 10 '21 at 00:24
You probably want to subtract $a-c$ from $a$ to get your result. — abiessu, Aug 10 '21 at 00:26
Got it @abiessu, how couldn't see that? Much thanks, should I answer my own question? — ranu, Aug 10 '21 at 00:27
@abiessu that's why you told me to differentiate the different k... — ranu, Aug 10 '21 at 00:28
Transitivity is, if $x\mid y$ and $y\mid z$, then $x\mid z$. — Shaun, Aug 10 '21 at 00:28
That's what I wrote in a very peculiar way? $b|a$ and $b|a-c$ then $b|c$ — ranu, Aug 10 '21 at 00:30
Since $x=a-c$ and $y=a$ are multiples of $b$ the so is there difference $x-y$ (an integral linear combination of $x$ and $y)$, by the linked dupe. — Bill Dubuque, Feb 18 '22 at 19:41

score 2 · Accepted Answer · answered Aug 10 '21 at 00:41

The answer to that follows, as explained in the comments:

$ \begin{align*} a = b \cdot k_1 && \text{*} \\ a - c = b \cdot k_2 && \text{**} \end{align*} $

Subtracting $*$ from $**$ we have the following solution

$ \begin{align*} a - c - a &= b \cdot k_2 - b \cdot k_1 \\ - c &= b \cdot (k_2 - k_1) \\ - c &= - b \cdot (- k_2 + k_1) && -k_2+k_1 =l , l\ \in \mathbb{Z} \\ -c &= -b \cdot l && \cdot(-1) \\ c &= b \cdot l \end{align*} $

So we have that indeed $b|c$.

Shaun · Answer 2 · 2021-08-10T15:22:25.423

1

Suppose $b\mid a$ and $b\mid a-c$. Then there exist integers $k,l$ such that $a=bk$ and $a-c=bl$. Now $bk-c=bl$, so that $c=bk-bl=b(k-l)$; but $k-l$ is an integer, so, by definition, $b\mid c$.

edited Aug 10 '21 at 15:22

answered Aug 10 '21 at 01:20

Shaun

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If $b\mid a$ and $b\mid a-c$ then $b\mid c$ proof

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