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This one is killing me, any help is greatly appreciated!

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8 Answers 8

up vote 2 down vote accepted

Go back to the definition of divisibility: $a\mid b$ means that there is an integer $m$ such that $b=ma$, and $a\mid b+c$ means that there is an integer $n$ such that $b+c=na$. You’re interested in $c$, so isolate it:

$$c=na-b=na-ma=(n-m)a\;.$$

Is $n-m$ an integer? Does this show that $a\mid c$?

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(In this answer all variables are integers, i.e., elements of $\mathbb{Z}$.)

By the definition of divisibility we are given that $\;n * a = b\;$ and $\;m * a = b+c\;$ for some $\;n\;$ and $\;m\;$. Now we are asked to find a $\;k\;$ which makes $\;k*a = c\;$:

\begin{align} & k*a = c \\ \equiv & \;\;\;\;\;\text{"use the only fact we know about $\;c\;$"} \\ & k*a = m*a - b \\ \equiv & \;\;\;\;\;\text{"use the other fact"} \\ & k*a = m*a - n*a \\ \equiv & \;\;\;\;\;\text{"factor out $\;a\;$ -- to make both sides more alike"} \\ & k*a = (m-n)*a \\ \Leftarrow & \;\;\;\;\;\text{"weaken using Leibniz' rule -- to achieve our goal"} \\ & k = m-n \\ \end{align}

Therefore we have found such a $\;k\;$, and hence proved $\;a|c\;$.

(Yes, this answer looks a lot like http://math.stackexchange.com/a/452159/11994 since both questions are a lot alike. Apologies if the duplication is against site policy.)

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$a|b$ mean that exists whole number $k$ such that $$b=ka \dots(1)$$ and $a|(b+c)$ mean that exists whole number $l$ such that $$(b+c)=la\dots(2)$$ replacing (1) in (2) we get $$ka+c=la$$ $$c=la-ka$$ $$c=a(l-k)$$ because $l-k=r$ is whole number that mean$$c|a$$

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Hint $\rm\ \ \dfrac{b+c}a,\,\dfrac{b}a\in\Bbb Z\ \Rightarrow\ \dfrac{c}a = \dfrac{b+c}a-\dfrac{b}a\in\Bbb Z\ $ since $\,\Bbb Z\,$ is closed under subtraction.

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Note that, by definition of divisibility, $a|b$ implies

$$b = ak$$

for some integer $k$. Also, we have that $a|(b + c)$ implies

$$b + c = al$$

for some integer $l$. So, substitute the first into the second and solve for c. You should get $c = a(l - k)$, which implies $a|c$.

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In general, if $a\mid m$ and $a\mid n$ then $a$ divides any linear combination of $n$ and $m$. That is, for all $x,\ y\in\mathbb{Z}$, we have $a\mid mx + ny$.

Given these facts, can you now find a linear combination of $b$ and $b+c$ which gives $c$?

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If $a|b$ then $b=a\cdot n$.
If $a|(b+c)$ then $b+c=a\cdot m$.
Hence $c=b+c-b=a\cdot m-a\cdot n=a\cdot(m-n)$, so $a|c$

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If $a\mid b$ and if $a\mid (b+c)$, there exist some integers $k,l$ such that $b=k\cdot a,b+c=l\cdot a $

So,$c=b+c-b=la-ka=a(l-k)\implies \frac c a=k-l$ some integer

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