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Provide a counterexample to show that the following is not true:

If $a$, $b$, and $c$ are integers for which $a|(b + c)$, then $a|b$ and $a|c$.

My Solution

To find a counterexample, I first find the contrapositive of the entire implication by negating the statement:

NOT [If a, b, and c are integers for which a|(b + c), then a|b and a|c].

$\equiv$ If $a$, $b$, and $c$ are integers for which $a$ does not divide $b$ and $c$, then $a$ does not divide $b + c$.

Since I am attempting to find the contrapositive, I (1) applied negation to each of the 'something that happens' of the statement (added NOT) and swapped $A \implies B$ to $\neg B \implies \neg A$.

It seems that I must have done something incorrectly. For instance, take $a = 2$, $b = 3$, and $c = 9$.

I would greatly appreciate it if people could please take the time to explain the error in my reasoning and what the correct reasoning should be.

EDIT

It seems that I confused contrapositive with negation. It is important to remember that these are two independent concepts. Unfortunately, this was an incorrect situation to use contrapositives.

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    The statement is the negation of $$\forall a, b, c \in \Bbb Z, \quad [a \mid (b+c) \implies a\mid b \text{ and } a \mid c]$$ – 2017-02-08
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    @Watson Thanks for the response. I thought that's what I did? NOT [ $\quad [a \mid (b+c) \implies a\mid b \text{ and } a \mid c$] $\equiv$ $\neg [a\mid b] \text{ and } \neg [a \mid c] \implies \neg [a \mid (b+c)]$? – 2017-02-08
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    The counterexample is correct: $2$ divides $3+9$ but $2$ does not divide $3$. This means that you have found a case where $a|(b+c)$ is *true* but $a|b$ **and** $a|c$ is *false*, and *true* $\to$ *false* is *false*. – 2017-02-08
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    @MauroALLEGRANZA Shouldn't that change to $a|b$ OR $a|c$? – 2017-02-08
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    Yes; in this case ($2$ does not divide $3$ nor $9$) also $a|b$ **or** $a|c$ is *false*; what I'm saying is that it is enough that one of the *conjuncts* is *false* to get : $a|b$ **and** $a|c$ *false*. – 2017-02-08

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Recall that the negation of $\forall x \; P(x)$ is $\exists x \; \neg P(x)$, and the negation of $A \implies B$ is $A \wedge \neg B$.

The statement is the negation of $$\forall a, b, c \in \Bbb Z, \quad [a \mid (b+c) \implies a\mid b \text{ and } a \mid c]$$ which is $$\exists a, b, c \in \Bbb Z, \quad [a \mid (b+c) \;\text{ and }\; \text{not(} a\mid b \text{ and } a \mid c)]$$ that is: $$\exists a, b, c \in \Bbb Z, \quad [a \mid (b+c) \;\text{ and }\; (a\not\mid b \text{ or } a \not\mid c)]$$

As you found: $a=2,b=3,c=9$ provides an example, so that the original sentence in the problem is false, i.e. its negation (the last formula I've just written above) is true.

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    @ThePointer : the statement $$\exists a, b, c \in \Bbb Z, \quad [a \mid (b+c) \;\text{ and }\; (a\not\mid b \text{ or } a \not\mid c)]$$ just means that you can find three integers $a,b,c$ such that $a$ divides the sum $b+c$, but $a$ doesn't divide $b$ or doesn't divide $c$. – 2017-02-08