Proof by contraposition for dividing integers

[tmt1]

For all integers $a$, $b$, and $c$, if $a \nmid bc$, then $a \nmid b$

I need to prove this by contraposition.

I get that by definition, $b = ak$ for some integer $k$. But I don't get the following step in the textbook:

$bc = (ak)c = a(kc)$

I'm guessing there is something very obvious I'm missing.

Thanks for any suggestions.

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For all integers $a$, $b$, and $c$, if $a \nmid bc$, then $a \nmid b$

I need to prove this by contraposition.

I get that by definition, $b = ak$ for some integer $k$. But I don't get the following step in the textbook:

$bc = (ak)c = a(kc)$

I'm guessing there is something very obvious I'm missing.

Thanks for any suggestions.

Oh, I just got it.

We just multiply both sides by $c$.

 

[jvicens]

Yes, you are correct. By multiplying both sides by $c$, we are essentially showing that if $a$ does not divide $bc$, then it also cannot divide $b$. This is because the product of $a$ and $c$ is equal to the product of $b$ and some other integer $kc$. This means that $a$ and $b$ have a common factor of $c$, and since $c$ is not a factor of $a$, it cannot be a factor of $b$. Therefore, $a$ cannot divide $b$.

In mathematical terms, we can write this as $a \nmid bc \implies a \nmid b$. This is the contrapositive of the original statement, and by proving it, we have also proven the original statement.