I'm stuck on this problem for my math homework and could really use some help.

Definition A: Let $a$ and $b$ be integers. We say that $a$ is divisible by $b$ provided there is an integer $c$ such that $bc = a$.

Definition B: Let $a$ and $b$ be integers. We say that $a$ is divisible by $b$ provided $a/b$ is an integer.

The question asks to find integers $a$ and $b$ that are divisible according to one definition, but are not divisible according to the other.

Any guidance would be really appreciated.

  • 2
    $\begingroup$ What happens when $a=b=0$? $\endgroup$
    – aduh
    Sep 12 '16 at 2:01
  • 3
    $\begingroup$ @aduh In def A, that would be 0c = 0, c=0. In the alternate definition, 0/0 is undefined. This looks the right solution. I was operating under the assumption that nothing is divisible by 0, even 0. I guess that is where my logic went wrong. Thank you very much. $\endgroup$ Sep 12 '16 at 2:07

Thanks to @aduh, I have solved the question:

Let $a= b=0$. In Definition A, $a$ is divisibile by $b$ since $0c = 0$, where $c=0$. In definition B, $a/b = 0/0$ is undefined.


In the following I'll use the abbreviation "A-divisible" (resp. "B-divisible") for "divisible according to definition A (resp. B)".

  1. If $a$ is B-divisible by $b$, then $a$ is A-divisible by $b$.

    Proof: If $a$ is B-divisible by $b$, it means that $a/b$ is an integer. Therefore $c=a/b$ is an integer such that $bc=a$, thus such an integer exists, and thus $a$ is A-divisible by $b$.

  2. If $a$ is A-divisible by $b$, then exactly one of the following are true:

    • $a$ is B-divisible by $b$.
    • $a=b=0$.

    Proof: If $a$ is A-divisible by $b$, then there exists an integer $c$ such that $bc=a$. Now either $b=0$ or $b\ne0$.

    If $b\ne 0$, then both sides of the equation can by divided by $b$ to get the equivalent equation $c=a/b$. Since $c$ is an integer, so is $a/b$, and thus $a$ is B-divisible by $b$.

    If $b=0$ then for any integer $c$, $bc=0$, and therefore the equation $bc=a$ can only be fulfilled if also $a=0$; on the other hand, if $a=b=0$, then any integer $c$ causes the equation $bc=a$ to be true.

    Obviously exactly one of $b=0$ or $b\ne 0$ is true.

Therefore there exists exactly one pair of integers $a$ and $b$ such that $a$ is divisible by $b$ only according to one of the definitions, and that is $a=b=0$. $0$ is A-divisible by $0$, but $0$ is not $b$-divisible by $0$.


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