A set $S$ of whole numbers is called stapled if and only if for every whole number $a$ which is in $S$ there exists a prime factor of $a$ which divides at least one other number in $S.$
Let $T$ be a set of whole numbers. Which of the following is true if and only if $T$ is not stapled?
Taking the domain as all the whole numbers in the set, $S$ is stapled iff
$\forall a \exists b$(there exists a prime factor in $a$ which divides $b$).
Taking the negation:
$\exists a \forall b$(there does not exist a prime factor in $a$ which divides $b$).
In normal words:
There exists a whole number $a$ where there does not exist a prime factor in $a$ which divides all $b.$
So my answer would be:
There exists an number $a$ which is in $T$ such that there is no prime factor of $a$ which divides every other number in $T.$
However, the correct answer is:
There exists a number a which is in $T$ such that there is no prime factor of $a$ which divides at least one other number in $T.$
What went wrong in my solution?
You neglected to account for the word "other" in the given definition. A correct translation (where $\mathbb P$ denotes the set of prime numbers) is:
$$\forall a\in S \;\;\exists p\in\mathbb P \;\;\exists b\in S \;( p|a \;\text{and}\; p|b \;\text{and}\; a \neq b)\tag#$$
Your natural-language translation of your symbolic negation is wrong; it ought to be instead be (note that the placement of every/all $b$ is crucial because the meaning of the sentence varies according to that): "There exists a whole number $a$ such that for each $b,$ there does not exist a prime factor in $a$ which divides $b.$"
This is correct, because it does correspond to the negation of $(\#):$ $$\exists a\in S \;\;\not\exists p\in\mathbb P \;\;\exists b\in S \;( p|a \;\text{and}\; p|b \;\text{and}\; a \neq b).$$ An equivalent negation of $(\#):$ $$\exists a\in T \;\;\forall p\in\mathbb P \;\;\forall b\in T \;( p\not|a \;\text{or}\; p\not|b \;\text{or}\; a = b).$$