Negation of "If $ l$ is line and P a point not on $l $, then every line through P intersects "$l$"

Question

To negate "If $ l$ is line and P a point not on $l $, then every line through P intersects "$l$":

I came up with "$l$ is a line and $P$ is a point that is not on $l$, and no line through $P$ intersects $l$"

Answer 1

Let $l$ be a line and $ P $ be a point that is not on $l$, then there exists a line through $P$ which does not intersect $l$.

Answer 2

Not quite. Towards the end it should be "some line through $P$ does not intersect $l$", or "not every line through $P$ intersects $l$", which means the same. Also this gives the correct negation only if $l$ and $P$ are fixed things known before the statement is given, but about which one somehow isn't sure whether they are actually a line respectively a point, so that the condition "If $l$ is a line and $P$ is a point" is necessary. This seems rather unlikely; almost certainly such a phrase should be read as "Whenever $l$ is a line and $P$ is a point...", in other words with an implicit "For all $l$ and all $P$" prefixed. The correct negation of that sentence then is: "There exists a line $l$ and there exists a point $P$ that is not on $l$ such that some line through $P$ does not intersect $l$". This negation has many equivalent forms, of which I would maybe prefer "There exist lines $l_1,l_2$ and a point $P$ such that $P$ does not lie on $l_1$, but $l_2$ does pass through $P$, and $l_2$ does not intersect $l_1$."

Answer 3

To negate a statement asserting every line intersects $l$, you assert that there must exist at least one line that does not intersect $l$. The negation of "every line..." is NOT "no line...".

The negation of your title question, then is as follows:

"$l$ is a line and $ P $ is a point that is not on $l$, and there exists a line through $P$ which does not intersect $l$."

When you want to negate a statement of the form $p \rightarrow \forall x(Q(x))$, you have to negate the entire conditional, not just the conclusion:

$$\lnot[p \rightarrow \forall x(\lnot Q(x))] \equiv \lnot[\lnot p \lor \forall x(Q(x))]$$ $$\equiv p \land \lnot\forall x(Q(x))$$ $$\equiv p \land \exists x(\lnot Q(x))$$

Note that for the assertion of the form $\forall x(Q(x)$, its negation is $$\lnot \forall x(Q(x))\equiv \exists x(\lnot Q(x))$$

Answer 4

Your confusion lies in that for "every line" having some property $p$ to be false, all you need is for one line to not have $p$. The other answerers have the correct negation.

Another example: if you have a set $S$ of integers, the negation of all $s \in S$ are odd is that there exists at least one $s \in S$ that isn't odd, because then not all of the integers are odd.

Answer 5

You have to negate that every line through $P$ intersects $l$. The quantifier opposite to $\forall$ is NOT $\;\nexists$ (it doesn't exist any line such that...), but $\exists$ (it exists at least one line such that...).