
NOTThe term "not" reverses the truth of a sentence, P. If P is true, not(P) is false. If P is false, not(P) is true. The table below formalizes this; T stands for true and F stands for false.
In technical sciences such as mathematics, but not in ordinary writing, it is OK to have more than one "not": not(not(P)), not(not(not(P))), etc. The table below gives the truth values for these. Not(not(P)) is evaluated by finding the truth values of not(P) first, and then reversing them. Likewise, one evaluates not(not(not(P))) by reversing the truth values of not(not(P)).
Here is an example of double negatives from high school geometry. Two lines in the Euclidean plane intersect if they have at least one point in common; they are parallel if they have no point in common. What does it mean that two lines are NOT parallel? That means, not(they have no point in common). Notice the double negative "not" and "no". They "cancel": two lines are NOT parallel means that they have at least one point in common (that is, they intersect). There is more happening in this example than just the cancelling of two negatives (a forall statement is being converted to a thereexist statementa discussion saved for elsewhere). Here is an example from linear algebra that bewilders many students. Some vectors v_{i}, 1 < = i < = n, are said to be linearly independent, if for all scalars c_{i}, if the sum of c_{i} v_{i} is the 0 vector then each scalar c_{i}=0 for all 1 < = i < = n. Note that the prefix "in" on independent usually means "not" dependent. It should be no surprise that linear algebra textbooks often define linearly dependent to be not linearly independent. That's a classic example of a double negative! Elsewhere, you may learn that not[(for all x) (P(x))], where P(x) is some sentence about x, has the same meaning as (there is at least one x) [not(P(x))] (see "for all"). So, in the example of the previous paragraph, the v_{i}'s not being linearly independent means for at least one choices of scalars c_{i}, 1 < = i < = n, not[ if the sum of c_{i} v_{i} is the 0 vector, then each c_{i}=0 for all 1 < = i < = n].Elsewhere, perhaps at the "not(if P, then Q)" page, you may learn that "not(if P, then Q)" means the same as "P and (not(Q))". If we use that above in our analysis of not being linearly independent, we discover that the v_{i}'s not being linearly independent means for at least one choices of scalars c_{i}, 1 < = i < = n,[( the sum of c_{i} v_{i} is the 0 vector) and {not(each c_{i}=0 for all 1 < = i < = n)}].However, the second sentence in the "and" just above is again the "not" of a "for all" sentence, and is equivalent to "there is at least one integer i in [1,n] for which c_{i} is not 0". Thus we discover that the v_{i}'s not being linearly independent means for at least one choices of scalars c_{i}, 1 < = i < = n, [(the sum of c_{i} v_{i} is the 0 vector) and (at least one of the c_{i}'s is not 0 for some integer i in [1,n])].In this situation, let j be an integer in [1,n] such that c_{j} is not 0. In the equation, the sum of c_{i} v_{i} is the 0 vector, move the term c_{j}v_{j} to the other side: one obtains the sum of c_{i} v_{i}, for all integers i in [1,n] except i=j, is equal to c_{j}v_{j}.Since c_{j} is not equal to zero, we may divide both sides of the equation by it: the sum of [c_{i}/(c_{j})] v_{i}, for all integers i in [1,n] except i=j, is equal to v_{j}. We have just proved that, when the v_{i}'s are not linearly independent, one of them can be written as a linear combination of the others. The converse is also true, and is left to the reader.
