Describe Failure Exactly
In Necessary and Sufficient Conditions, failure of an implication had a precise form: its hypothesis held while its conclusion failed. We now examine the operation underlying that description. To negate a statement is to express exactly that the statement is false.
This is more precise than finding something that sounds opposite. For a fixed real number \(x\), the statements \(x>2\) and \(x<2\) cannot both be true, but they can both be false: take \(x=2\). Consequently, \(x<2\) is not the negation of \(x>2\). The correct negation is \(x\leq2\).
We continue to use \(p,q,r\) for proposition letters and \(A,B\) for formulas. In examples involving variables, we fix an allowed input and negate the condition at that input. Negating a condition such as \(x>2\) is a different task from negating a whole assertion beginning “for every \(x\).” Here our focus is the structure of statements built from the logical connectives.
Negation and Its Truth Table
| \(A\) | \(\neg A\) | \(\neg\neg A\) |
|---|---|---|
| True | False | True |
| False | True | False |
The last column records the Double Negation Law:
The table proves the law by covering both possible truth values of \(A\). As in Logical Equivalence, \(\equiv\) means that the formulas agree under every truth assignment.
A proposed negation must pass two tests: it cannot be true together with the original statement, and it cannot be false together with the original statement. Exactly one must be true in each allowed case. Merely being incompatible with the original is not enough.
Negating Equalities and Inequalities
For fixed real numbers \(x\) and \(a\), the following pairs describe complementary possibilities.
| Original condition | Its negation |
|---|---|
| \(x=a\) | \(x\neq a\) |
| \(x\neq a\) | \(x=a\) |
| \(x>a\) | \(x\leq a\) |
| \(x\geq a\) | \(x<a\) |
| \(x<a\) | \(x\geq a\) |
| \(x\leq a\) | \(x>a\) |
The inequality rows use the order of the real numbers: exactly one of \(x<a\), \(x=a\), and \(x>a\) holds. Thus failure of \(x>a\) includes both \(x<a\) and \(x=a\), which together mean \(x\leq a\). Failure of \(x\geq a\), on the other hand, leaves only \(x<a\). The other two inequality rows follow by the same three possibilities.
Worked Example: Negate Before Solving
Fix a real \(x\). Negate the statement \(3x+1>7\).
First negate the comparison itself:
For the last equivalence, subtract \(1\) and divide by \(3>0\); the steps are reversible and preserve the inequality direction. This agrees with the previous tutorial, where \(3x+1>7\) was shown to be equivalent to \(x>2\).
At \(x=2\), the original statement is \(7>7\), which is false, while its negation is \(7\leq7\), which is true. Replacing the original inequality by \(3x+1<7\) would miss this boundary case.
Negating “And” and “Or”
When a statement has more than one component, negation must account for the connective joining them. Recall that mathematical “or” is inclusive: \(A\lor B\) is true when at least one component is true, including when both are true.
Proof. Fix any truth assignment. The conjunction \(A\land B\) is true exactly when both components are true. Its negation is therefore false when both are true and true in each remaining case: only \(A\) is false, only \(B\) is false, or both are false. These are exactly the cases in which \(\neg A\lor\neg B\) is true, proving the first equivalence.
The disjunction \(A\lor B\) is false exactly when both components are false. Its negation is therefore true exactly in that case. The conjunction \(\neg A\land\neg B\) is also true exactly when both \(A\) and \(B\) are false; in every other case at least one of its components is false. This proves the second equivalence.
Worked Example: Outside a Range
For a fixed real \(x\), consider \(1<x\leq4\). This is shorthand for the conjunction
By the first of De Morgan’s Laws, its negation is
At \(x=1\), the original condition fails and the first alternative of the negation holds. At \(x=4\), the original condition holds and both alternatives of the negation fail. At \(x=5\), the original condition fails and the second alternative holds.
Writing \((x\leq1)\land(x>4)\) would be incorrect: no real number satisfies that conjunction. Failure to meet both requirements does not mean that both requirements must fail.
Worked Example: Neither of Two Values
Fix a real \(x\). The negation of “\(x=0\) or \(x=1\)” is
Both exclusions are required. The proposed alternative “\(x\neq0\) or \(x\neq1\)” is not a negation: at \(x=0\), it is true because \(0\neq1\), while the original statement is also true.
Negating an Implication
The conditional truth table gives a particularly important negation rule.
Proof. If \(A\) is false, \(A\to B\) is true regardless of \(B\), so its negation is false; \(A\land\neg B\) is also false. If \(A\) and \(B\) are both true, the implication is true and its negation is false, while \(A\land\neg B\) is false. Finally, if \(A\) is true and \(B\) is false, the implication is false, so its negation is true; \(A\land\neg B\) is also true. These cases exhaust all truth assignments.
This rule explains why the counterexamples in the previous tutorial required a true hypothesis and a false conclusion. It also separates negation from the transformations studied in Converse Inverse and Contrapositive.
| Statement related to \(A\to B\) | Formula |
|---|---|
| Negation | \(A\land\neg B\) |
| Converse | \(B\to A\) |
| Inverse | \(\neg A\to\neg B\) |
| Contrapositive | \(\neg B\to\neg A\) |
The contrapositive is equivalent to the original implication by the Contraposition Theorem; it is not its negation. Negating both components while leaving the implication arrow in place produces the inverse, which is not the negation either.
Worked Example: A Failed Conditional
Fix a real \(x\) and consider “If \(x^2>4\), then \(x>2\).” Its negation at that input is
At \(x=-3\), both parts of the negation hold: \(9>4\) and \(-3\leq2\). Thus the conditional is false at that input.
At \(x=0\), the hypothesis \(x^2>4\) is false, so the conditional is true. Its negation is false because its first component fails. Merely having \(x\leq2\) is not enough to make the conditional false.
Negating “If and Only If”
The previous tutorial established that \(A\leftrightarrow B\) is true exactly when \(A\) and \(B\) have the same truth value. Its negation therefore asserts disagreement:
To verify this equivalence, there are two kinds of cases. If \(A,B\) are both true or both false, the biconditional is true, so its negation is false; each conjunction on the right is also false. If only \(A\) is true, the first conjunction is true. If only \(B\) is true, the second conjunction is true. In both disagreement cases the biconditional is false and its negation is true.
In particular, \(\neg A\leftrightarrow\neg B\) is not the negation of \(A\leftrightarrow B\). Negating both components preserves whether they agree.
Worked Example: Failure of Agreement
For a fixed real \(x\), negate “\(x=2\) if and only if \(x^2=4\).” The result is
The first alternative cannot hold, because \(x=2\) gives \(x^2=4\). In the second alternative, \(x^2=4\) gives \((x-2)(x+2)=0\), so \(x=2\) or \(x=-2\). The requirement \(x\neq2\) leaves precisely \(x=-2\).
Thus, on the real numbers, this negated condition holds exactly at \(x=-2\). At \(x=0\), both components of the original biconditional are false, so they agree: the biconditional is true and its negation is false.
Scope, Parentheses, and the Domain
The scope of a negation is the formula to which “not” applies. Parentheses make that scope explicit. For example, \(\neg(A\land B)\) negates a whole conjunction, whereas \((\neg A)\land B\) negates only \(A\). When \(A\) is true and \(B\) is false, the first formula is true and the second is false.
Keep the same allowed inputs and make any hidden “and,” “or,” or implication explicit.
Use parentheses so that the scope is unambiguous.
Then work inward, simplifying each remaining negation and preserving the grouping.
Test equality cases and representative inputs. The original and its negation must always have opposite truth values.
Worked Example: A Nested Statement
Negate \(A\land(B\lor\neg C)\). Start with the outer conjunction, then the inner disjunction:
For a concrete instance, fix a real \(x\) and let \(A\) be \(x>0\), \(B\) be \(x<2\), and \(C\) be \(x=3\). The original condition is “\(x>0\), and either \(x<2\) or \(x\neq3\).” Its negation becomes
which is equivalent to “\(x\leq0\) or \(x=3\),” since \(x=3\) already implies \(x\geq2\). At \(x=3\), the inner disjunction of the original is false; at \(x=1\), both the outer positivity requirement and the inner disjunction hold. The negation gives the opposite truth value in each case.
Forming a negation is a logical task; simplifying it may require mathematical facts. The rule for negating a biconditional works for any formulas, but simplifying the worked example to \(x=-2\) used algebra. Keeping these two steps separate makes both the logic and the calculation easier to check.
Check Your Understanding
For each variable-based question, keep the stated domain unchanged. Write the negation first, then simplify it if possible.
- For real \(x\), negate \(x\geq5\), \(2x-1<7\), and \(x\neq0\). Explain why the equality case matters in the first two.
- For real \(x\), negate \(-2\leq x<3\). Check the original and its negation at \(x=-2\), \(x=3\), and \(x=0\).
- Negate \(A\lor(B\land\neg C)\). Justify each step using De Morgan’s Laws or the Double Negation Law.
- For a fixed real \(x\), negate “If \(x>0\), then \(x>2\).” Give one input where the negation is true and one where it is false. Why is “If \(x\leq0\), then \(x\leq2\)” not the requested negation?
- For a fixed real \(x\), negate “\(x=1\) if and only if \(x^2=1\).” Simplify your result using \((x-1)(x+1)=0\). What happens to its allowed solutions if the domain is restricted to nonnegative real numbers?
- A learner proposes \(x<0\) as the negation of \(x>0\) on the real numbers. Identify the input that disproves this proposal. Would the proposal become correct on the domain of nonzero real numbers? Explain.