From Failure at One Input to Failure of “Every”
In Negating a Mathematical Statement, we negated conditions at a fixed input. For example, the negation of \(x>2\) is \(x\leq2\). A statement beginning “for every \(x\)” makes a different claim: it requires the condition to hold throughout a specified domain.
Consider “Every real number is greater than \(2\).” To deny this assertion, we do not need every real number to be at most \(2\). We need only one real number that is at most \(2\). The number \(0\) supplies such an example, even though many other real numbers are greater than \(2\).
The essential distinction is between failure everywhere and failure somewhere. A universal assertion requires success everywhere; its exact negation requires failure somewhere.
The Universal Statement and Its Domain
Let \(D\) be a set of allowed inputs, called the domain, and let \(A(x)\) be a condition that is either true or false for each \(x\in D\). Here \(A(x)\) uses the same formula notation as before, with the variable displayed explicitly.
We will express its negation using the symbol \(\exists\), read “there exists.” The assertion
means “There is at least one element \(x\) of \(D\) for which \(A(x)\) is false.” It does not assert that there is exactly one such element. There may be one, several, or even no elements where the original condition holds.
The domain is part of the assertion, not an optional detail. “Every positive real number is greater than \(0\)” and “Every real number is greater than \(0\)” have different truth values because they concern different allowed inputs.
The Rule for Negating a Universal Statement
Proof. Fix a domain \(D\) and an interpretation of \(A(x)\). Suppose first that the universal statement is false. If no element of \(D\) made \(A(x)\) false, then \(A(x)\) would be true for every element of \(D\), since each instance is either true or false. That would make the universal statement true, a contradiction. Therefore at least one element of \(D\) makes \(A(x)\) false. The statement \(\exists x\in D,\ \neg A(x)\) is true.
Conversely, suppose \(\exists x\in D,\ \neg A(x)\) is true. Then some element \(c\in D\) satisfies \(\neg A(c)\), so \(A(c)\) is false. Consequently, \(A(x)\) does not hold for every element of \(D\); the universal statement is false and its negation is true. The two formulas are therefore true in exactly the same circumstances, proving the equivalence.
If \(D\) has no elements, the universal statement is true: there is no element at which its requirement fails. Its negation is false. The existential statement on the right is also false, because no element of \(D\) can be supplied. Thus the rule includes the empty domain.
As before, \(\equiv\) indicates logical equivalence. Here the agreement holds for every choice of domain and condition, not just for one numerical example.
“Not Every” Does Not Mean “Every One Does Not”
The position of the negation determines its scope. Compare the following statements.
| Formula | Meaning |
|---|---|
| \(\forall x\in D,\ A(x)\) | Every allowed input satisfies the condition. |
| \(\neg(\forall x\in D,\ A(x))\) | Not every allowed input satisfies the condition. |
| \(\exists x\in D,\ \neg A(x)\) | At least one allowed input fails the condition. |
| \(\forall x\in D,\ \neg A(x)\) | Every allowed input fails the condition. |
The middle two rows are equivalent. The last row is not the negation of the first: it demands failure at every input rather than at least one.
Worked Example: Mixed Success and Failure
Let \(D=\{-1,0,2\}\), and let \(A(x)\) mean \(x>0\). The universal statement is
Its negation is
This negation is true: \(x=-1\) belongs to \(D\) and satisfies \(-1\leq0\). The value \(x=0\) also works. But “Every \(x\in D\) satisfies \(x\leq0\)” is false, because \(2\in D\) and \(2>0\).
Thus the universal claim fails without the condition failing everywhere. Two failures rather than exactly one also cause no difficulty: the existential quantifier requires at least one.
Counterexamples: Membership and Failure
A complete counterexample checks two things: the proposed element lies in the domain, and the condition fails there. A value outside the domain cannot disprove the statement, regardless of how the condition behaves at that value.
Worked Example: An Inequality That Fails Between 0 and 1
Negate the assertion “For every positive real number \(x\), \(x^2\geq x\).” Write \(D\) for the set of positive real numbers. The rule gives
Choose \(c=\tfrac12\). This is a positive real number, so \(c\in D\). Moreover,
Both required checks succeed, so \(c=\tfrac12\) is a counterexample. The universal statement is false.
The value \(c=1\) would not work: \(1^2=1\), so the original non-strict inequality holds. Negating \(\geq\) produces \(<\), not \(\leq\).
Forming the negation and proving it are separate tasks. For instance, the negation of “Every real \(x\) satisfies \(x+1>x\)” is “There exists a real \(x\) with \(x+1\leq x\).” This is the correct negation even though it is false: subtracting \(x\) would give \(1\leq0\).
Negating a Compound Condition After “Every”
The quantifier rule handles the outermost structure. After applying it, use the rules from Negating a Mathematical Statement to negate the condition at the chosen input.
Write the assertion as \(\forall x\in D,\ A(x)\), making any hidden connectives explicit.
Write \(\exists x\in D,\ \neg A(x)\). Keep the same domain.
Apply De Morgan’s Laws, the implication rule, or the appropriate inequality rule.
Check membership in \(D\) and every requirement of the negated condition.
Worked Example: A Universal Range Claim
Let \(D=\{0,2,4\}\). Negate “Every \(x\in D\) satisfies \(1<x\leq4\).” The range condition is a conjunction. Applying the quantifier rule and then De Morgan’s Laws gives
The value \(x=0\) is a witness because \(0\in D\) and \(0\leq1\). It is not necessary for both range requirements to fail. At \(x=0\), the upper bound \(x\leq4\) still holds.
The endpoint \(x=4\) is not a counterexample: it satisfies both original requirements. The negated upper-bound condition is \(x>4\), so it correctly excludes that endpoint.
Worked Example: Both Alternatives Must Fail at the Same Input
Negate “Every real number \(x\) satisfies \(x=0\) or \(x=1\).”
The real number \(2\) satisfies both exclusions and is therefore a counterexample.
Both exclusions concern the same input. It is not enough to note that \(1\neq0\) and, separately, \(0\neq1\): neither \(1\) nor \(0\) makes the original disjunction false.
Universal Statements with a Hypothesis
Many mathematical assertions say that every input satisfying one condition must satisfy another. Their form is
First negate the universal quantifier. Then use the previously established rule \(\neg(A\to B)\equiv A\land\neg B\):
Thus a counterexample must satisfy the hypothesis and fail the conclusion. An input at which the hypothesis is false cannot serve as a counterexample.
Worked Example: A True Hypothesis and a False Conclusion
Consider the universal version of a conditional examined in the previous tutorial:
Its negation is
Choose \(x=-3\). It is real, its square is \(9>4\), and \(-3\leq2\). This verifies the entire negation and disproves the universal statement.
Choosing \(x=0\) would not work. Although \(0\leq2\), its square does not exceed \(4\). The conclusion fails there, but the hypothesis fails too, so the conditional is true at that input.
A phrase such as “Every real \(x\) with \(x>0\) satisfies \(B(x)\)” has this same conditional structure: \(\forall x\in\mathbb R,\ (x>0\to B(x))\). Its negation requires a real \(x\) with \(x>0\) and \(\neg B(x)\). The restriction \(x>0\) remains a requirement for the counterexample; it is not negated.
Empty Domains and Vacuous Truth
A universal assertion does not, by itself, assert that its domain contains any elements. If \(D=\varnothing\), the empty set, then \(\forall x\in D,\ A(x)\) is true. This is called vacuous truth: there is no allowed input violating the condition.
For a concrete example, let
No real number meets both restrictions, so \(D\) is empty. The statement “Every \(x\in D\) equals \(7\)” is true. Its negation, “There exists \(x\in D\) with \(x\neq7\),” is false because \(D\) contains no possible witness.
Likewise, a universal conditional is true if no input satisfies its hypothesis. Its negation would require an input where that hypothesis holds, so no counterexample is available.
Check Your Understanding
Write each negation as a complete quantified statement. When a counterexample is requested, verify both its domain membership and the failure of the original condition.
- Negate “Every real number \(x\) satisfies \(x>5\).” Give a counterexample at the boundary. Why is “Every real number is at most \(5\)” not the exact negation?
- Let \(D=\{1,2,3\}\). Negate “Every \(x\in D\) satisfies \(x^2>1\).” Which elements of \(D\) witness the negation? Does the existential statement require exactly one witness?
- Negate “Every real \(x\) satisfies \(x<0\) or \(x>1\).” Simplify the inner condition and give a witness. Why must both parts of your negated condition hold at the same input?
- Negate “For every real \(x\), if \(x^2=4\), then \(x=2\).” Explain why \(x=-2\) is a counterexample but \(x=0\) is not.
- Negate “Every positive real \(x\) satisfies \(x+1>1\).” Is the negation true? Why is \(x=-1\) not an admissible counterexample?
- Suppose \(D=\varnothing\). Determine the truth values of \(\forall x\in D,\ A(x)\), its negation, and \(\forall x\in D,\ \neg A(x)\). What does this reveal about moving “not” inside a universal statement without changing the quantifier?