# Quantifiers and Predicates

Mathematics I · Mathematical Language and Reasoning · https://tryals.app/learn/mathematics-i/quantifiers-and-predicates

## Predicates and Quantifiers

A **predicate** is a statement with a variable, like $P(x): x^2 \ge 0$, which becomes true or false only when $x$ is specified. **Quantifiers** bind variables over a **domain** to make definite claims.

| Quantifier | Symbol | Meaning | True Condition |
| --- | --- | --- | --- |
| **Universal** | $\forall x \, P(x)$ | For all $x$, $P(x)$ | Holds for every element |
| **Existential** | $\exists x \, P(x)$ | There exists $x$ | At least one element satisfies $P$ |

**Order matters** for mixed quantifiers. $\forall x \, \exists y \, P(x, y)$ means for every $x$ there is some $y$. This is weaker than $\exists y \, \forall x \, P(x, y)$, where one single $y$ works for every $x$.

**Example:** Over integers, "for every $x$ there is a larger $y$" is true, but "there is a $y$ larger than every $x$" is false. Swapping order can change truth value.

## Negating Quantifiers

**Negation** of quantifiers requires pushing the negation inward, flipping the quantifier and negating the predicate:

$$\neg \forall x \, P(x) \equiv \exists x \, \neg P(x)$$
$$\neg \exists x \, P(x) \equiv \forall x \, \neg P(x)$$

The negation of "all are $P$" is "some is not $P$." This is why a single **counterexample** disproves a universal claim.

> **Common pitfall:** $\neg \forall x \, P(x)$ means at least one is not $P$, **not** that none are $P$. Never swap $\forall x \, \exists y$ with $\exists y \, \forall x$, as order completely changes logical meaning.

## Practice questions

6 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Order the steps that negate 'for every integer x there exists an integer y with y > x'.

**Answer:**

1. Write the statement as: for all x, (there exists y, y > x)
2. Negate the outer quantifier: there exists x, NOT (there exists y, y > x)
3. Negate the inner quantifier: there exists x, for all y, NOT (y > x)
4. Simplify the inner predicate: there exists x, for all y, y is at most x

**Why:** Push the negation through one layer at a time: each quantifier it crosses flips, and only at the end does it act on the predicate itself. The result claims a largest integer exists, which is false, correctly making the original statement true.

Page: https://tryals.app/practice/mathematics-i/quantifiers-and-predicates/order-the-steps-that-negate-for-every-integer-x-there-exists-an

### 2. If 'there exists x with P(x)' is true, then 'for all x, P(x)' must also be true.

**Answer:** False

**Why:** False. Over the integers, 'there exists x with x = 0' is true while 'for all x, x = 0' is plainly false. An existential claim needs one witness; a universal claim needs the entire domain. The implication runs the other way, over a non-empty domain, universal implies existential.

Page: https://tryals.app/practice/mathematics-i/quantifiers-and-predicates/if-there-exists-x-with-p-x-is-true-then-for-all-x-p-x-must

### 3. Sort each statement by the quantifier it genuinely needs.

**Answer:**

- Universal (for all): Every prime above 2 is odd, No integer is both even and odd, Each square is non-negative
- Existential (there exists): Some integer is its own square, At least one even number is prime

**Why:** 'Every', 'each' and 'no' are universal, note that 'no integer is both even and odd' means 'for all x, NOT(...)'. 'Some' and 'at least one' are existential, each satisfied by a single witness (x = 0 or 1, and 2 respectively).

Page: https://tryals.app/practice/mathematics-i/quantifiers-and-predicates/sort-each-statement-by-the-quantifier-it-genuinely-needs

### 4. The negation of "for all $x$ there exists $y$ with $P(x,y)$" is "there exists $x$ such that for all $y$, not $P(x,y)$". Why do the quantifiers flip as well as the predicate?

A. Because only the innermost quantifier changes and the outer one is unaffected
B. Because negation reverses the order in which the variables were introduced
C. Because denying 'every' needs one failure and denying 'some' needs all
D. Because the predicate’s negation makes the quantifiers redundant

**Answer:** C. Because denying 'every' needs one failure and denying 'some' needs all

**Why:** Push the negation inward and each quantifier dualises. This is the mechanical rule behind every proof by counterexample: one witness refutes a universal claim.

Page: https://tryals.app/practice/mathematics-i/quantifiers-and-predicates/the-negation-of-for-all-x-there-exists-y-with-p-x-y-is-there

### 5. The domain is {1, 2, 3, 4, 5, 6}. Let P(x) be 'x is even'. Select every x that makes P(x) TRUE.

A. 4
B. 3
C. 6
D. 2
E. 5
F. 1

**Answer:** A. 4; C. 6; D. 2

**Why:** P(2), P(4) and P(6) are true; P(1), P(3), P(5) are false. Because P holds for some but not all of the domain, 'there exists x, P(x)' is true here while 'for all x, P(x)' is false, the same predicate, two different quantified verdicts.

Page: https://tryals.app/practice/mathematics-i/quantifiers-and-predicates/the-domain-is-1-2-3-4-5-6-let-p-x-be-x-is-even-select

### 6. Continuity is defined with ∀ε ∃δ, and *uniform* continuity swaps them to ∃δ ∀x. Why is that swap the whole difference between the two ideas?

A. Because ε and δ change roles between the two definitions
B. Because in the second, one δ must work everywhere at once
C. Because uniform continuity applies to a larger class of functions
D. Because the second version is stronger only for infinite domains

**Answer:** B. Because in the second, one δ must work everywhere at once

**Why:** Later variables may depend on earlier ones, so moving δ before x forbids it from adapting. 1/x is the standard witness: continuous at every point, and no single δ survives the approach to zero. Two symbols in a different order state genuinely different theorems.

Page: https://tryals.app/practice/mathematics-i/quantifiers-and-predicates/continuity-is-defined-with-and-uniform-continuity-swaps-them