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Before you can prove a statement, you must know exactly what it claims. This lesson separates definitions, propositions and theorems, reads a conditional “if P then Q” as a single promise, distinguishes its converse, inverse and contrapositive, translates necessary and sufficient conditions, and negates statements with quantifiers such as “for all” and “there exists”, including a full definition. These are the skills every later proof in H3 relies on.
Definitions, propositions and theorems
H3 Mathematics asks you to read and write statements as a mathematician does. A definition fixes the meaning of a word: an integer n is even when n = 2k for some integer k, and odd when n = 2k + 1 for some integer k. A proposition is a statement that is either true or false. A theorem is an important proposition that has been proved; a smaller one used on the way is often called a lemma.
A statement needs every symbol to be accounted for. “x² > 4” is neither true nor false until you say which x is meant. Adding a quantifier, “for all real x” or “there is a real x”, turns it into a proposition.
A conditional makes one promise
“If P, then Q” (written P ⇒ Q) promises that Q holds whenever P holds. It says nothing about what happens when P fails. So it is false in exactly one situation: P true and Q false.
| P | Q | P ⇒ Q |
|---|---|---|
| true | true | true |
| true | false | false |
| false | true | true |
| false | false | true |
Example: for integers n, “if 6 divides n, then 3 divides n” is true: n = 6k = 3(2k). The case n = 9 has P false and Q true, and does not break the promise.
A conditional whose hypothesis can never hold is vacuously true. “If n is an integer with n² = 2, then n = 7” is true, because no integer case makes the hypothesis true.
Converse, inverse and contrapositive
From P ⇒ Q we form three related statements.
| Name | Form |
|---|---|
| Converse | Q ⇒ P |
| Inverse | not P ⇒ not Q |
| Contrapositive | not Q ⇒ not P |
The contrapositive is false exactly when not Q is true and not P is false, which is the same situation that makes the original false. So a conditional and its contrapositive are logically equivalent. The converse and the inverse are contrapositives of each other, so they are equivalent to each other, but not, in general, to the original.
Worked example 1
Four statements, two truth values
Worked example 1
Four statements, two truth values
For integers n, consider “if 6 divides n, then 3 divides n”. Write its converse, inverse and contrapositive, and decide whether each is true.
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Write the three forms
Method
Swap, negate, or swap and negate the hypothesis and conclusion.Reason
Each form is a new statement with its own falsifying case.Working
Answer
- Converse: if 3 divides n, then 6 divides n.
- Inverse: if 6 does not divide n, then 3 does not divide n.
- Contrapositive: if 3 does not divide n, then 6 does not divide n.
Decide each truth value
Method
The contrapositive shares the original’s truth value; the converse and inverse share theirs.Reason
One verified counterexample is enough to show a universal statement is false.Working
Answer
The original and contrapositive are true. The converse is false: n = 3 is divisible by 3 but not by 6. The same n = 3 shows the inverse is false.
Quick check
Each counterexample must make the hypothesis true and the conclusion false. n = 4 does not disprove the converse, because 3 does not divide 4.
Necessary and sufficient conditions
When P ⇒ Q is true, P is sufficient for Q (having P is enough), and Q is necessary for P (you cannot have P without Q). “P if and only if Q”, written P ⇔ Q, means both P ⇒ Q and Q ⇒ P: each condition is necessary and sufficient for the other.
For real x, x > 2 is sufficient for x² > 4, but not necessary: x = -3 has x² > 4 without x > 2. For real numbers, x² > 4 ⇔ (x > 2 or x < -2).
Check your understanding 1
Necessary, sufficient, both or neither
Check your understanding 1
Necessary, sufficient, both or neither
In each pair, is the first condition necessary, sufficient, both or neither for the second? Give a counterexample for each direction that fails.
- For an integer n: “n is divisible by 4” and “n is even”.
- For real a, b, c with a ≠ 0: “b²-4ac ≥ 0” and “ax² + bx + c = 0 has a real root”.
- For real x: “x² = x” and “x = 1”.
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- Sufficient, not necessary: 4k = 2(2k) is even, but n = 2 is even and not divisible by 4.
- Both: the quadratic formula gives a real root exactly when the discriminant is nonnegative.
- Necessary, not sufficient: x = 1 gives x² = x, but x = 0 satisfies x² = x without x = 1.
Quantifiers and their negations
“For all” (∀) and “there exists” (∃) say how many cases a statement covers. “There exists a unique” (∃!) claims existence and that any two such objects are equal.
To negate, swap the quantifier and negate what follows:
Connectives follow De Morgan’s laws: the negation of “P and Q” is “not P or not Q”, and the negation of “P or Q” is “not P and not Q”. The negation of P ⇒ Q is “P and not Q”, not another conditional.
Quantifier order matters. Over the real numbers, “∀ x ∃ y: y > x” is true (take y = x + 1), but “∃ y ∀ x: y > x” is false: no single real number exceeds every real number.
Spot the mistake 1
The opposite is not the negation
Spot the mistake 1
The opposite is not the negation
A student writes that the negation of “every prime number is odd” is “every prime number is even”, and concludes that the original statement is true because the negation is false. Explain the error, write the correct negation, and decide which of the two is true.
Write your answer, then compare it with the solution
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Swap the quantifier
Method
The original is “∀ primes p, p is odd”.Reason
Its negation must be true exactly when the original fails, and the original fails as soon as one prime is even.Working
Answer
The negation is “there exists a prime number that is even”. It is true, since 2 is prime and even, so the original statement is false.
Common mistake
“Every prime is even” is the opposite claim, not the negation. Both the statement and its opposite can be false at once, as here: 3 is an odd prime and 2 is an even prime. A statement and its negation always have different truth values.
Nested quantifiers are negated one at a time from the outside. “For every ε > 0 there is an N such that |aₙ| < ε for all n > N” says that aₙ → 0. Its negation is: there is an ε > 0 such that for every N there is some n > N with |aₙ| ≥ ε.
Check your understanding 2
One statement, four forms
Check your understanding 2
One statement, four forms
Consider “for all real x, if x² < 1 then x < 1”. Write its converse, its contrapositive and its negation, and decide whether each of the four statements is true.
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- Original: true, because x² < 1 means -1 < x < 1.
- Converse, “for all real x, if x < 1 then x² < 1”: false, for example x = -2.
- Contrapositive, “for all real x, if x ≥ 1 then x² ≥ 1”: true, as it must be.
- Negation, “there is a real x with x² < 1 and x ≥ 1”: false, because the original is true.
The quantifier “for all” stays at the front of the converse and contrapositive; only the negation changes it.
Negating a definition
Common mistakes
- Treating the converse of a true conditional as automatically true.
- Negating “for all x, P(x)” as “for all x, not P(x)”; the negation is “there exists x with not P(x)”.
- Negating “P and Q” as “not P and not Q” instead of “not P or not Q”.
- Confusing “P is necessary for Q” (Q ⇒ P) with “P is sufficient for Q” (P ⇒ Q).
- Changing the order of nested quantifiers, which can change the meaning completely.
Before you move on
Try it yourself 1
What it means not to be increasing
Try it yourself 1
What it means not to be increasing
A function f:R → R is called strictly increasing when, for all real a and b, a < b ⇒ f(a) < f(b).
- Write the negation of this definition without using the word “not” in front of a quantifier.
- Use your negation to decide whether f(x) = x³-3x is strictly increasing.
- Decide whether “there exists a unique real x with x³ = x” is true, and justify your answer.
Write your answer, then compare it with the solution
Hint 1
Negate the conditional inside the quantifiers
“For all a, b” becomes “there exist a, b”. The conditional a < b ⇒ f(a) < f(b) becomes “a < b and f(a) ≥ f(b)”.
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Negate the definition
Method
Swap the quantifiers and negate the conditional.Reason
The negation of P ⇒ Q is “P and not Q”.Working
Answer
f is not strictly increasing exactly when there exist real a and b with a < b and f(a) ≥ f(b).
Find a witness
Method
Look for a < b with f(a) ≥ f(b).Reason
An existence statement is proved by exhibiting one example and checking it.Working
Answer
Take a = -1 and b = 1. Then f(-1) = -1 + 3 = 2 and f(1) = 1-3 = -2, so a < b and f(a) ≥ f(b). So f is not strictly increasing.
Test the uniqueness claim
Method
“Unique” needs existence and no second solution.Reason
Two different solutions are enough to refute uniqueness.Working
Answer
x³-x = x(x-1)(x + 1) = 0 has the three solutions -1, 0 and 1. A solution exists, but it is not unique, so the statement is false.
Quick check
A sketch that “looks increasing” over part of the line is not enough either way: the negation needs one checked pair. Compare your three answers with the solution; written reasoning here is self-checked, not marked.
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