When you negate a statement, ask: what would make this statement false? Keep the same domain of values and change the claim precisely. A contrapositive is different: it rewrites an implication as an equivalent implication.

Printable versions Questions PDF · Questions and solutions PDF · Download the complete ZIP (PDFs and editable LaTeX source).

1. Start with one condition

For a real number x, “x > 3” fails when x ≤ 3. Include equality: x = 3 does not satisfy the original. Likewise, the negation of x ≥ 3 is x < 3, and the negation of x ≠ 3 is x = 3.

2. Make “and” or “or” fail

If a statement requires both parts, breaking either part makes it false. If either part is enough, both must fail:

  • Negation of “P and Q”: “not P or not Q”.
  • Negation of “P or Q”: “not P and not Q”.

For example, “n is positive and even” is false when n ≤ 0 or n is odd.

3. Turn “every” into “some”

One exception defeats “every”. “Some” fails only when there are no examples. Keep the domain unchanged:

  • Negation of “Every x in the domain has property P”: “There is an x in the domain without P”.
  • Negation of “There is an x in the domain with P”: “Every x in the domain lacks P”.

With two quantifiers, change each in its original order. “For every x there is a y satisfying R” negates to “There is an x such that every y fails R”. A correct negation can itself be false.

4. Separate negation and contrapositive

“If P, then Q” fails precisely when P happens and Q does not. Its negation is “P and not Q”. Its contrapositive is “if not Q, then not P”, which has the same truth value as the original.

For every integer n, consider “if 4 divides n, then n is even”. Its negation says some integer is divisible by 4 and odd. Its contrapositive says for every integer, if n is odd, then 4 does not divide n. The converse, “if n is even, then 4 divides n”, fails at n = 2.

Before answeringIdentify the domain, check boundary values, and decide whether you need a negation or a contrapositive. “P only if Q” means if P, then Q.

Practice questions

Try each question before opening the hint. Then open the solution and check why the logical form works. The questions-only PDF has space for written answers.

A. Basic negations

Question 1

For a real number x, negate each statement: (a) x > 7; (b) x ≠ 2.

Hint

Include equality when a strict inequality fails. What is the only way for “not equal to 2” to fail?

Show solution

(a) x ≤ 7. The boundary x = 7 makes the original false. (b) x = 2.

Question 2

For an integer n, negate: “n is positive and even.”

Hint

How could either of the two requirements fail?

Show solution

n ≤ 0 or n is odd. Breaking either part of an “and” statement is enough.

Question 3

For real numbers a, b, negate: “a < 1 or b ≥ 4.”

Hint

An “or” statement fails only when both alternatives fail.

Show solution

a ≥ 1 and b < 4.

Question 4

For an integer n, negate: “n is divisible by 3 but not by 5.”

Hint

“But” joins the two conditions as “and” does.

Show solution

n is not divisible by 3 or n is divisible by 5.

B. Every, some and quantifiers

Question 5

Negate: “Every integer n ≥ 1 satisfies n2 > n.” If the negation is true, give a value of n that shows it.

Hint

One allowed exception is enough. Test the smallest positive integer and include equality.

Show solution

There exists an integer n ≥ 1 with n2 ≤ n. Take n = 1, since 12 = 1. The original claim is false.

Question 6

Negate: “There is a real number x such that x2 + 1 = 0.” Decide which statement is true.

Hint

“There is” becomes “for every”. Can a real square be negative?

Show solution

For every real x, x2 + 1 ≠ 0. This negation is true because x2 ≥ 0, so x2 + 1 ≥ 1.

Question 7

Negate: “For every integer n, there is an integer m > n.” Decide which statement is true.

Hint

Change both quantifiers in order. After choosing n, could m = n + 1 defeat the negation?

Show solution

There exists an integer n such that every integer m satisfies m ≤ n. This is false: for any n, m = n + 1 is greater. The original is true.

Question 8

Negate: “Some student solved every question on the worksheet.”

Hint

For the original to fail, what must be true of each student?

Show solution

Every student has at least one question on the worksheet that they did not solve. “Some student, every question” becomes “every student, some question” when negated.

C. Implications and contrapositives

Question 9

For every integer n: “If 8 divides n, then n is even.” Write (a) the negation and (b) the contrapositive.

Hint

For the negation, keep the hypothesis and make the conclusion fail. For the contrapositive, swap and negate both parts.

Show solution

(a) There exists an integer n divisible by 8 that is odd. This is false. (b) For every integer n, if n is odd, then 8 does not divide n. This is true, like the original: 8k = 2(4k) is even.

Question 10

For every real x: “If x2 > 9, then x > 3.” (a) Find a counterexample. (b) Negate the statement. (c) Write its contrapositive.

Hint

Look at negative numbers whose square exceeds 9. The negation needs one example with the hypothesis true and conclusion false.

Show solution

(a) x = −4: its square is 16 > 9, but −4 ≤ 3. (b) There exists a real x with x2 > 9 and x ≤ 3. (c) For every real x, if x ≤ 3, then x2 ≤ 9. This contrapositive is also false at −4.

Question 11

“An integer is divisible by 12 only if it is divisible by 3.” (a) Rewrite this as an if-then statement. (b) Give its contrapositive. (c) Is its converse true?

Hint

“P only if Q” means P ⇒ Q. Try 3 when testing the converse.

Show solution

(a) For every integer n, if 12 divides n, then 3 divides n. (b) If 3 does not divide n, then 12 does not divide n. (c) The converse is false: 3 divides 3, but 12 does not.

D. Short proofs

Question 12

For an integer n, prove by contrapositive: “If n2 is odd, then n is odd.”

Hint

Start with n even. Write n = 2k and examine its square.

Show solution

The contrapositive is: if n is even, then n2 is even. Write n = 2k for an integer k. Then n2 = 4k2 = 2(2k2), which is even. Thus the original implication follows.

Question 13

For integers a, b, prove by contrapositive: “If ab is odd, then both a and b are odd.”

Hint

Negate the conclusion first: at least one factor is even. What happens to the product?

Show solution

The contrapositive is: if a is even or b is even, then ab is even. If a = 2k, then ab = 2(kb); if b = 2k, then ab = 2(ka). In either case the product is even. This proves the original.

Final checkDid you keep the domain? Did equality move to the correct side? Did “every” change to “some”, and “some” to “every”? For an implication, did you distinguish its negation from its contrapositive?

Open the complete questions and solutions PDF or download the ZIP with PDFs and editable LaTeX.

These are original teaching notes and practice questions, not official TMUA questions.