Mathematical-Reasoning · Unit 3 · Video 1 · Interactive Practice

And, Or, Not, If-Then: One Truth Rule for Each Connective

IKey Formulas

Operator	Symbol	Truth rule	Key row
Conjunction	$P \wedge Q$	true only when both $P$ and $Q$ are true	$TT$ : true
Disjunction	$P \vee Q$	false only when both $P$ and $Q$ are false	$FF$ : false
Negation	$\neg P$	the opposite truth value of $P$	none
Conditional	$P \to Q$	false only when $P$ is true and $Q$ is false	$TF$ : false

Key Insight: In mathematics “or” is the inclusive or, so $P \vee Q$ is true when both parts are true. The exclusive or of everyday speech is false in that row.

IIOne Rule, One Exceptional Row

Conjunction, disjunction and conditional each single out one row of the table; negation simply flips $P$ .

Operator

IIIBuilding Statements About an Integer

A compound statement's truth value is computed from its parts: negations first, then the connective joining them.

n Left part Connective Right part

IVNegating an Inequality

The negation of $a < 9$ must be true exactly where $a < 9$ is false, and that includes $a = 9$ .

a 12 Proposed negation

💡 Challenge: one proposed negation fails at a single value of $a$ only; find that value.

VQuiz Questions

Problem 1 · Evaluate the Connectives

Given: $P$ : “ $21$ is odd” and $Q$ : “ $21$ is prime.” Which of these statements is true?

✅ Correct!

Q

is false because

21 = 3 \cdot 7

, so

Q \to \neg P

has a false hypothesis, and a conditional with a false hypothesis is true.

❌ Not quite. A conjunction needs both parts true, but

Q

is false:

21 = 3 \cdot 7

is not prime.

❌ Not quite. Here

P

is true and

Q

is false, which is exactly the one row that makes a conditional false.

❌ Not quite.

\neg Q

is true, but

\neg P

is false because

21

is odd, and a conjunction needs both parts true.

Show solution

$P$ is true, since $21 = 2 \cdot 10 + 1$ . $Q$ is false, since $21 = 3 \cdot 7$ . So $\neg P$ is false and $\neg Q$ is true.

$P \wedge Q$ : one part ( $Q$ ) is false, so the conjunction is false.
$P \to Q$ : true hypothesis, false conclusion, so the conditional is false.
$\neg P \wedge \neg Q$ : the first part ( $\neg P$ ) is false, so the conjunction is false.
$Q \to \neg P$ : the hypothesis $Q$ is false, so the conditional is true (whatever $\neg P$ is).

Only (d) is true.

Problem 2 · The Negation of “Less Than”

Given: the statement $17 < 3 \cdot 6$ . Which statement is its negation, written without the word “not”?

✅ Correct! “Less than” fails when the first number is larger and also when the two are equal, so the negation is “greater than or equal to.”

❌ Close, but…

17 > 3 \cdot 6

leaves out equality. If the two sides were equal, both

17 < 3 \cdot 6

and

17 > 3 \cdot 6

would be false, so they cannot be negations of each other.

❌ Not quite.

17 \le 3 \cdot 6

says

17 \le 18

, which is true, just like the original

17 < 18

. A negation must have the opposite truth value.

❌ Not quite.

17 \ne 3 \cdot 6

says

17 \ne 18

, which is true, the same truth value as

17 < 18

. “Not equal” is the negation of “equal,” not of “less than.”

Show solution

Since $3 \cdot 6 = 18$ , the statement is $17 < 18$ , which is true. Its negation must be false.

For numbers $x$ and $y$ , the statement $x < y$ is false exactly when $x > y$ or $x = y$ , that is, when $x \ge y$ . So the negation of $17 < 3 \cdot 6$ is

17 \ge 3 \cdot 6,

which is false, as it should be. Options (b) and (d) say $17 \le 18$ and $17 \ne 18$ , which are both true; option (a) misses the case of equality.

Problem 3 · Build a Truth Table Column

Given: the compound statement $(P \vee Q) \wedge \neg(P \wedge Q)$ . Find its truth values in the row order $P, Q = TT,\ TF,\ FT,\ FF$ .

✅ Correct! The statement is true exactly when one of

P

Q

is true and the other false: it is the exclusive or, built from

\vee

\wedge

and

\neg

❌ Not quite. That is the column of

P \vee Q

alone. In row

TT

the second part

\neg(P \wedge Q)

is false, so the conjunction is false there.

❌ Not quite. That is the column of

P \wedge Q

. The statement negates

P \wedge Q

before using it.

❌ Not quite. That is the column of

\neg(P \wedge Q)

alone. In row

FF

the first part

P \vee Q

is false, so the conjunction is false there.

Show solution

Work column by column, innermost operators first.

$P$	$Q$	$P \vee Q$	$P \wedge Q$	$\neg(P \wedge Q)$	whole
$T$	$T$	$T$	$T$	$F$	$F$
$T$	$F$	$T$	$F$	$T$	$T$
$F$	$T$	$T$	$F$	$T$	$T$
$F$	$F$	$F$	$F$	$T$	$F$

The last column is a conjunction, so it is true only where both $P \vee Q$ and $\neg(P \wedge Q)$ are true: rows $TF$ and $FT$ . The column is $F, T, T, F$ , the same as the exclusive or.

Problem 4 · From English to Symbols

Given: $P$ : “ $9$ is odd” and $R$ : “ $9 < 5$ .” Translate the sentence “ $9$ is even or $9 \ge 5$ ” into symbols, then find its truth value.

Which symbolic form?

What is its truth value?

✅ Correct! The sentence is

\neg P \vee \neg R

; its second part

\neg R

(“

9 \ge 5

”) is true, and one true part makes a disjunction true.

❌ Check the first part. For an integer, “even” means “not odd,” so “

9

is even” is

\neg P

, not

P

❌ Check the connective. The sentence joins its parts with “or,” which is

\vee

, not

\wedge

❌ Check the second part.

R

is “

9 < 5

,” but the sentence says “

9 \ge 5

,” which means

9

is not less than

5

: that is

\neg R

, not

R

❌ Check the translation. “

9

is even” is

\neg P

and “

9 \ge 5

” is

\neg R

❌ Not quite. Here

\neg P

is false and

\neg R

is true, so exactly one part is true, and both kinds of “or” give true. The two meanings differ only when both parts are true; in mathematics “or” is inclusive in any case.

❌ Check the truth value.

\neg R

is “

9 \ge 5

,” which is true, and a disjunction with a true part is true.

Show solution

Translate. “ $9$ is even” means $9$ is not odd, which is $\neg P$ . “ $9 \ge 5$ ” means $9$ is not less than $5$ , which is $\neg R$ . The parts are joined by “or,” so the sentence is

\neg P \vee \neg R.

Evaluate. $P$ is true ( $9 = 2 \cdot 4 + 1$ ), so $\neg P$ is false. $R$ : $9 < 5$ is false, so $\neg R$ is true. A disjunction is false only when both parts are false; here one part is true, so the sentence is true.

Solved: 0 / 4

$P$	$Q$	$P \vee Q$	$P \wedge Q$	$\neg(P \wedge Q)$	whole
$T$	$T$	$T$	$T$	$F$	$F$
$T$	$F$	$T$	$F$	$T$	$T$
$F$	$T$	$T$	$F$	$T$	$T$
$F$	$F$	$F$	$F$	$T$	$F$

$P$	$Q$	$P \vee Q$	$P \wedge Q$	$\neg(P \wedge Q)$	whole
$T$	$T$	$T$	$T$	$F$	$F$
$T$	$F$	$T$	$F$	$T$	$T$
$F$	$T$	$T$	$F$	$T$	$T$
$F$	$F$	$F$	$F$	$T$	$F$

$P$	$Q$	$P \vee Q$	$P \wedge Q$	$\neg(P \wedge Q)$	whole
$T$	$T$	$T$	$T$	$F$	$F$
$T$	$F$	$T$	$F$	$T$	$T$
$F$	$T$	$T$	$F$	$T$	$T$
$F$	$F$	$F$	$F$	$T$	$F$