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

Conditional Statements and the Truth Table for P → Q

IKey Formulas

Formula	Name	Meaning
$P \to Q$	Conditional statement	Read “if $P$ , then $Q$ ”; $P$ is the hypothesis, $Q$ the conclusion
$P$ true, $Q$ false $\;\Rightarrow\;$ $P \to Q$ false	The only false row	An example of this kind is a counterexample
$P$ false $\;\Rightarrow\;$ $P \to Q$ true	Rows three and four	A false hypothesis can never make the conditional false

Key Insight: Down the $P \to Q$ column the truth table reads T, F, T, T. An example with a true hypothesis and a true conclusion supports a conditional without proving it; one counterexample proves it false.

IIVisualization 1 — Four Cases, One Lie

Each weather-and-Daisy situation is one row of the truth table, and only one row makes the promise a lie.

IIIVisualization 2 — Which Row Does an Example Land In?

For “if $x$ is positive, then $x^2 + 8x$ is positive,” every value of $x$ lands in exactly one row.

x −3

💡 No value of $x$ lands in row two, yet the examples alone prove nothing; the proof is that for $x > 0$ both $x^2$ and $8x$ are positive, so their sum is too.

IVVisualization 3 — Forty Primes, Then a Counterexample

Forty true cases of “ $n^2 - n + 41$ is prime” prove nothing; one false case disproves the conditional.

n 1

VQuiz Questions

Problem 1 · Reading the Truth Table

Given: the conditional “If $3 > 5$ , then $7$ is even.” — find its truth value.

✅ Correct! The hypothesis

3 > 5

is false, so this is row four of the truth table, where

P \to Q

is true.

❌ Not quite. A false conclusion makes

P \to Q

false only when the hypothesis is true (row two). Here

3 > 5

is false.

❌ Not quite.

P

false and

Q

false is row four, and the last column of that row is T, not F.

❌ Not quite.

P

and

Q

are statements, so

P \to Q

has its own truth value in every one of the four cases, including those with a false hypothesis.

Show solution

Let $P$ be “ $3 > 5$ ” and $Q$ be “ $7$ is even.” Both are false.

A conditional is false only when $P$ is true and $Q$ is false. That is not the case here: the hypothesis is false, so we are in row four ( $P$ false, $Q$ false), where $P \to Q$ is true. The conditional makes no claim about what happens when $3 > 5$ is false, so nothing it said is contradicted.

Problem 2 · Spotting the Counterexample

Given: “If $x$ is a real number with $x > 2$ , then $x^2 > 9$ .” — find the value of $x$ that is a counterexample.

✅ Correct!

\tfrac{5}{2} > 2

makes the hypothesis true, but

\left(\tfrac{5}{2}\right)^2 = \tfrac{25}{4} = 6.25

is not greater than

9

: row two.

❌ Not quite. This value is not greater than

2

, so the hypothesis is false. An example with a false hypothesis sits in row three or four and can never make the conditional false.

❌ Not quite.

5 > 2

and

5^2 = 25 > 9

: hypothesis true, conclusion true. That is row one, which supports the conditional rather than disproving it.

Show solution

A counterexample needs a true hypothesis and a false conclusion.

$x = 1$ and $x = -4$ : $x > 2$ is false, so the hypothesis fails; these cannot be counterexamples (even though $(-4)^2 = 16 > 9$ ).
$x = 5$ : $5 > 2$ and $25 > 9$ , so both parts are true (row one).
$x = \tfrac{5}{2}$ : $\tfrac{5}{2} > 2$ is true, while $\tfrac{25}{4} > 9$ is false (row two).

So $x = \tfrac{5}{2}$ is a counterexample, and this single example proves the conditional false.

Problem 3 · Placing Examples on the Table

Given: the video’s conditional “If $x$ is a positive real number, then $x^2 + 8x$ is a positive real number.” — find the truth-table row each example occupies.

Which row does $x = -10$ occupy?

Which row does $x = -8$ occupy?

✅ Correct! Both examples have a false hypothesis, so neither can make the conditional false, whatever the value of

x^2 + 8x

❌ Not quite. A negative

x

is not a positive real number, so the hypothesis is false. Rows one and two need a true hypothesis.

❌ Not quite.

(-8)^2 + 8(-8) = 64 - 64 = 0

, and

0

is not positive, so the conclusion is false too.

❌ Not quite. Compute

(-10)^2 + 8(-10)

and check its sign;

-10

itself is not positive.

❌ Not quite.

P

is false, but

(-10)^2 + 8(-10) = 100 - 80 = 20 > 0

, so

Q

is true: that is row three, not row four.

❌ Not quite.

-8

is not positive; now compute

(-8)^2 + 8(-8)

and decide whether it is positive.

Show solution

Let $P$ be “ $x$ is a positive real number” and $Q$ be “ $x^2 + 8x$ is a positive real number.”

$x = -10$ : $P$ is false. $(-10)^2 + 8(-10) = 100 - 80 = 20 > 0$ , so $Q$ is true. Row three: $P \to Q$ is true.

$x = -8$ : $P$ is false. $(-8)^2 + 8(-8) = 64 - 64 = 0$ , which is not positive, so $Q$ is false. Row four: $P \to Q$ is true.

With a false hypothesis the conditional is true whatever the conclusion, so neither example tells us anything against it.

Problem 4 · A Prime-Generating Polynomial

Given: “If $n$ is a positive integer, then $n^2 + n + 17$ is a prime number.” — find the choice that settles whether it is true.

✅ Correct!

16^2 + 16 + 17 = 289 = 17 \times 17

is not prime, and

16

is a positive integer: a counterexample, so the conditional is false.

❌ Not quite.

15^2 + 15 + 17 = 257

, which is prime. A true hypothesis with a true conclusion supports the conditional but settles nothing.

❌ Not quite.

0

is not a positive integer, so the hypothesis is false, and a false hypothesis can never disprove a conditional (besides,

17

is prime).

❌ Not quite. Fifteen cases in a row are evidence, not proof, just as forty cases of

n^2 - n + 41

were. Keep going to

n = 16

Show solution

For $n = 1, 2, \dots, 15$ the values are $19$ , $23$ , $29$ , $37$ , $47$ , $59$ , $73$ , $89$ , $107$ , $127$ , $149$ , $173$ , $199$ , $227$ , $257$ , all prime. That supports the conditional but cannot prove it.

At $n = 16$ the hypothesis is true, and

16^2 + 16 + 17 = 16(16 + 1) + 17 = 16 \cdot 17 + 17 = 17 \cdot 17 = 289,

which has the factor $17$ other than $1$ and itself, so it is not prime. Hypothesis true, conclusion false: row two. So $n = 16$ is a counterexample and the conditional is false. (This mirrors $41^2 - 41 + 41 = 41 \cdot 41$ in the video.)

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