Mathematical-Reasoning · Unit 7 · Video 3 · Interactive Practice

Proving Biconditionals: Two Conditionals, Each Marked

IKey Formulas

Formula	Name	What it tells you
$(P \leftrightarrow Q) \equiv (P \to Q) \wedge (Q \to P)$	Biconditional as two conditionals	Prove $P \leftrightarrow Q$ by proving both $P \to Q$ and $Q \to P$
$P$ only if $Q$ $\;\equiv\;$ $P \to Q$	The "only if" part	$P$ cannot hold without $Q$
$P$ if $Q$ $\;\equiv\;$ $Q \to P$	The "if" part	$Q$ is enough to guarantee $P$
"For the first part, we assume …" / "This completes the first part."	Part markers	Where each part begins and ends, and which assumption is in force

Key Insight: Each part is an ordinary conditional statement, so each may use its own method: a direct proof, a proof by contrapositive, or a citation of a theorem already proved.

IIVisualization 1 — Which Direction Does the Sentence Say?

" $P$ if $Q$ " and " $P$ only if $Q$ " point in opposite directions between the same two statements.

IIIVisualization 2 — What Does Each Direction Prove About the Solutions?

For real $x$ , substitution and factoring each prove one half of $x = 2 \iff x^3 - 2x^2 + x = 2$ .

💡 Part 2 needs $x \in \mathbb{R}$ : over $\mathbb{C}$ , $x = i$ gives $i^3 - 2i^2 + i = -i + 2 + i = 2$ , so the equation has solutions other than $2$ .

IVVisualization 3 — Writing the Proof With Each Part Marked

For each integer $n$ : $n$ is even $\iff$ $n^2$ is even, written as two marked parts.

VQuiz Questions

Problem 1 · Name the "If" Part

Given: For each integer $n$ , $n$ is even if and only if $n^2$ is even. Let $P$ be " $n$ is even" and $Q$ be " $n^2$ is even". Which statement is the "if" part?

✅ Correct! "

P

Q

" says

Q

is enough for

P

, which is

Q \to P

: if

n^2

is even, then

n

is even.

❌ That is the other direction.

P \to Q

is "

P

only if

Q

", the "only if" part.

❌ Not quite.

P \wedge Q

is a conjunction, not a conditional; it is false for

n = 3

, while each part of the biconditional is a conditional statement.

❌ Not quite. The "if" part has hypothesis

Q

and conclusion

P

; this option changes the conclusion to "not

P

Show solution

Split $P \leftrightarrow Q$ into its two conditionals:

P \leftrightarrow Q \;\equiv\; (P \to Q) \wedge (Q \to P).

" $P$ if $Q$ " means $Q$ guarantees $P$ , so it is $Q \to P$ : if $n^2$ is even, then $n$ is even. This is the if part.

" $P$ only if $Q$ " means $P$ cannot hold without $Q$ , so it is $P \to Q$ : if $n$ is even, then $n^2$ is even. This is the only if part.

Problem 2 · A Proof That Stops Halfway

Given: A student "proves" that for real $x$ , $x = 2 \iff x^3 - 2x^2 + x = 2$ , by writing: "Assume $x = 2$ . Then $2^3 - 2 \cdot 2^2 + 2 = 8 - 8 + 2 = 2$ , so the equation holds. Therefore $x = 2$ if and only if $x^3 - 2x^2 + x = 2$ ." What is wrong?

✅ Correct! Substitution proves

x = 2 \Rightarrow

equation, so

2

is a solution. It leaves room for other solutions.

❌ Check the direction. The student assumed

x = 2

and concluded the equation: that is

P \to Q

, "

x = 2

only if the equation holds".

❌ Not quite. Substitution shows

2

is a solution, but not that no other real number is. That needs the converse.

❌ The arithmetic is fine.

8 - 8 + 2 = 2

. The problem is which conditional was proved.

Show solution

Let $P$ be " $x = 2$ " and $Q$ be " $x^3 - 2x^2 + x = 2$ ". The student assumed $P$ and derived $Q$ , which proves $P \to Q$ , the only if part.

The if part, $Q \to P$ , is never proved: assume $x^3 - 2x^2 + x = 2$ and show $x = 2$ . That requires solving:

x^3 - 2x^2 + x - 2 = x^2(x - 2) + (x - 2) = (x - 2)(x^2 + 1) = 0,

and since $x^2 + 1 \ge 1$ for real $x$ , the factor $x - 2$ must be $0$ . Only with both parts is the biconditional proved, and the write-up should mark where each part begins and ends.

Problem 3 · The Second Part by Factoring

Given: For real $x$ , prove $x = 3 \iff x^3 - 3x^2 + 4x = 12$ . In the second part, assume $x^3 - 3x^2 + 4x = 12$ and subtract $12$ from both sides.

Factoring by grouping gives which equation?

Why does that equation force $x = 3$ ?

✅ Correct!

(x - 3)(x^2 + 4) = 0

and

x^2 + 4 \ge 4

, so

x = 3

: every real solution is

3

❌ Check the grouping.

x^3 - 3x^2 = x^2(x - 3)

and

4x - 12 = 4(x - 3)

; expand your choice to test it.

❌ Not quite. A product is

0

only if a factor is

0

, so rule out the factor

x^2 + 4

❌ Check the signs.

(x - 3)(x^2 - 4)

expands to

x^3 - 3x^2 - 4x + 12

; the second group is

4x - 12 = 4(x - 3)

, so the factor is

x^2 + 4

❌ Check the signs.

(x + 3)(x^2 - 4)

expands to

x^3 + 3x^2 - 4x - 12

; grouping gives

x^2(x - 3) + 4(x - 3)

, with the common factor

x - 3

❌ Check the common factor.

(x + 3)(x^2 + 4)

expands to

x^3 + 3x^2 + 4x + 12

; since

x^3 - 3x^2 = x^2(x - 3)

, the common factor is

x - 3

, not

x + 3

❌ Not allowed here. Dividing by

x - 3

assumes

x - 3 \neq 0

, which is exactly what is unknown, and it leads to

x^2 + 4 = 0

, which no real number satisfies.

❌ Check that equation.

(\pm 2)^2 + 4 = 8

; for real

x

x^2 + 4 = 0

has no solution at all, because

x^2 \ge 0

❌ That is the first part. Substituting

3

shows

3

is a solution; it does not show that no other real number is.

Show solution

We prove two conditional statements: if $x = 3$ , then $x^3 - 3x^2 + 4x = 12$ ; and if $x^3 - 3x^2 + 4x = 12$ , then $x = 3$ .

First part. Assume $x = 3$ . Then $3^3 - 3 \cdot 3^2 + 4 \cdot 3 = 27 - 27 + 12 = 12$ . This completes the first part.

Second part. Assume $x^3 - 3x^2 + 4x = 12$ . Then

x^3 - 3x^2 + 4x - 12 = 0

x^2(x - 3) + 4(x - 3) = 0

(x - 3)(x^2 + 4) = 0.

If a product of two real numbers is $0$ , one factor is $0$ . Since $x^2 \ge 0$ , we have $x^2 + 4 \ge 4$ , so $x^2 + 4 \neq 0$ . Hence $x - 3 = 0$ and $x = 3$ . This completes the second part.

Both conditional statements are proved, so the biconditional is proved. Dividing by $x - 3$ (option d) is not allowed when $x - 3$ might be $0$ , and it throws away exactly the solution $x = 3$ .

Problem 4 · Test Each Part Separately

Given: The statement "For each real number $x$ , $x = 2$ if and only if $x^2 - 2x = 0$ ." Which description is correct?

✅ Correct!

x = 0

satisfies

x^2 - 2x = 0

but is not

2

, so the "if" part fails and the biconditional is false.

❌ Directions swapped. The "only if" part is

x = 2 \Rightarrow x^2 - 2x = 0

, which is true:

4 - 4 = 0

❌ Not quite. Factor:

x^2 - 2x = x(x - 2)

, so the equation has a second solution.

❌ Check the hypothesis. For

x = -2

x^2 - 2x = 4 + 4 = 8 \neq 0

, so the hypothesis is false and

x = -2

is not a counterexample.

Show solution

Let $P$ be " $x = 2$ " and $Q$ be " $x^2 - 2x = 0$ ". Test each part on its own.

Only if part, $P \to Q$ : if $x = 2$ , then $x^2 - 2x = 4 - 4 = 0$ . True.

If part, $Q \to P$ : $x^2 - 2x = x(x - 2) = 0$ gives $x = 0$ or $x = 2$ . For $x = 0$ the hypothesis $Q$ is true and the conclusion $P$ is false, so $x = 0$ is a counterexample.

Since $(P \leftrightarrow Q) \equiv (P \to Q) \wedge (Q \to P)$ and $Q \to P$ fails, the biconditional is false.

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