# Biconditional Statement

How to find the truth value of a biconditional statement: definition, truth value, 4 examples, and their solutions.

## Definition

### Definition

[p ↔ q] is the biconditional statement.

It is the conjunction

of a conditional [p → q] and its converse [q → p].

It looks like below:

p if and only if q,

p iff. q.

## Truth Value

### Truth Table

[p ↔ q] is

the conjunction of [p → q] and [q → p].

So if both [p → q] and [q → p] are all true,

then [p ↔ q] is true.

Otherwise, [p ↔ q] is false.

## Example 1

### Example

### Solution

The structure of the given statement is

[... if and only if ...].

It's a biconditional statement.

So the former statement is p:

2 is a prime number.

And the latter statement is q:

2 is an even number.

p and q seems to have one truth value.

So start from p and q.

p: 2 is a prime number.

This is true.

q: 2 is an even number.

This is true.

p is true.

q is true.

Then p → q is true.

And q → p is true.

Conditional Statement: Truth Value

p → q is true.

q → p is true.

Then the conjunction of these two,

p ↔ q,

is true.

So true is the answer.

## Example 2

### Example

### Solution

The structure of the given statement is

[... iff. ...].

It's a biconditional statement.

So the former statement is p:

∠A is a right angle.

And the latter statement is q:

m∠A = 90.

p and q can be either true or false.

So start from p → q and q → p.

p → q:

If ∠A is a right angle,

then m∠A = 90.

This is true.

q → p:

If m∠A = 90,

then ∠A is a right angle.

This is also true.

p → q is true.

q → p is true.

Then the conjunction of these two,

p ↔ q,

is true.

So true is the answer.

Just like this example,

a biconditional statement can be used

to show the definition of something.

## Example 3

### Example

### Solution

The structure of the given statement is

[... iff. ...].

It's a biconditional statement.

So the former statement is p:

x + 2 = 3.

And the latter statement is q:

x = 1.

p and q can be either true or false.

So start from p → q and q → p.

p → q:

If x + 2 = 3,

then x = 1.

From p, x + 2 = 3,

move +2 to the right side.

Then x = 3 - 2 = 1.

So q, x = 1, is true.

So p → q is true.

q → p:

If x = 1,

then x + 2 = 3.

From q, x = 1,

add +2 to both sides.

Then x + 2 = 3.

So p, x + 2 = 3, is true.

So q → p is true.

p → q is true.

q → p is true.

Then the conjunction of these two,

p ↔ q,

is true.

So true is the answer.

Just like this example,

a biconditional statement can also be used

to show the solution of an equation.

## Example 4

### Example

### Solution

The structure of the given statement is

[... iff. ...].

It's a biconditional statement.

So the former statement is p:

x^{2} = 4.

And the latter statement is q:

x = 2.

p and q can be either true or false.

So start from p → q and q → p.

p → q:

If x^{2} = 4,

then x = 2.

If x^{2} = 4,

x = ±2.

Quadratic Equation: Square Root

So q, x = 2, is false.

q doesn't include x = -2.

So p → q is false.

q → p:

If x = 2,

then x^{2} = 4.

From x = 2,

square both sides.

Then x^{2} = 4.

So p, x^{2} = 4, is true.

So q → p is true.

p → q is false.

q → p is true.

Then the conjunction of these two,

p ↔ q,

is false.

So false is the answer.