# 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

### 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.

## Example 2

### 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.

Just like this example,
a biconditional statement can be used
to show the definition of something.

## Example 3

### 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,
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.

Just like this example,
a biconditional statement can also be used
to show the solution of an equation.

## Example 4

### Solution

The structure of the given statement is
[... iff. ...].

It's a biconditional statement.

So the former statement is p:
x2 = 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 x2 = 4,
then x = 2.

If x2 = 4,
x = ±2.

So q, x = 2, is false.
q doesn't include x = -2.

So p → q is false.

q → p:
If x = 2,
then x2 = 4.

From x = 2,
square both sides.
Then x2 = 4.

So p, x2 = 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.