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The biconditional statement “\(p\) if and only if \(q\),” denoted \(p \Leftrightarrow q\), is true when both \(p\) and \(q\) carry the same truth value, and is false otherwise. It is sometimes abbreviated as “\(p\) iff \(q\).” Its truth table is depicted below.
| \(p\) | \(q\) | \(p \Leftrightarrow q\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | T |
Example \(\PageIndex{1}\label{eg:bicond-01}\)
The following biconditional statements
\(2x - 5 = 0 \Leftrightarrow x = 5/2\),
\(x > y \Leftrightarrow x - y > 0\),
are true, because, in both examples, the two statements joined by \(\Leftrightarrow\) are true or false simultaneously.
A biconditional statement can also be defined as the compound statement
\[(p \Rightarrow q) \wedge (q \Rightarrow p).\]
This explains why we call it a biconditional statement. A biconditional statement is often used to define a new concept.
Example \(\PageIndex{2}\label{eg:bicond-02}\)
A number is even if and only if it is a multiple of 2. Mathematically, this means \[n \mbox{ is even} \Leftrightarrow n = 2q \mbox{ for some integer $q$}.\] It follows that for any integer \(m\), \[mn = m\cdot 2q = 2(mq).\] Since \(mq\) is an integer (because it is a product of two integers), by definition, \(mn\) is even. This shows that the product of any integer with an even integer is always even.
hands-on exercise \(\PageIndex{1}\label{he:bicond-01}\)
Complete the following statement: \[n \mbox{ is odd} \Leftrightarrow \hskip1.25in.\] Use this to prove that if \(n\) is odd, then \(n^2\) is also odd.
Example \(\PageIndex{3}\label{eg:bicond-03}\)
The operation “exclusive or” can be defined as \[p\veebar q \Leftrightarrow (p\vee q) \wedge \overline{(p\wedge q)}.\] See Problem [ex:imply-10] in Exercises 1.2.
When we have a complex statement involving more than one logical operation, care must be taken to determine which operation should be carried out first. The precedence or priority is listed below.
| Connectives | Priority |
|---|---|
| \(\neg\) | Highest |
| \(\wedge\) | |
| \(\vee\) | \(\vdots\) |
| \(\Rightarrow\) | |
| \(\Leftrightarrow\) | Lowest |
This is the order in which the operations should be carried out if the logical expression is read from left to right. To override the precedence, use parentheses.
Example \(\PageIndex{4}\label{eg:bicond-04}\)
The precedence of logical operations can be compared to those of arithmetic operations.
| Operations | Priority |
|---|---|
| \(-\) (Negative) | Highest |
| Exponentiation | \(\vdots\) |
| Multiplication/Division | \(\vdots\) |
| Addition/Subtraction | Lowest |
For example, \(yz^{-3} \neq (yz)^{-3}\). To evaluate \(yz^{-3}\), we have to perform exponentiation first. Hence, \(yz^{-3} = y\cdot z^{-3} = \frac{y}{z^3}\).
Another example: the notation \(x^{2^3}\) means \(x\) raised to the power of \(2^3\), hence \(x^{2^3}=x^8\); it should not be interpreted as \((x^2)^3\), because \((x^2)^3=x^6\).
Example \(\PageIndex{5}\label{eg:bicond-05}\)
It is not true that \(p \Leftrightarrow q\) can be written as “\(p \Rightarrow q \wedge q \Rightarrow p\),” because it would mean, technically, \[p \Rightarrow (q \wedge q) \Rightarrow p.\] The correct notation is \((p \Rightarrow q) \wedge (q \Rightarrow p)\).
hands-on exercise\(\PageIndex{2}\label{he:bicond-02}\)
Insert parentheses in the following formula \[p\Rightarrow q\wedge r\] to identify the proper procedure for evaluating its truth value. Construct its truth table.
hand-on exercise \(\PageIndex{3}\label{he:bicond-03}\)
Insert parentheses in the following formula \[p\wedge q \Leftrightarrow \overline{p}\vee\overline{q}.\] to identify the proper procedure for evaluating its truth value. Construct its truth table.
We close this section with a justification of our choice in the truth value of \(p\Rightarrow q\) when \(p\) is false. The truth value of \(p\Rightarrow q\) is obvious when \(p\) is true.
| \(p\) | \(q\) | \(p \Rightarrow q\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | ? |
| F | F | ? |
We want to decide what are the best choices for the two missing values so that they are consistent with the other logical connectives. Observe that if \(p \Rightarrow q\) is true, and \(q\) is false, then \(p\) must be false as well, because if \(p\) were true, with \(q\) being false, then the implication \(p\Rightarrow q\) would have been false. For instance, if we promise
“If tomorrow is sunny, we will go to the beach”
but we do not go to the beach tomorrow, then we know tomorrow must not be sunny. This means the two statements \(p\Rightarrow q\) and \(\overline{q} \Rightarrow \overline{p}\) should share the same truth value.
When both \(p\) and \(q\) are false, then both \(\overline{p}\) and \(\overline{q}\) are true. Hence \(\overline{q} \Rightarrow \overline{p}\) should be true, consequently so is \(p\Rightarrow q\). Thus far, we have the following partially completed truth table:
| \(p\) | \(q\) | \(p \Rightarrow q\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | ? |
| F | F | T |
If the last missing entry is F, the resulting truth table would be identical to that of \(p \Leftrightarrow q\). To distinguish \(p\Leftrightarrow q\) from \(p\Rightarrow q\), we have to define \(p \Rightarrow q\) to be true in this case.
Summary and Review
- A biconditional statement \(p\Leftrightarrow q\) is the combination of the two implications \(p\Rightarrow q\) and \(q\Rightarrow p\).
- The biconditional statement \(p\Leftrightarrow q\) is true when both \(p\) and \(q\) have the same truth value, and is false otherwise.
- A biconditional statement is often used in defining a notation or a mathematical concept.
Example \(\PageIndex{1}\label{ex:bicond-01}\)
Let \(p\), \(q\), and \(r\) represent the following statements:
| \(p\): | Sam had pizza last night. |
| \(q\): | Chris finished her homework. |
| \(r\): | Pat watched the news this morning. |
Give a formula (using appropriate symbols) for each of these statements.
- Sam had pizza last night if and only if Chris finished her homework.
- Pat watched the news this morning iff Sam did not have pizza last night.
- Pat watched the news this morning if and only if Chris finished her homework and Sam did not have pizza last night.
- In order for Pat to watch the news this morning, it is necessary and sufficient that Sam had pizza last night and Chris finished her homework.
Example \(\PageIndex{2}\label{ex:bicond-02}\)
Define the propositional variables as in Problem 1. Express in words the statements represented by the following formulas:
(a) \(q\Leftrightarrow r\) & (b) \(p\Leftrightarrow(q\wedge r)\)
(c) \(\overline{p}\Leftrightarrow (q\vee r)\) & (d) \(r\Leftrightarrow(p\vee q)\)
Example \(\PageIndex{3}\label{ex:bicond-03}\)
Consider the following statements:
| \(p\): | Niagara Falls is in New York. |
| \(q\): | New York City is the state capital of New York. |
| \(r\): | New York City will have more than 40 inches of snow in 2525. |
The statement \(p\) is true, and the statement \(q\) is false. Represent each of the following statements by a formula. What is their truth value if \(r\) is true? What if \(r\) is false?
- Niagara Falls is in New York if and only if New York City is the state capital of New York.
- Niagara Falls is in New York iff New York City will have more than 40 inches of snow in 2525.
- Niagara Falls is in New York or New York City is the state capital of New York if and only if New York City will have more than 40 inches of snow in 2525.
Example \(\PageIndex{4}\label{ex:bicond-04}\)
Express each of the following compound statements symbolically:
- The product \(xy=0\) if and only if either \(x=0\) or \(y=0\).
- The integer \(n=4\) if and only if \(7n-5=23\).
- A necessary condition for \(x=2\) is \(x^4-x^2-12=0\).
- A sufficient condition for \(x=2\) is \(x^4-x^2-12=0\).
- For \(x^4-x^2-12=0\), it is both sufficient and necessary to have \(x=2\).
- The sum of squares \(x^2+y^2>1\) iff both \(x\) and \(y\) are greater than 1.
Example \(\PageIndex{5}\label{ex:bicond-05}\)
Determine the truth values of the following statements (assuming that \(x\) and \(y\) are real numbers):
- The product \(xy=0\) if and only if either \(x=0\) or \(y=0\).
- The sum of squares \(x^2+y^2>1\) iff both \(x\) and \(y\) are greater than 1.
- \(x^2-4x+3-0 \Leftrightarrow x=3\).
- \(x^2>y^2 \Leftrightarrow x>y\).
Example \(\PageIndex{6}\label{ex:bicond-06}\)
Determine the truth values of the following statements (assuming that \(x\) and \(y\) are real numbers):
- \(u\) is a vowel if and only if \(b\) is a consonant.
- \(x^2+y^2=0\) if and only if \(x=0\) and \(y=0\).
- \(x^2-4x+4=0\) if and only if \(x=2\).
- \(xy\neq0\) if and only if \(x\) and \(y\) are both positive.
Example \(\PageIndex{7}\label{ex:bicond-07}\)
We have seen that a number \(n\) is even if and only if \(n=2q\) for some integer \(q\). Accordingly, what can you say about an odd number?
Example \(\PageIndex{8}\label{ex:bicond-08}\)
We also say that an integer \(n\) is even if it is divisible by 2, hence it can be written as \(n=2q\) for some integer \(q\), where \(q\) represents the quotient when \(n\) is divided by 2. Thus, \(n\) is even if it is a multiple of 2. What if the integer \(n\) is a multiple of 3? What form must it take? What if \(n\) is not a multiple of 3?
