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if and only if

iff
if and only if
Biconditional

A way of writing two conditionals at once: both a conditional and its converse.

For example, the statement "A triangle is equilateral iff its angles all measure 60°" means both "If a triangle is equilateral then its angles all measure 60°" and "If all the angles of a triangle measure 60° then the triangle is equilateral".

Biconditionals can be written using the ⇔ symbol:

A triangle is equilateral ⇔ its angles all measure 60°

 

See also

Inverse, contrapositive

Key Formula

P⇔Q  ≡  (P⇒Q)  ∧  (Q⇒P)P \Leftrightarrow Q \;\equiv\; (P \Rightarrow Q) \;\land\; (Q \Rightarrow P)
Where:
  • PP = A statement (proposition)
  • QQ = Another statement (proposition)
  • ⇒\Rightarrow = "implies" (conditional arrow)
  • ∧\land = "and" (logical conjunction)
  • ⇔\Leftrightarrow = "if and only if" (biconditional arrow)

Example

Problem: Determine whether the following biconditional is true: "An integer n is even if and only if n² is even."
Step 1: Identify the two directions. The biconditional claims both: (1) If n is even, then n² is even, and (2) If n² is even, then n is even.
P⇔Q   means   (P⇒Q)  ∧  (Q⇒P)P \Leftrightarrow Q \;\text{ means }\; (P \Rightarrow Q) \;\land\; (Q \Rightarrow P)
Step 2: Check direction 1: If n is even, then n = 2k for some integer k, so n² = 4k², which is even. Direction 1 is true.
n=2k  ⇒  n2=(2k)2=4k2=2(2k2)n = 2k \;\Rightarrow\; n^2 = (2k)^2 = 4k^2 = 2(2k^2)
Step 3: Check direction 2: Suppose n² is even but n is odd. Then n = 2k + 1, so n² = 4k² + 4k + 1, which is odd — a contradiction. Therefore if n² is even, n must be even. Direction 2 is true.
n=2k+1  ⇒  n2=4k2+4k+1=2(2k2+2k)+1  (odd)n = 2k+1 \;\Rightarrow\; n^2 = 4k^2 + 4k + 1 = 2(2k^2+2k) + 1 \;\text{(odd)}
Step 4: Since both directions hold, the biconditional is true.
Answer: The statement "n is even if and only if n² is even" is true, because each condition implies the other.

Another Example

Problem: Is the following biconditional true or false? "A number x is positive if and only if x > 2."
Step 1: Break it into two conditionals. Direction 1: If x is positive, then x > 2. Direction 2: If x > 2, then x is positive.
Step 2: Test direction 1. Consider x = 1. It is positive, but 1 > 2 is false. So direction 1 fails.
Step 3: Since one direction is false, the entire biconditional is false — even though direction 2 happens to be true.
Answer: The biconditional is false. "If and only if" requires both directions to hold, and x = 1 is a counterexample to one direction.

Frequently Asked Questions

What is the difference between 'if' and 'if and only if'?
"If P then Q" is a one-way conditional: P guarantees Q, but Q does not necessarily guarantee P. "P if and only if Q" is two-way: P guarantees Q and Q also guarantees P. In other words, P and Q are either both true or both false.
What does 'iff' mean in math?
"Iff" is a standard abbreviation for "if and only if." It indicates a biconditional — two statements that are logically equivalent, each implying the other. You will see it in textbooks, proofs, and problem sets.

If … then (conditional) vs. If and only if (biconditional)

A conditional P⇒QP \Rightarrow Q is one direction: P implies Q. A biconditional P⇔QP \Leftrightarrow Q is both directions: P implies Q and Q implies P. For example, "If a shape is a square, then it is a rectangle" is a true conditional. But "A shape is a square if and only if it is a rectangle" is false, because a rectangle does not have to be a square.

Why It Matters

"If and only if" appears throughout mathematics whenever two conditions are exactly equivalent. Definitions in geometry, algebra, and logic are often biconditionals — for instance, a quadrilateral is a parallelogram if and only if its opposite sides are parallel. Understanding the biconditional also matters in proofs, because proving a statement of the form P⇔QP \Leftrightarrow Q requires you to argue in both directions.

Common Mistakes

Mistake: Assuming "if P then Q" automatically means "if Q then P."
Correction: A conditional only goes one way. To claim equivalence, you need to verify both directions separately. That is what "if and only if" demands.
Mistake: Proving only one direction of a biconditional and declaring the proof complete.
Correction: A biconditional proof has two parts. You must show P ⇒ Q and also Q ⇒ P (or use an equivalent method such as a chain of biconditionals).

Related Terms