Biconditional
Two statements always turn out true together and false together. What is the one connective that says exactly that, no more and no less?
Where you are 9 in Logic
Why this exists
And and or join two statements into one. If-then joins them a third way, but only in one direction, and the two ends of an if-then do not play the same part. “If a number is divisible by 6, then it is even.” Learn that a number is divisible by 6 and you know for certain that it is even. Learn that a number is even and you know nothing certain about 6: the number 4 is even and is not a multiple of 6. Two words name the two ends: divisibility by 6 is sufficient for being even, since it is enough on its own, and being even is necessary for divisibility by 6, since no number manages that without it.
Often one direction is not what you mean. “A number is divisible by 10 if and only if it ends in 0” is not claiming a one-way link; it claims that the two facts always travel together, so that knowing either one tells you the other. Each condition is both necessary and sufficient for the other. Writing that out as two separate if-then sentences works, but it is clumsy, and it hides the fact that the two directions are being asserted as a single package. The biconditional is the connective that says it in one piece, and its four-row truth table is a small tool for testing when two statements say the same thing.
Definition
For statements P and Q, the biconditional P \leftrightarrow Q is read “P if and only if Q,” often shortened to “P iff Q.” It is true exactly when P and Q have the same truth value: both true, or both false. It is false exactly when they disagree.
Put another way, P \leftrightarrow Q is nothing more than the two implications P \to Q and Q \to P, asserted together. P \to Q is called the forward direction, and Q \to P the backward direction (it is the converse of the forward direction). To assert P \leftrightarrow Q is to claim both directions hold. When the claim is made for every value of a variable at once, the double arrow is often written \iff, as in 2x + 3 = 11 \iff x = 4 for every real x. It is the same connective.
Necessary and sufficient conditions
The same idea has a second vocabulary. Take two conditions p and q about the same collection of objects (numbers, shapes, functions), and suppose “if p, then q” is true for every object in that collection. Write it p \to q. Lower-case letters mark conditions such as “x > 3”, which hold for some objects and fail for others; for any one object they become statements, true or false.
- p is a sufficient condition for q: whenever p holds, q holds too.
- q is a necessary condition for p: p never holds without q. If q fails, p fails with it.
Both sentences describe the one fact p \to q from its two ends. The sufficient condition is the one after “if”; the necessary condition is the one after “then”.
When the arrow runs both ways, so that p \to q and q \to p are both true, p is necessary and sufficient for q. That is exactly the biconditional p \leftrightarrow q.
All of these say exactly the same thing as p \to q:
- If p, then q.
- q if p.
- p only if q.
- p is sufficient for q.
- q is necessary for p.
“p only if q” says p happens only where q does, so q is necessary for p. On its own, “if” introduces a sufficient condition and “only if” introduces a necessary one, which is why “p if and only if q” claims both. Some books write the arrow as p \Rightarrow q when the if-then is claimed for every object at once; it is the same arrow.
Intuition
Think of P and Q as two switches wired so that one can never move without the other: flip P on and Q comes on with it, flip P off and Q goes off too. P \leftrightarrow Q is the claim that the two switches are wired that way. It says nothing about which switch causes which to move, only that the two of them are locked to agree.
That is a stronger claim than a single if-then. “If P then Q” only rules out P true with Q false; it says nothing at all about what happens when P is false. The biconditional rules on that case too: when P is false, Q must be false as well.
Conditions as regions
Necessary and sufficient have a picture. Think of each condition as a region: all the objects for which it holds. In Figure 1, every square is a rectangle, so the region of squares sits wholly inside the region of rectangles.
Land anywhere in the inner region and you are certainly in the outer one: being a square is sufficient for being a rectangle. Stand outside the outer region and you cannot be in the inner one: being a rectangle is necessary for being a square. The long thin rectangle is in the outer region only, so being a rectangle is not sufficient for being a square.
So the sufficient condition is the smaller region, a strong demand, and the necessary condition is the larger region around it, a weaker one. In the language of sets, the first region is a subset of the second.
Numbers work the same way. In Figure 2 the numbers greater than 5 lie among the numbers greater than 3.
So x > 5 is sufficient for x > 3, and x > 3 is necessary for x > 5. The number 4 lies on the lower ray only, which is why x > 3 is not sufficient for x > 5. When the two regions are the same region, each condition is both necessary and sufficient for the other, and the biconditional between them is true for every object: the two switches again.
Results
Truth table for the biconditional
Like every connective, P \leftrightarrow Q has a truth value that depends only on the truth values of its parts, so its whole behavior fits in a list of four cases. P is true or false, and for each of those Q is true or false, so there are 2 \times 2 = 4 rows. They are written in the fixed order used for every truth table: P true in the top half, false in the bottom half, Q alternating. Mark each row by asking whether the two parts match.
| Row | P | Q | P \leftrightarrow Q |
|---|---|---|---|
| 1 | T | T | T |
| 2 | T | F | F |
| 3 | F | T | F |
| 4 | F | F | T |
The row to notice is the last one. P \leftrightarrow Q is true when P and Q are both false, not only when they are both true. Agreeing on “no” counts exactly as much as agreeing on “yes.”
Read the column as a test of agreement. The column is T, F, F, T. The tables for and, or and if-then each have one odd row out; this one splits two and two. Inside a larger table, run down the two input columns side by side and write T wherever the entries match, F wherever they differ.
Build it from two conditionals, column by column. A conditional is false only when its hypothesis is true and its conclusion false, so P \to Q fails on row 2 alone and Q \to P fails on row 3 alone.
| Row | P | Q | {P \to Q} | {Q \to P} |
|---|---|---|---|---|
| 1 | T | T | T | T |
| 2 | T | F | F | T |
| 3 | F | T | T | F |
| 4 | F | F | T | T |
Joining the two with “and” gives a column that is F wherever either one fails. Set beside the biconditional’s own column, it matches row for row:
| Row | {(P \to Q) \land (Q \to P)} | {P \leftrightarrow Q} |
|---|---|---|
| 1 | T | T |
| 2 | F | F |
| 3 | F | F |
| 4 | T | T |
Both columns read T, F, F, T. Each direction rules out one of the two mismatch rows, and it takes both directions to rule out both.
Flip it and you get exclusive or. “P or Q but not both”, written P \oplus Q, is true exactly when the parts differ: rows 2 and 3. That is the biconditional’s column with every entry flipped. Exclusive or is the negation of the biconditional, and the biconditional is the negation of exclusive or.
Read it backwards. Unlike “and” or “or”, the biconditional loses nothing. If P \leftrightarrow Q is true, Q has the same value as P; if it is false, Q has the opposite value. So the value of P together with the value of P \leftrightarrow Q always pins down Q exactly.
Other ways to write it
Four facts follow from the table:
The second fact says the biconditional is symmetric: unlike an if-then statement, which has a converse that may or may not be true, a biconditional and its “reverse” are not two different claims requiring two different checks. They are the same claim, written twice. The third writes the connective with \land, \lor and \lnot alone, for when you would rather not introduce a new symbol.
Facts about necessary and sufficient conditions
One fact, two names. For conditions p and q, these three say the same thing:
Necessity read backwards. q is necessary for p exactly when “not q” implies “not p”: take q away and p goes with it. This is the contrapositive of p \to q.
Both at once. p is necessary and sufficient for q exactly when both directions hold:
Chains. If p is sufficient for q and q is sufficient for r, then p is sufficient for r. Read from the other end, r is necessary for p. In the same way, if p \leftrightarrow q and q \leftrightarrow r are both true, so is p \leftrightarrow r (Exercise 28).
Why it is true
Every fact about the biconditional in Results is checked by going through the four rows of the table one at a time, so nothing beyond the definitions of the connectives is needed.
P \leftrightarrow Q matches (P \to Q) \land (Q \to P). Recall that an implication is false only when its hypothesis is true and its conclusion is false.
- P true, Q true: P \to Q is true, Q \to P is true, so the conjunction is true. P and Q agree, so P \leftrightarrow Q is true. Both sides agree.
- P true, Q false: P \to Q is false (true hypothesis, false conclusion), so the conjunction is false, regardless of Q \to P. P and Q disagree, so P \leftrightarrow Q is false. Both sides agree.
- P false, Q true: the row above with the roles swapped. Q \to P is now the false one, so the conjunction is false, and P \leftrightarrow Q is false. Both sides agree.
- P false, Q false: P \to Q is true (false hypothesis), and Q \to P is true for the same reason, so the conjunction is true. P and Q agree (both false), so P \leftrightarrow Q is true. Both sides agree.
All four rows match, so the two statements are logically equivalent.
P \leftrightarrow Q matches Q \leftrightarrow P. “Same truth value” does not care which statement is named first: P and Q having the same truth value is exactly the same fact as Q and P having the same truth value. The table for Q \leftrightarrow P has the identical four rows, only relabeled.
P \leftrightarrow Q matches (P \land Q) \lor (\lnot P \land \lnot Q). P \leftrightarrow Q is true exactly on the rows where P and Q agree: both true, or both false. P \land Q is true exactly on the “both true” row, and \lnot P \land \lnot Q exactly on the “both false” row. Their disjunction is therefore true on precisely those two rows and false on the other two, which is the same four-row pattern as P \leftrightarrow Q.
\lnot(P \leftrightarrow Q) matches P \oplus Q. Negating flips T, F, F, T to F, T, T, F, which is true on rows 2 and 3, the two rows where exactly one part is true.
Reading backwards. For each value of P, the two rows with that value have different values of P \leftrightarrow Q (rows 1 and 2 read T, F; rows 3 and 4 read F, T). So the pair of values picks out a single row, and hence the value of Q.
Sufficient and necessary are one claim. “p is sufficient for q” promises that whenever p holds, q holds. “q is necessary for p” promises that p never happens without q. The only case that could break either promise is one where p holds and q fails, so the two are true together and false together. That is also the one row in which a conditional is false, as the truth table for the conditional shows:
| p | q | p \to q |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
The contrapositive forbids the same case once more. “Both at once” is the first result used twice, since p necessary for q is q \to p. For chains, wherever p holds, q holds, and so r holds: the p region sits inside the q region, which sits inside the r region.
Worked examples
Finding the row
For each statement, name the row of the table it falls on and give its truth value.
(a) \displaystyle \sqrt{49} = 7 if and only if 7^2 = 49.
(b) 0.5 > 0.6 if and only if \displaystyle \frac{1}{2} = 0.5.
(c) 2^{10} < 1000 if and only if 10^3 < 1000.
SolutionDecide each part on its own, then find the row with that pair of values.
(a) \displaystyle \sqrt{49} = 7 is true and 7^2 = 49 is true. The pair is T, T, row 1, so the statement is true.
(b) 0.5 > 0.6 is false, and \displaystyle \frac{1}{2} = 0.5 is true. The pair is F, T, row 3, so the statement is false.
(c) 2^{10} = 1024, which is not less than 1000, and 10^3 = 1000, which is not less than itself. Both parts are false, the pair is F, F, row 4, and the statement is true. Neither part holds, but that is exactly the agreement the biconditional asks for.
One direction holding is not enough
The statement “if an integer is divisible by 6, then it is even” is true. Which condition is sufficient, and which is necessary? Is “an integer is divisible by 6 if and only if it is even” true for every integer?
SolutionThe if-then is true because a multiple of 6 has the form 6k = 2(3k), which is even.
The sufficient condition is the one after “if”: being divisible by 6 is sufficient for being even. The necessary condition is the one after “then”: being even is necessary for being divisible by 6. Put the other way round, an odd number is never a multiple of 6.
The biconditional also needs the backward direction, “if an integer is even, then it is divisible by 6”, for every even integer. Test n = 4: “4 is divisible by 6” is false, since 4 \div 6 is not a whole number, while “4 is even” is true. The two parts disagree, so the biconditional is False at n = 4, and one such counterexample defeats the “for every integer” claim.
So divisibility by 6 is sufficient but not necessary for being even, and being even is necessary but not sufficient for divisibility by 6. A biconditional needs both directions, and a true one is not a substitute for the other. (Being even and divisible by 3 is necessary and sufficient; see divisibility rules.)
Turning the words around
Rewrite each sentence as an if-then statement and decide whether it is true.
- (a) A triangle is equilateral only if it is isosceles.
- (b) For an integer to be prime, it is sufficient that it be odd.
- (c) Having four right angles is necessary for a quadrilateral to be a square.
(a) “Only if” introduces the necessary condition, which goes after “then”: if a triangle is equilateral, then it is isosceles. True, since three equal sides include two equal sides.
(b) The sufficient condition goes after “if”: if an integer is odd, then it is prime. False, since 9 is odd and 9 = 3 \times 3.
(c) The necessary condition goes after “then”: if a quadrilateral is a square, then it has four right angles. True, by the definition of a square.
Classifying conditions on a number
Let x be a real number. For each condition, decide whether it is sufficient, necessary, both or neither for x > 3.
- (a) x > 5
- (b) x > 0
- (c) 2x - 6 > 0
- (d) x \ne 4
Ask two questions of each. Does it force x > 3? Then it is sufficient. Does x > 3 force it? Then it is necessary.
(a) If x > 5 then x > 3, because 5 > 3. But x = 4 has x > 3 and not x > 5. Sufficient but not necessary, as Figure 2 shows.
(b) x = 1 has x > 0 and not x > 3, so it is not sufficient. Every number above 3 is above 0, so it is necessary. Necessary but not sufficient.
(c) Add 6 to both sides, then divide by 2:
Each link holds in both directions, because adding 6 is undone by subtracting 6, and dividing by 2 is undone by multiplying by 2. Two biconditionals that share a middle statement give a third (Exercise 28), so the two conditions hold for exactly the same numbers. Necessary and sufficient.
(d) x = 0 has x \ne 4 and not x > 3, so it is not sufficient. x = 4 has x > 3 and not x \ne 4, so it is not necessary. Neither.
Part (c) is what every solution of an equation or inequality quietly claims. Writing
says that 2x + 3 = 11 is necessary and sufficient for x = 4: the value 4 works, and nothing else does. Each step is reversible, and that reversibility is what licenses the double arrow.
A condition from an equation
Let x be a real number. Is “x^2 = 4 if and only if x = 2” true for every x? Is x = 2 sufficient for x^2 = 4, and is it necessary? Then repair the statement in two ways: once by changing the right side, and once by adding a condition to the left.
SolutionCheck x = 2 first: 2^2 = 4 is true and x = 2 is true, so the biconditional holds there. But at x = -2, (-2)^2 = 4 is true while x = 2 is false. The parts disagree, so the biconditional is False at x = -2, and the “for every x” claim fails.
In the other vocabulary: x = 2 is sufficient for x^2 = 4, since 2^2 = 4, but not necessary, since x = -2 also has x^2 = 4. Squaring is not a step that can be undone the way adding 6 can.
Repair 1. The solutions of x^2 = 4 are exactly x = 2 and x = -2, so “x^2 = 4 if and only if x = 2 or x = -2” is true for every real x.
Repair 2. Add a condition that rules out -2: try “x^2 = 4 and x > 0”, and test both directions. If x = 2, then x^2 = 4 and x > 0, so the new condition is necessary for x = 2. If x^2 = 4 and x > 0, then x is 2 or -2 and is positive, so x = 2; the new condition is sufficient.
In regions: the solutions of x^2 = 4 form the set \{-2, 2\}, and adding x > 0 trims it to \{2\}.
Working back from the value
For a whole number n from 1 to 20, the statement “n is odd if and only if n is prime” is false. What can n be?
SolutionA false biconditional sits on row 2 or row 3, where the two parts disagree. So there are two ways for it to happen.
Row 2: n is odd and not prime. From 1 to 20 these are 1, 9 and 15. (1 is not prime, 9 = 3 \times 3 and 15 = 3 \times 5.)
Row 3: n is not odd, that is even, and prime. The only even prime is 2, since every other even number has 2 as a factor.
So n is one of 1, 2, 9, 15; on the other sixteen the statement is true.
Three ways to write “the parts differ”
Show that \neg(P \leftrightarrow Q), P \leftrightarrow \neg Q and P \oplus Q all have the same column.
SolutionP \leftrightarrow Q is T, F, F, T, and negating flips each entry, so \neg(P \leftrightarrow Q) is F, T, T, F.
| Row | P | Q | {P \leftrightarrow Q} | {\neg(P \leftrightarrow Q)} |
|---|---|---|---|---|
| 1 | T | T | T | F |
| 2 | T | F | F | T |
| 3 | F | T | F | T |
| 4 | F | F | T | F |
For P \leftrightarrow \neg Q, first write \neg Q, which is F, T, F, T. Compare it with P (T, T, F, F) entry by entry: row 1 T against F, a mismatch; row 2 T against T, a match; row 3 F against F, a match; row 4 F against T, a mismatch. That gives F, T, T, F.
| Row | P | Q | \neg Q | {P \leftrightarrow \neg Q} |
|---|---|---|---|---|
| 1 | T | T | F | F |
| 2 | T | F | T | T |
| 3 | F | T | F | T |
| 4 | F | F | T | F |
P \oplus Q is true exactly when one part is true, rows 2 and 3, so it is F, T, T, F as well.
Negating one side of a biconditional negates the whole thing. In particular P \leftrightarrow Q and P \leftrightarrow \neg Q are never true together: one is the negation of the other.
Testing two statements for equivalence
Use a biconditional column to decide whether P \to Q says the same thing as its contrapositive \neg Q \to \neg P, and whether it says the same thing as its converse Q \to P.
SolutionThe biconditional between two statements is T on every row exactly when their columns match, which is what it means for them to say the same thing.
Call the three statements A for P \to Q, B for \neg Q \to \neg P and C for Q \to P. A is T, F, T, T. For B: \neg Q is F, T, F, T and \neg P is F, F, T, T, and the conditional fails only where the first is T and the second F, which is row 2. So B is T, F, T, T. Compare A with B:
| Row | P | Q | A | B | {A \leftrightarrow B} |
|---|---|---|---|---|---|
| 1 | T | T | T | T | T |
| 2 | T | F | F | F | T |
| 3 | F | T | T | T | T |
| 4 | F | F | T | T | T |
A \leftrightarrow B is T on every row, a tautology, so a conditional and its contrapositive are logically equivalent, the fact taken up in contrapositive equivalence.
The converse C fails only on row 3, so it is T, T, F, T. Compare A with C:
| Row | P | Q | A | C | {A \leftrightarrow C} |
|---|---|---|---|---|---|
| 1 | T | T | T | T | T |
| 2 | T | F | F | T | F |
| 3 | F | T | T | F | F |
| 4 | F | F | T | T | T |
A \leftrightarrow C is F on rows 2 and 3, so a conditional and its converse are not equivalent, and row 2 is a case that separates them: P true, Q false.
Proving a condition necessary and sufficient
Let n be an integer. Show that “n is even” is necessary and sufficient for “n^2 is even”.
Solution“Necessary and sufficient” is two claims, so prove two conditionals, one for each direction, as in proving an if and only if.
Sufficient: if n is even, then n^2 is even. Write n = 2k for an integer k. Then
which is 2 times an integer, so it is even.
Necessary: if n^2 is even, then n is even. Knowing n^2 = 2m says little about n directly, so prove the contrapositive: if n is odd, then n^2 is odd. Write n = 2k + 1. Then
which is one more than an even number, so it is odd. An odd n never has an even square, so an even square needs an even n. Both directions hold, so for every integer n, n \text{ is even} \iff n^2 \text{ is even}.
Knights and knaves
On an island every person is a knight, who always tells the truth, or a knave, who always lies. Islander A says: “B and I are the same type.” Islander B says: “A is a knave.” Which is which?
SolutionLet A stand for “A is a knight” and B for “B is a knight”. A speaker is a knight exactly when the claim is true, so for every speaker, “speaker is a knight \leftrightarrow the claim” must be true.
A’s claim is A \leftrightarrow B; call it S. So A \leftrightarrow S must be true. B’s claim is \neg A, so B \leftrightarrow \neg A must be true. S is a biconditional, so its column is T, F, F, T, and A \leftrightarrow S compares that with A (T, T, F, F). Build both conditions over the four cases:
| Row | A | B | {A \leftrightarrow S} | {B \leftrightarrow \neg A} |
|---|---|---|---|---|
| 1 | T | T | T | F |
| 2 | T | F | F | T |
| 3 | F | T | T | T |
| 4 | F | F | F | F |
Only row 3 makes both conditions true, so A is a knave and B is a knight.
Check. A, a knave, says they are the same type; they are not, so A has lied, as a knave must. B, a knight, says A is a knave, which is true.
Boundaries
A biconditional needs both directions, checked separately. Confirming the forward direction tells you nothing about the backward one, as in Example 2. In the same way, “sufficient” means every case, not one case: showing a condition is sufficient needs an argument about all cases, and showing it is not needs one counterexample.
The table needs statements. Each part must be true or false. “x^2 = 4 if and only if x = 2” has no row until x is given a value, and then each value picks its own row: true at x = 2, false at x = -2. The claim “for every x” is a different, stronger statement, and a single case where the sides disagree defeats it (Example 5). Agreeing cases, however many, never prove it; an argument that covers every case does (Example 4).
The answer depends on the collection of objects. Over all real numbers, x^2 = 4 is not sufficient for x = 2. Over the positive numbers alone it is, because -2 is no longer in play.
As a formula, a chain of \leftrightarrow does not say that everything agrees. Read as the compound statement (P \leftrightarrow Q) \leftrightarrow R, the string P \leftrightarrow Q \leftrightarrow R is true on four of its eight rows, including rows where the parts disagree: it is true when P is true while Q and R are false, and false when all three are false (Exercise 24, which also shows that grouping it as P \leftrightarrow (Q \leftrightarrow R) gives the same column). A chain of working such as Example 4’s means something else: it is shorthand for the separate links P \leftrightarrow Q and Q \leftrightarrow R, and it is the links that force all three to share one truth value. A theorem listing several equivalent conditions says “the following are equivalent” for the same reason.
The table compares truth values, not meanings. “2 + 2 = 4 if and only if 7 is prime” sits on row 1 and is true, though the two parts have nothing to do with each other.
The biconditional does not say why the two sides agree, and the words do not say when. P \leftrightarrow Q records that P and Q rise and fall together; it is silent on which one, if either, is the cause. Likewise x > 0 is necessary for x > 3, but it neither comes first nor makes x > 3 happen. “Necessary” does not mean earlier and “sufficient” does not mean the cause; the words say which cases lie inside which, and nothing about time or cause.
Impossible and certain conditions. For real x the condition x^2 < 0 never holds, so no case can break “if x^2 < 0, then q”: an impossible condition counts as sufficient for anything. Likewise a condition that always holds, such as x^2 \ge 0, counts as necessary for anything. Both are correct, and neither tells you anything.
The table is built for exactly two truth values. In a logic with a third value such as “unknown”, two parts have nine combinations, and the four-row table does not cover them.
Everyday speech often means more than it says. “You can go out if you finish your homework” is usually heard as “if and only if”. In mathematics a stated sufficient condition claims only that, and the converse needs its own proof: see converses and biconditional definitions.
Where this goes next
To prove a biconditional, split it into its two directions and prove each on its own, usually with two separate arguments: that is the entire content of proving an if and only if. The necessary direction is often easiest by contraposition, as in Example 9.
Mathematical definitions almost always mean “if and only if” even when they are written with a plain “if,” because a definition is meant to pin a word down exactly, in both directions at once: why definitions work both ways takes that up directly.
The biconditional column is one of the building blocks for tables of compound statements, alongside not, and, or and if-then. Placed between two statements it becomes the test for logical equivalence, which generalizes the rewriting in Results to any two statements that always agree: an all-T column is a tautology, and an all-F column, such as the one for P \leftrightarrow \neg P, is a contradiction.
The same words run through calculus. A function with a derivative at a point is continuous there, so continuity is necessary for differentiability. It is not sufficient: |x| is continuous at 0 and has no derivative there, as corners, cusps and piecewise functions shows. They also guide the hunt for a proof: to prove q, look for something sufficient for it.
Exercises
Reading the table
Name the row of the table for P \leftrightarrow Q that each statement falls on, and give its truth value.
- 3^2 = 9 if and only if 9 is odd.
- \displaystyle \frac{1}{3} = 0.3 if and only if \displaystyle \frac{1}{2} = 0.5.
- 100 is a perfect square if and only if 100 is a multiple of 3.
- 5^2 = 10 if and only if 5 + 5 = 25.
The two directions, and the words for them
- “A triangle is isosceles if and only if two of its angles are equal.” Write out the forward direction and the backward direction as two separate if-then sentences.
- “If a triangle is equilateral, then it is isosceles.” Name the sufficient condition and the necessary condition.
- Rewrite as an if-then statement: “An integer is prime only if it is greater than 1.”
- Which of these say the same thing as “if p, then q”? (a) q is necessary for p. (b) p is necessary for q. (c) q if p. (d) p if q. (e) p only if q.
- P \leftrightarrow Q is true and P is false. What is the truth value of Q? If instead you knew only that P \to Q is true and P is false, what could you say about Q?
Sufficient, necessary, both or neither
Decide whether the first condition is sufficient but not necessary, necessary but not sufficient, necessary and sufficient, or neither, for the second. In Exercises 10 to 12, x is a real number.
- x \ne 3 for x > 2
- |x| = 3 for x^2 = 9
- x < 2 for x^2 < 4
- For an integer n: “n is odd” for “n^2 + n is even”.
- For a quadrilateral: “all four sides are equal” for “it is a square”.
- For integers a and b: “a and b are both even” for “a + b is even”.
True for every value?
- True or false: “A quadrilateral is a rectangle if and only if its four angles are all equal.” Explain using both directions.
- True or false: “For every integer n, n^2 > 0 if and only if n > 0.” If false, give a value of n that shows it.
- True or false: “For every real number x, |x| = x if and only if x \ge 0.” Explain.
- Find a condition on x, written as an inequality, that is necessary and sufficient for x^2 - 5x + 6 < 0.
Columns and working back
- Two columns of a table read, top to bottom, A: F, T, T, F and B: F, F, T, T. Fill in the column for A \leftrightarrow B.
- Fill in the column for (P \lor Q) \leftrightarrow Q. Which conditional’s column does it match?
- For which whole numbers n from 1 to 12 is “n is a multiple of 4 if and only if n is a multiple of 6” true?
- On the island of knights and knaves, A says “B is a knave” and B says “A and I are both knaves.” Which is which?
- On which rows of the eight-row table is (P \leftrightarrow Q) \leftrightarrow R true, and how many letters are false on each of those rows? Then fill in the column for P \leftrightarrow (Q \leftrightarrow R) and compare.
- Fill in the columns for \neg P \leftrightarrow \neg Q and for (P \to Q) \land (\neg P \to \neg Q), and compare each with the column of P \leftrightarrow Q.
Reasoning
- Explain why “p is sufficient for q” and “q is necessary for p” are the same claim.
- Sam says: “4 is even and 4 is divisible by 4, so being even is sufficient for being divisible by 4.” What is wrong with the argument, and is the conclusion true?
- Suppose P \leftrightarrow Q is true and Q \leftrightarrow R is true. Explain why P \leftrightarrow R must be true as well.
- Knowing P and the value of P \leftrightarrow Q always fixes Q. Explain why, and explain why knowing P and the value of P \lor Q does not always fix Q.
- On the island of knights and knaves, explain why nobody can say “I am a knave.”
- Show that for an integer n, “n is even” is necessary and sufficient for “n^2 is divisible by 4”.
- For ax^2 + bx + c = 0 with real numbers a, b, c and a \ne 0, the condition b^2 - 4ac \ge 0 is necessary and sufficient for a real solution (see the discriminant). Show that “a and c have opposite signs” is sufficient for a real solution, but not necessary.
Answers
- Row 1, true. 3^2 = 9 holds and 9 is odd.
- Row 3, false. \displaystyle \frac{1}{3} = 0.333\ldots \ne 0.3, while \displaystyle \frac{1}{2} = 0.5 holds.
- Row 2, false. 100 = 10^2 is a perfect square, but 100 = 3 \times 33 + 1 is not a multiple of 3.
- Row 4, true. 5^2 = 25, not 10, and 5 + 5 = 10, not 25; both parts are false, so they agree.
- Forward: if a triangle is isosceles, then two of its angles are equal. Backward: if two of a triangle’s angles are equal, then the triangle is isosceles.
- Sufficient: the triangle is equilateral. Necessary: the triangle is isosceles. The sufficient condition follows “if” and the necessary one follows “then”.
- If an integer is prime, then it is greater than 1. “Only if” introduces the necessary condition, which goes after “then”.
- (a), (c) and (e). Each makes p the sufficient condition and q the necessary one. (b) and (d) both say q \to p, the converse.
- Q is false, since a true biconditional forces Q to share P’s value. From P \to Q alone, nothing: with P false the conditional is true whether Q is true or false.
- Neither. x = 0 has x \ne 3 but not x > 2, so it is not sufficient. x = 3 has x > 2 but not x \ne 3, so it is not necessary.
- Necessary and sufficient. Both conditions hold for exactly two numbers, x = 3 and x = -3.
- Necessary but not sufficient. x^2 < 4 means -2 < x < 2, and every such x is below 2. But x = -5 has x < 2 and x^2 = 25.
- Sufficient but not necessary. n^2 + n = n(n + 1) is a product of two neighboring integers, one of which is even, so it is even for every n. An odd n guarantees it, but n = 2 gives 2^2 + 2 = 6, even, with n not odd.
- Necessary but not sufficient. A square has four equal sides, so the condition is necessary. A rhombus with angles of 60^\circ and 120^\circ has four equal sides and is not a square.
- Sufficient but not necessary. If a = 2m and b = 2k, then a + b = 2(m + k). But 3 + 5 = 8 is even with both numbers odd.
- True. Forward: a rectangle’s four angles are all 90^\circ, so they are equal. Backward: the angles of a quadrilateral add to 360^\circ, so four equal angles are each 360^\circ \div 4 = 90^\circ, and a quadrilateral with four right angles is a rectangle.
- False. At n = -2: n^2 = 4 > 0 is true, but n > 0 is false, so they disagree.
- True. When x \ge 0, |x| = x by the definition of absolute value, so both sides are true. When x < 0, |x| = -x \ne x (since x \ne 0), so both sides are false. Every case agrees.
- 2 < x < 3. x^2 - 5x + 6 = (x - 2)(x - 3), which is negative exactly when one factor is positive and the other negative, that is, when x lies strictly between 2 and 3.
- T, F, T, F. The entries match on rows 1 and 3 and differ on rows 2 and 4.
- T, F, T, T, the column of P \to Q. P \lor Q is T, T, T, F and Q is T, F, T, F; they differ only on row 2.
- 1, 2, 3, 5, 7, 9, 10, 11, 12: nine numbers. The parts disagree only at 4 and 8 (multiples of 4, not of 6) and at 6 (a multiple of 6, not of 4); 12 is a multiple of both.
- A is a knight and B is a knave. The conditions are A \leftrightarrow \neg B and B \leftrightarrow (\neg A \land \neg B), and only row 2 (A true, B false) makes both true. Check: B’s claim that both are knaves is false, as a knave’s must be.
- Rows 1, 4, 6 and 7. P \leftrightarrow Q is T, T, F, F, F, F, T, T, and comparing it with R (T, F, T, F, T, F, T, F) gives T, F, F, T, F, T, T, F. Row 1 has no false letters, and rows 4, 6 and 7 have two each: the chain is true exactly when an even number of letters are false. Q \leftrightarrow R is T, F, F, T, T, F, F, T; P \leftrightarrow (Q \leftrightarrow R) copies it on rows 1 to 4, where P is T, and flips it on rows 5 to 8, where P is F, giving T, F, F, T, F, T, T, F again. The grouping makes no difference.
- Both are T, F, F, T, the column of P \leftrightarrow Q. \neg P is F, F, T, T and \neg Q is F, T, F, T; they match on rows 1 and 4. \neg P \to \neg Q fails only where \neg P is T and \neg Q is F, which is row 3, so it is T, T, F, T; joined by “and” with P \to Q (T, F, T, T) it gives T, F, F, T. Negating both sides of a biconditional changes nothing, while negating one side flips it (Example 7). The second form is the plan of Example 9’s proof: show “if p, then q” and “if not p, then not q”.
- Each claim says there is no case in which p holds and q fails. “Sufficient” says p always brings q with it; “necessary” says p never occurs without q. Both are the statement p \to q.
- One case where both conditions hold says nothing about every case. Sufficiency needs every even number to be divisible by 4, and 2 is even but not divisible by 4. The conclusion is false. The true statement runs the other way: divisibility by 4 is sufficient for being even.
- P \leftrightarrow Q true means P and Q share a truth value; Q \leftrightarrow R true means Q and R share a truth value. Whichever value Q has, both P and R are forced to share it, so they match each other, and P \leftrightarrow R is true.
- For each value of P, the two rows with that value have different values of P \leftrightarrow Q (rows 1 and 2 read T, F; rows 3 and 4 read F, T), so the pair picks out one row and hence Q. For P \lor Q, rows 1 and 2 both read T, so P true and P \lor Q true leave Q undecided.
- A speaker who claims S makes “speaker is a knight \leftrightarrow S” true. Here S is “speaker is a knave”, the negation of “speaker is a knight”, and K \leftrightarrow \neg K is F on both rows, since \neg K always has the opposite value to K. No speaker can make it true, so nobody can say it.
- Sufficient. If n is even, n = 2k for an integer k, and n^2 = 4k^2 is divisible by 4. Necessary, by the contrapositive: if n is odd, n = 2k + 1 and n^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1, which leaves remainder 1 on division by 4. So an odd n never has a square divisible by 4, and n^2 divisible by 4 forces n to be even.
- Sufficient. If a and c have opposite signs, then ac < 0, and multiplying by -4 gives -4ac > 0. Since b^2 \ge 0 for every real b, b^2 - 4ac > 0, so there are real solutions. For example, 2x^2 + 7x - 3 = 0 has 7^2 - 4(2)(-3) = 49 + 24 = 73 > 0. Not necessary. x^2 - 5x + 6 = 0 has a = 1 and c = 6, both positive, yet it factors as (x - 2)(x - 3) = 0 and has the real solutions x = 2 and x = 3.