Applications: Mathematical Statements

A conditional has a hypothesis and a conclusion. The converse swaps them, the inverse negates both, and the contrapositive swaps them and negates both. A statement about every visitor is true only if no visitor breaks it, so each one is tested against every age the policy covers.

1. The hypothesis is "the visitor is under 12" and the conclusion is "the visitor enters free". Every visitor either enters free or pays, so their negations are "the visitor is 12 or over" and "the visitor pays".
2. (a) The converse swaps the two parts: if a visitor enters free, the visitor is under 12. The inverse negates both: if a visitor is 12 or over, the visitor pays. The contrapositive swaps and negates: if a visitor pays, the visitor is 12 or over.
3. Under the full policy every visitor under 12 enters free, so the statement on the board is true. The contrapositive is true exactly when the original is, and indeed the visitors who pay are those aged 12 to 64, each of whom is 12 or over.
4. Test the converse on a visitor aged 65. That visitor enters free but is not under 12, so the converse is false. The same visitor is 12 or over and does not pay, so the inverse is false too.
5. (b) The statement on the board and its contrapositive are true, and the converse and the inverse are false. Check: the inverse is the contrapositive of the converse, so those two must share a truth value, and they do.

answer(a) Converse: if a visitor enters free, the visitor is under 12. Inverse: if a visitor is 12 or over, the visitor pays. Contrapositive: if a visitor pays, the visitor is 12 or over. (b) The statement on the board and the contrapositive are true; the converse and the inverse are false, as a visitor aged 65 shows.

techniqueThe Converse and the Inverse · The Contrapositive · Conditional Statements

Give each fact a letter: I for the traveler being ill, A for the airline canceling and B for a trip booked less than 7 days ahead. The clause then says a claim is refunded exactly when (I or A) and (not B), and a refusal is the negation of that.

1. Translate the clause. "Unless B" means the refund also needs B to be false, so a claim is refunded exactly when (I or A) and (not B). The "or" is true when either or both of I and A hold.
2. Test each claim. Claim 1 has I true and was booked 30 days ahead, so it is refunded. Claim 2 was booked 3 days ahead, so B is true and it is refused, even though the airline canceled.
3. Claim 3 has I and A both true, so the inclusive "or" is true, and the trip was booked 10 days ahead, so it is refunded. Claim 4 has neither I nor A, so it is refused. (a) Claims 1 and 3 are refunded.
4. A refusal is the negation of "(I or A) and (not B)". By De Morgan's laws the negation of an "and" is an "or" of the negations: not (I or A), or B. Then not (I or A) is (not I) and (not A). So a claim is refused when the traveler was well and the airline did not cancel, or the trip was booked less than 7 days ahead.
5. (b) Count the combinations. B is true in 4 of the 8, and each of those is refused. Of the 4 with B false, only the one with I and A both false is refused. That makes 4 + 1 = 5 refused combinations. Check: 3 combinations are refunded, and 3 + 5 = 8.

answer(a) Claims 1 and 3; (b) refused exactly when the traveler was well and the airline did not cancel, or the trip was booked less than 7 days ahead: 5 of the 8 combinations

techniqueLogical Connectives and Truth Tables · Negating a Statement

Work out the bracket first in every row, then the connective outside it. Two circuits do the same job only if their output columns agree in all eight rows, so compare them one row at a time.

1. In the rule, the bracket A and D is true only when both are true, which is rows 1 and 2. P is true in rows 1, 3, 5 and 7.
2. "Or" is true when at least one side is true, so the rule is true in rows 1 and 2 from the bracket and in rows 3, 5 and 7 from P. Row 1 has both.
3. (a) The siren should sound in rows 1, 2, 3, 5 and 7, which is 5 of the 8 rows.
4. In the wired circuit the outer connective is "and", so it needs A true and is false in rows 5 to 8. In rows 1 to 4, D or P is true in rows 1, 2 and 3.
5. (b) The two columns differ in rows 5 and 7. In both the system is disarmed and the panic button is pressed: the rule sounds the siren, but the wired circuit stays silent. Check: in rows 1 to 4 A is true, so both circuits reduce to D or P and agree there.

answer(a) Rows 1, 2, 3, 5 and 7; (b) rows 5 and 7, where the system is disarmed and the panic button is pressed: the wired siren stays silent

techniqueLogical Connectives and Truth Tables

A condition is sufficient for passing when every length that meets it passes, and necessary when every length that passes meets it. One length that meets a condition and fails the check shows it is not sufficient, and one length that passes without meeting it shows it is not necessary.

1. Every rod that passes has L ≥ 199.5, so the condition L ≥ 199.5 is necessary. A rod of 201 mm meets it and still fails the check, so it is not sufficient.
2. Every length with 199.8 ≤ L ≤ 200.2 lies inside 199.5 ≤ L ≤ 200.5, so that condition is sufficient. A rod of 200.4 mm passes without meeting it, so it is not necessary.
3. (a) L ≥ 199.5 is necessary but not sufficient, and 199.8 ≤ L ≤ 200.2 is sufficient but not necessary.
4. "Passing the length check is necessary for shipping" means: if a rod is shipped, it passed the length check. Its contrapositive is: if a rod fails the length check, it is not shipped. Rod X is 200.6 mm, which is more than 200.5, so it fails the length check and is not shipped.
5. (b) Rod Y passes the length check, but a necessary condition that holds promises nothing. It fails the straightness check, so the sufficient condition does not hold, and a sufficient condition that fails says nothing either. Rod X is not shipped, and the rule leaves Rod Y undecided.

answer(a) L ≥ 199.5 is necessary but not sufficient; 199.8 ≤ L ≤ 200.2 is sufficient but not necessary. (b) Rod X is not shipped; the rule leaves Rod Y undecided.

techniqueNecessary and Sufficient Conditions · The Contrapositive

"P if and only if Q" claims two conditionals at once: if Q then P, and if P then Q. A result breaks a conditional only when the hypothesis is true and the conclusion is false.

1. Let P be "the team is promoted" and Q be "the team finishes in the top two". "P if Q" is the conditional: if a team finishes in the top two, it is promoted. "P only if Q" is the conditional: if a team is promoted, it finished in the top two.
2. Ashford finished in the top two and was promoted, so both parts of each conditional are true for Ashford. Dunmore was outside the top two and was not promoted, so the hypothesis of each conditional is false and neither makes a claim about Dunmore.
3. Brampton finished in the top two and was not promoted: the hypothesis of the first conditional is true and its conclusion false, so Brampton breaks it. Carlton was promoted from position 3: the hypothesis of the second is true and its conclusion false, so Carlton breaks it.
4. (a) "If a team finishes in the top two, it is promoted" is broken by Brampton. "If a team is promoted, it finished in the top two" is broken by Carlton.
5. (b) The new rule keeps only the first conditional. Brampton still breaks it. For Carlton the hypothesis "finishes in the top two" is false, so the new rule makes no claim about Carlton. Only Brampton's result breaks the new rule.

answer(a) "If a team finishes in the top two, it is promoted", broken by Brampton (position 2, not promoted); "if a team is promoted, it finished in the top two", broken by Carlton (position 3, promoted). (b) Only Brampton's result: 1 of the four.

techniqueIf and Only If · Conditional Statements

Access is P and (K or C). Denial is its negation, found by De Morgan's laws: the negation of an "and" is an "or" of the negations, and the negation of an "or" is an "and" of the negations.

1. Write the rule: access is granted exactly when P and (K or C).
2. Negate the outer "and": (not P) or not (K or C). Then negate the inner "or": not (K or C) is (not K) and (not C).
3. (a) Access is denied when the password is wrong, or when the device is not recognized and the one-time code is wrong: (not P) or ((not K) and (not C)).
4. The correct deny test is true in the 4 combinations with P false, and in 1 more with P true and K and C both false: 5 in all. The developer's test is true only when all three are false, which is 1 combination, and that one is among the 5.
5. (b) The developer's code grants access in the other 5 − 1 = 4 combinations that the rule denies. One of them is a wrong password on a recognized device with a correct one-time code. Check: the developer's test is true only where the correct one is, so it never denies access the rule grants.

answer(a) Denied when (not P) or ((not K) and (not C)): the password is wrong, or the device is not recognized and the one-time code is wrong. (b) 4 combinations

techniqueNegating a Statement · Logical Connectives and Truth Tables

"Every train that stops at Airport also stops at Central" claims, for every train: if it stops at Airport, then it stops at Central. Its negation says that for at least one train that conditional fails, which means a train that stops at Airport and does not stop at Central.

1. The negation of "for every train, if it stops at Airport then it stops at Central" is "for some train, it stops at Airport and it does not stop at Central".
2. The trains that stop at Airport are 1, 2, 3 and 5. Trains 1 and 2 also stop at Central; trains 3 and 5 do not.
3. (a) The negation is "Some train stops at Airport and does not stop at Central". It is true, and trains 3 and 5 make it true.
4. "Every train stops at Harbor" is false as soon as one train does not stop there, so its negation is "at least one train does not stop at Harbor", which is what "Not every train stops at Harbor" says. "No train stops at Harbor" claims that every train fails to stop there, which denies far more.
5. (b) The first passenger states the negation, and it is true: trains 2, 5 and 6 do not stop at Harbor. "No train stops at Harbor" is false, because train 1 stops there.

answer(a) "Some train stops at Airport and does not stop at Central"; it is true, because of trains 3 and 5. (b) "Not every train stops at Harbor" is the negation, and it is true (trains 2, 5 and 6); "No train stops at Harbor" is false (train 1 stops there).

techniqueNegating a Quantified Statement · For All and There Exists · Conditional Statements

Put the results in a grid with one row per trainee and one column per assessment. "Some trainee passed every assessment" needs one row passed all the way across. "Every assessment was passed by some trainee" needs each column to hold at least one pass, possibly by different trainees.

1. Draw the grid of passes. For "Some trainee passed every assessment", look for a full row. Ana and Chen each passed 3 of the 4, Ben passed 2 and Dev passed 1, so no row is full.
2. For "Every assessment was passed by some trainee", check each column. Swimming was passed by Ana, rescue by Ana, first aid by Ben and the written test by Ana, so every column holds a pass.
3. (a) "Some trainee passed every assessment" is false, and "Every assessment was passed by some trainee" is true. The same two quantifiers in the other order make a different claim.
4. To negate "for every trainee there is some assessment the trainee passed", change "every" to "some" and "some" to "every", and negate what is claimed: "Some trainee failed every assessment."
5. (b) The negation needs a row with no pass at all. Every trainee passed at least one assessment (Dev passed the written test), so the negation is false. Check: the original statement is then true, and exactly one of a statement and its negation is true.

answer(a) "Some trainee passed every assessment" is false (the most any trainee passed is 3 of the 4); "Every assessment was passed by some trainee" is true. (b) "Some trainee failed every assessment"; it is false, because each trainee passed at least 1 assessment.

techniqueFor All and There Exists · Negating a Quantified Statement

The sign is a conditional: if a bag weighs more than 23 kg, then it carries a HEAVY tag. It is broken only by a bag that weighs more than 23 kg and has no tag, so a record needs checking only if its hidden fact could complete that pair.

1. The only bag that breaks the sign is one that weighs more than 23 kg and carries no HEAVY tag.
2. Bag 1 weighs 27 kg, so if it has no tag it breaks the sign: look at its tag. Bag 2 weighs 15 kg, so it follows the sign whatever its tag.
3. Bag 3 has a tag, so it follows the sign whatever it weighs. Bag 4 has no tag, so if it weighs more than 23 kg it breaks the sign: weigh it. (a) The supervisor must look at bags 1 and 4.
4. The converse is: if a bag carries a HEAVY tag, it weighs more than 23 kg. It is broken by a tagged bag of 23 kg or less, so the handler weighs bag 3 and looks for a tag on bag 2.
5. (b) The handler looks at bags 2 and 3. A tagged bag of any weight and a 15 kg bag with or without a tag both follow the real sign, so 0 of the handler's 2 checks can reveal a breach, and bags 1 and 4 go unchecked.

answer(a) Bags 1 and 4; (b) bags 2 and 3, and 0 of those checks can reveal a breach

techniqueConditional Statements · The Converse and the Inverse · The Contrapositive

Written as a single condition, a year is a leap year when it is divisible by 4 and not by 100, or when it is divisible by 400. A claim about every year in a range is false if one year in the range breaks it, and that year is a counterexample.

1. Write the definition as one condition: a year is a leap year when it is divisible by 4 and not by 100, or when it is divisible by 400.
2. 2024 = 4 × 506 and is not divisible by 100, so it is a leap year. 1900 and 2100 are divisible by 100 but not by 400, since 1900 ÷ 400 = 4.75 and 2100 ÷ 400 = 5.25, so neither is a leap year. 2000 = 400 × 5, so it is a leap year.
3. (a) 2000 and 2024 are leap years; 1900 and 2100 are not.
4. The proposition is false: 2100 is in the range and divisible by 4, but it is not a leap year. One counterexample is enough to make a claim about every year false.
5. (b) The multiples of 4 from 2004 to 2100 number 2100 − 20044 + 1 = 25. The only one divisible by 100 is 2100, which is not a leap year, so there are 25 − 1 = 24 leap years. Check: 2000 is outside the range, so no year divisible by 400 is added back.

answer(a) 2000 and 2024; (b) the proposition is false (2100 is a counterexample), and there are 24 leap years

techniqueDefinitions, Propositions and Theorems · For All and There Exists