"Mathematical Reasoning" — Practice

Question 1

Which of these are statements? (i) Close the window. (ii) $9$ is a prime number. (iii) $x + 5 = 9$. (iv) Every rectangle is a square.

Accepted Answer

(ii) and (iv) are statements. (ii) is false, since 9 is 3 times 3, but being false does not disqualify a sentence; it is definitely one or the other. (iv) is also false, since a rectangle need not have equal sides. (i) is a command and has no truth value at all. (iii) is an open sentence: it is true for $x=4$ and false otherwise, so it becomes a statement only once $x$ is fixed or quantified.

Question 2

Write the negation of: "All students in the class passed the examination."

Accepted Answer

Negating a universal statement produces an existential one, so the negation is: at least one student in the class did not pass the examination. Equivalently, there exists a student who failed. The common wrong answer, no student passed or all students failed, is a much stronger claim: it would be false in a class where all but one passed, whereas the correct negation is true there.

Question 3

Write the negation of "$\sqrt2$ is rational or $\pi$ is rational".

Accepted Answer

By De Morgan, the negation of a disjunction is the conjunction of the negations: $\sqrt2$ is not rational AND $\pi$ is not rational, that is, both are irrational. Note the connective flips from or to and. Since the original statement is false, the negation must be true, and indeed both numbers are irrational, which is a useful consistency check on the answer.

Question 4

For "if a triangle is equilateral, then it is isosceles", write the converse, inverse and contrapositive, and state which are true.

Accepted Answer

Original: equilateral implies isosceles, true, since equal on all three sides certainly means equal on two. Converse: isosceles implies equilateral, false, as a triangle with sides 5, 5, 8 shows. Inverse: not equilateral implies not isosceles, false, and the same triangle refutes it, which is expected since converse and inverse are equivalent. Contrapositive: not isosceles implies not equilateral, true, and necessarily so because it says exactly what the original says.

Question 5

Write the negation of "if a number is divisible by 6, then it is divisible by 3".

Accepted Answer

The negation of an implication is not another implication. Using the equivalence with not-p or q and applying De Morgan, the negation is p and not-q: there exists a number divisible by 6 that is not divisible by 3. As it happens this is false, since the original statement is true, and that is the point: negation must produce something false when applied to a true statement, which the inverse would not have done.

Question 6

Show that $p \Rightarrow q$ and its contrapositive have identical truth tables.

Accepted Answer

For $p\Rightarrow q$ the values across the four rows TT, TF, FT, FF are T, F, T, T. For the contrapositive, not-q implies not-p, evaluate row by row: with $p$ and $q$ both true, not-q is false so the implication is true; with $p$ true and $q$ false, not-q is true and not-p is false, so the implication is false; with $p$ false and $q$ true, not-q is false so it is true; with both false, not-q is true and not-p is true, so it is true. The pattern T, F, T, T matches exactly, so the two are equivalent.

Question 7

Determine whether $(p \lor q) \land (\sim p)$ is a tautology, a contradiction, or neither.

Accepted Answer

Evaluate across the four rows. With $p$ true, not-p is false and the conjunction is false, covering the first two rows. With $p$ false and $q$ true, the disjunction is true and not-p is true, so the conjunction is true. With both false, the disjunction is false and the conjunction is false. The column reads F, F, T, F: true in one row and false in three, so it is neither a tautology nor a contradiction but a contingency. In words it says q holds and p does not.

Question 8

Write the negation of "there exists a real number $x$ such that $x^{2}+1=0$".

Accepted Answer

Negating an existential statement produces a universal one: for every real number $x$, $x^2 + 1 
e 0$. The original is false, since a real square is never negative and $x^2+1$ is at least 1, so the negation must be true, and it is. Note that the negation quantifies over every real number, which is why proving it requires an argument while disproving the original would have required only one example had one existed.

Question 9

Prove by contrapositive: if $n^{3}$ is odd, then $n$ is odd.

Accepted Answer

The contrapositive is: if $n$ is even then $n^3$ is even. Let $n = 2m$. Then $n^3 = 8m^3 = 2(4m^3)$, which is even. The contrapositive is proved, and since it is logically identical to the original, the original is proved. The direct route would require deducing the parity of $n$ from a fact about $n^3$, which needs a cube root and is awkward, whereas the contrapositive starts from a clean substitution.

Question 10

Disprove: "for every positive integer $n$, the number $n^{2}+n+41$ is prime."

Accepted Answer

The claim survives a surprising number of tests: $n=1$ gives 43, $n=2$ gives 47, and it continues to produce primes for a long stretch. But at $n = 40$ the expression is $1600 + 40 + 41 = 1681 = 41^2$, which is not prime. One counterexample disproves a universal claim outright. This example is worth remembering as a caution: dozens of confirming cases prove nothing about a for-all statement, since a single failure anywhere destroys it.

Question 11

Prove by contradiction that there is no largest prime number.

Accepted Answer

Suppose there were, and call it $P$. Form $N = (2 	imes 3 	imes 5 	imes \cdots 	imes P) + 1$, the product of every prime up to $P$, plus one. $N$ leaves remainder 1 on division by each of those primes, so none of them divides it. But every integer above 1 has at least one prime factor, so $N$ has a prime factor, and that factor must exceed $P$. That contradicts $P$ being the largest. The assumption fails, so no largest prime exists. Note that $N$ itself need not be prime; only that it forces a prime larger than $P$ to exist.

Question 12

Write "$p$ if and only if $q$" as a conjunction of two implications, and say when it is true.

Accepted Answer

It is $(p\Rightarrow q) \land (q\Rightarrow p)$. Building the truth table, the biconditional is true when both statements are true and also when both are false, and false in the two rows where they disagree. The second case surprises people: two false statements still agree with each other, so the biconditional holds. In practice, if and only if is a promise that both directions have been separately checked, which is why a definition is a biconditional while a one-way theorem is not.

Practice — "Mathematical Reasoning"

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