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Math 215 Worksheet 1: Conditional statements
- Determine which of the following sentences are statements. For each that is a statement,
- determine if it can be rephrased as a conditional statement or not; if so, identify the hypothesis and the conclusion,
- circle or highlight the quantifiers,
- search for a counterexample,
- either establish its falsity, or conjecture its truth/falsity.
(a) The product of two consecutive positive integers is even.
Solution: This is a conditional statement because we can rephrase it as “if n and m are consecutive positive integers, then nm is even.” Equivalently, “for all consecutive positive integers n and m, nm is even,” which involves the universal quantifier “for all.” The hypothesis is n and m being consecutive positive integers, and the conclusion is nm being even. Let’s try some examples:
1 · 2 = 2, 2 · 3 = 6, 3 · 4 = 12, 4 · 5 = 20, 5 · 6 = 30, ,... ,
This statement appears to be true.
(b) The equation 3x + 8 = 0 has a solution.
Solution: This is not a statement, because “a solution” is ambiguous. If we replaced this with “a rational solution” or ”a real solution” the statement would be true, but if we used “a positive rational solution” or ”an integer solution” it would be false.
(c) The set of even numbers is closed under addition.
Solution: This is a statement; it can be rephrased to “if n and m are even numbers, then n + m is even,” which is a conditional statement. The hypothesis is n and m being even, and the conclusion is n + m being even. We could also rephrase it as a conditional statement involving the universal quantifier “for all” as in part (a). Moreover, since even/odd only makes sense for integers, it is implied that the numbers involved are integers.
This statement appears to be true; adding several combinations of small even numbers gives
2 + 2 = 4, 2 + 4 = 6, 4 + 4 = 8, 2 + 6 = 8, 4 + 6 = 10, 6 + 6 = 12,... ,
which are all even.
(d) The set of odd numbers is closed under addition.
Solution: This is conditional statement as in part (c). The hypothesis is n and m being odd, and the conclusion is n + m being odd. This statement is false, since 1 + 1 = 2 forms a counterexample. In fact, every pair of odd numbers appears to form a counterexample!
1 + 3 = 4, 3 + 3 = 6, 1 + 5 = 6, 3 + 5 = 8, 5 + 5 = 10,... ,
Based on this evidence, we can make an opposite conjecture: that the sum of any two odd numbers is even.
(e) There exist positive integers x and y such that 2x + 3y = 16.
Solution: This is a statement involving the existential quantifier “there exists.” It is not a conditional statement. Let’s search for positive integer solutions:
2(1) + 3(1) = 5, 2(1) + 3(2) = 8, 2(1) + 3(3) = 10, 2(1) + 3(4) = 13, 2(1) + 3(5) = 17, 2(2) + 3(1) = 7, 2(2) + 3(2) = 10, 2(2) + 3(3) = 13, 2(2) + 3(4) = 16
and we’ve found a solution, x = 2, y = 4! Is this the only one (is the solution unique)? Continuing the search yields x = 5, y = 2 as another solution. This appears to be all of them...
(f) The sum of any three consecutive positive integers is divisible by 3.
Parts (a) and (b) fit in the last line of the truth table; the hypothesis and conclusion are both false, and the statement is vacuously true. Part (c) fits with the third line; the hypothesis is false and the conclusion is true, and again the statement is vacuously true. Part (d) is an example of the first line; the hypothesis and conclusion are both true in this case, which suggests that the statement may be true. We don’t have any example of the second line; this would be a counterexample, showing that the statement is false. That’s what we’ll look for in the last part.
(f) For convenience, let p(n) = n^2 + n − 1. Listing the first five values of p(n) for integers n = 2, 3 , 4 , 5 , 6 gives
{p(n) : 2 ≤ n ≤ 6 } = { 5 , 11 , 19 , 29 , 41 }.
Since these are all prime numbers, the statement appears to be true. These are all examples of the first line to the truth table; the hypothesis and conclusion for each of these cases are true. Now, notice that the next case is p(7) = 55 which is not prime, so in fact the statement is false. Here we finally have an example of the second row of the truth table; to disprove the statement we only need a single counterexample. This is a good example showing that to make a plausible conjecture, a large data set of examples is necessary.
