Investigation: Constructing a Proof
This was a 45 minute session, led by the tutor in each breakout room. Here is the problem we gave them. Prove the following theorem:
Fourth powers of prime numbers have exactly 5 factors, and these are the only natural numbers that have this property. Notes: You will recall that:
1. If an integer x is a factor of an integer n, then x divides n, n = xy, where y is also an integer, and y = x/n is also a factor of n. 2. The number 12 has exactly 6 factors, 1, 2, 3, 4, 6 and 12, i.e. 1 and the number itself are counted. 3. A prime is a positive integer having exactly 2 factors. 4.
So 1 is not a prime number.
And here are the notes we gave tutors beforehand: Purpose To extend John's presentation (Logic & Proof Structures) into their own experience To encourage a structured approach to finding proofs To show them that formal logic is actually applicable to everyday problems To give them a chance to lie back and watch others do the work if they want to - some of them were very tired at the final session last year. Tutor notes • Please give them the hand-out summarising John's presentation at the start of the session, so that they can refer to it. • Give them a chance to read the notes given in the statement, and think about the problem for a bit • Let them try a few examples, to give them a feel for what's going on. • Get them to express the statement in formal logic terms: Perhaps start with the first bit; Fourth powers of primes have exactly 5 factors Nudge them towards setting up the propositions, so they get as far as a → b Then let them bring in the second part, and these are the only such numbers, to change the → to ↔ • Link with John's presentation Try and get them to find the logical equivalence a ↔ b ≡ (a → b) ∧ (b → a) in John's notes As you draw the proof out of them, please encourage them to link what they're doing with the presentation methods and notation. It may even offer them a starting point for a particular bit of the proof. (In the handout, I've summarised what we did in logic notation at the end − but your group may not use the same approach as the handout) • If you have a bunch of number theorists in your room, they may go for the jugular in the second part, i.e. (b → a), by using the following: k
n = p1 1 p2 2 ! pmm where the pi are distinct primes, and ki ≥ 1 r
Then any factor s of n is of the form s = p11 p22 ! pmm , where 0 ≤ ri ≤ ki. The total number of factors n is (k1 + 1)( k2 + 1) … (km + 1) = 5, which is prime. So we must have m = 1 and k1 = 4, giving the result. This is fine, but may be a bit difficult for the others to follow, so you'll have to make the NTs convince them. • Please give them the hand-out of the proof at the end of the session If your group goes for a different approach, it may be worth mentioning the alternative method in the handout, if you have time at the end. Shirleen Stibbe
Constructing a Proof (handout for students)
Prove the following theorem:
Fourth powers of prime numbers have exactly 5 factors, and these are the only natural numbers that have this property. Proof
Let a be the proposition 'n is the fourth power of a prime' Let b be the proposition 'n has exactly 5 factors' First we must prove that if a natural number n can be expressed as the fourth power of a prime, then n has exactly 5 factors (a → b), and then we must prove that the only natural numbers having exactly 5 factors are the fourth powers of primes (b → a).
We begin by proving a → b. Suppose n can be expressed as the fourth power of a prime number. Then n = p4, where p is prime. The factors of n are 1, p, p2, p3 and p4 and we see that n has exactly 5 factors. So a → b.
Next, prove b → a, i.e. if n has exactly 5 factors, then n is the fourth power of a prime - harder! It is easier to deal with this if we label the factors, and order them by size so we know exactly what we are dealing with. Suppose the factors of n are 1, r, s, t and n, where 1 < r < s < t < n. We know that n/n (= 1), n/t, n/s, n/r and n/1 (= n) are also factors of n, where 1 < n/t < n/s < n/r < n, so these must also be the 5 factors of n. It follows that n/t = r, n/s = s and n/r = t. In other words, n = rt = s2. So far, we have proved b → (n is the square of an integer). To complete the proof, we need to show that the integer s must be the square of a prime, i.e. s = p2, where p is prime, so that n = s2 = (p2)2 = p4. If s is not the square of a prime, then either: a) s is a prime number, or b) s has two different factors, x and y, such that s = xy and 1 < x < y < s. In the first case, if s = p, where p is prime, then n = s2 = p2 and n has only 3 factors, 1, p and p2. In the second case, if s = xy, where x and y are different, then n = s2 = (xy)2 = x2y2, and n has at least 7 distinct factors; 1, x, y, xy, x2y, xy2 and x2y2. The two cases show that if s is not the square of a prime, then n does not have 5 factors - a contradiction. So s = p2 and n = s2 = p4, where p is a prime. We have succeeded in proving that b → a. Therefore, fourth powers of prime numbers have exactly 5 factors, and these are the only natural numbers that have this property. Shirleen Stibbe
Summary What we actually did (in logic notation)
• We defined the following propositions: a
n = p4, p a prime
n has exactly 5 factors
so that we could (implicitly) express the statement of the theorem as a ↔ b. • To prove a ↔ b, we proved the logically equivalent statement (a → b) ∧ (b → a). • First, we proved a → b using a direct proof (we assumed a and deduced b). • In order to prove b → a, we needed some intermediate propositions: c
n = s2, s an integer
s = p2, p a prime
• To prove b → a, We first proved b → c, using a direct proof, by assuming b and deducing c. Then we proved b → d using a contrapositive argument, i.e. we proved the logically equivalent statement ¬d → ¬b, by assuming d was false and deducing that b was false, although we disguised the method as a proof by contradiction. We used, without stating or proving, ((b → c) ∧ (b → d)) → (b → (c ∧ d)) (is this proof by osmosis?) We finally stated that (c ∧ d) → a (i.e. n = s2 = (p2)2 = p4), hence b → a so that we had proved a ↔ b