Math 3303 Module 1B Chapter 3 pages 25 – 38 Page 25 The Fundamental Theorem of Arithmetic* If you have a natural number, you can write it factored to primes** in essentially one way. *Arithmetic is adding, subtracting, multiplying, or dividing numbers and has no variables! **Prime numbers have no proper divisors. Example: 12 Factored: 12(1), 3(4), 2(6) Factored to primes: 22 (3) Further, If you have two numbers, factored to primes, and you compare the factors only two outcomes are possible: A. They share no factors (we call this situation: relatively prime) B. They share at least one factor (the product of the shared factors is the Greatest Common Divisor) Example: (12, 30) = 6 (check out the notation!) This says the GCD of 12 and 30 is 6. 12 = 2 2 3 30 = 2 3 5 1 Example: (5, 21) = 1 This says 5 and 21 are relatively prime. It does NOT claim that both numbers are primes; 21 is composite! It says they share no prime factors. Pages 27 – 29 explain the Euclidean Algorithm. Please note, you need to explain this in a homework problem. In fact, you need several ways to find the GCD – the Euclidean Algorithm is one. Pages 29 – 35 You are excused from delving into these pages – they go WAY WAY beyond where we need to be in this class. Note, though, that the conclusion is interesting. The probability that two randomly chosen natural numbers are relatively prime is 61% Now, on to page 36. Here we have a Big, Old, Famous Theorem. Originally proved by Euclid (the famous geometer!). The number of primes is infinite. Now before we look at Euclid’s proof, we need to run through some preliminaries. Preliminary proofs in mathematics are called “lemmas”. They are very common. This is a proof “by contradiction”. In general, a proof by contradiction works like this: You have a claim. You form a contradiction to the claim. You take the contradiction and, using standard logic and basic math facts, you prove that it cannot be. Therefore, the original claim is the true statement. So, moving into some known math facts: Let’s review “!” – this symbol is called “factorial” and it means a special calculation. 3! = 3(2)(1) = 6 4! = 4(3)(2)(1) = 24 = 4(3!) 2 5! = 5(4!) = 5(4)(3)(2)(1) = 120 … n! = n(n−1)! = n(n − 1)(n – 2)(n – 3)…(1) Note that n! is divisible by EVERY natural number less than or equal to n. This is a consequence of how we made n!. Now, we’re interested in n! + 1. A new number, one larger than n!. Here is an important fact about this number: This number is either prime or composite. If it is a composite it is divisible ONLY by a prime factor LARGER than n. All the numbers less than or equal to n divide n! and this number is bigger than n. Example: 5! + 1 = 121 = 11(11) when factored to primes. Note that 11 is a prime LARGER than 5! Ok. Now we’re ready to start our proof. Here’s what we’re trying to prove: There are an infinite number of primes. (note: this means that if you identify a prime number, there is ALWAYS another prime bigger than the one you found) Here’s our contradiction: There is a largest prime and no primes greater than this largest one. Proof: There is a largest prime and I’m going to call it P. Now there is no prime number larger than P. I’m going to calculate P! + 1. Now, here’s what I know about P! + 1. It is prime or it is composite. 3 Now, it’s not prime because it’s bigger than P, the largest prime. Unfortunately, that means it is composite and divisible by a prime larger than P. Which contradicts our contradiction. So, really, there are an infinite number of primes. Done. Popper 2, Question 1 Fill in the blank: 3! _____ 4! A. = B. < C. > Now, on page 36 we’re back to talking more about primes. Back in the days before calculators, a famous mathematician named Fermat thought he’d come up with a formula for primes n 22 + 1 n ≥ 0 Note the complicated superscript! Let’s calculate a few of these 0 For n = 0 22 + 1213 = + = a prime number 1 For n = 1, 22 + 1 = 41 + = 5 a prime number 2 For n = 2, 22+= 1 2 4 += 1 17 another prime Now note: he didn’t claim it made all the primes in order, just that it produced primes. 4 It works up to 5 which is 232 + 1 a very very large composite number… Here’s another try at a formula: fx( )= x2 − x + 41 Note that this is a parabola. And it works up to f (41) is a composite number! And here’s another one: Px()= x2 − 79 x + 1601 The tries went on and one until early in the 20 th century when Dr. Kurt Godel proved that there is NO formula! Popper 2, Question 2 n Given Fermat’s formula: 22 + 1 Which of the following is the correct number for n = 4? A. 28 + 1 B. 15 C. 216 + 1 D. 17 Now for an ASIDE. This will show up in a couple of weeks and I want to start you on this type of number early. There is another famous type of number. We can partition the natural numbers into: {repunits}U{the rest} A “repunit” is a number made up of repeated 1’s. The word itself is a contraction of 5 “repeated units”…”rep…unit” where a single one is a “unit”. So if R is the set of repunits, then R = {1, 11, 111, 1111, …} And we call each repunit Rn where n is the number of ones in the repunit. Now repunits turn out to be quite interesting. There’s a formula for them: 10n − 1 R = n 9 101 − 1 9 R = = = 1 1 9 9 102 − 1 100 − 1 99 R = = == 11 2 9 9 9 And so on. Here’s a nice theorem about repunits: If n is divisible by a, then Rn is divisible by Ra . Now, on the mid-semester, I’ll have a theorem like this (ie one you’ve never seen before) and I’ll ask you to “illustrate the theorem”….NOT prove it, just illustrate it. Let’s illustrate this theorem. I need an n and an a and n is divisible by a. I’ll choose n = 6 and a = 2. And to illustrate the theorem I need to show that R6 is divisible by R2 . 6 Let’s look at 111, 111 and divide it by 11. This means that R6 is composite and divisible by R2 which is a prime. Note that some repunits are composite and some are primes. How can we tell which are which? Well, another theorem says: if the index (the subscript) is composite, then the repunit is always composite and if the index is prime it is sometimes a prime…theorems are pretty handy sometimes. = = Example: R3 111 3(37) while R19 is a prime number. Let’s look at a set diagram of the repunits, the primes and the composites as subsets of the natural numbers: 7 Now, there are numbers called palindromes. A palindromic number looks like this: abba or abcba. 12321 is a palindrome . A palindromic sentence is “Madam, I’m Adam.” Popper 2, Question 3 A repunit squared is ALWAYS a palindromic number. ALWAYS. A. True B. False Now since most mathematics are people, you know that they just keep messing with the definition of repunits. So they moved into repunits in bases other than 10…naturally. So let’s look at repunits base 2 = 12 1 = 112 3 = 1112 7 And just as naturally, a pattern was observed. If the repunit base 2 has n ones, the base 10 formula for the number is 22 − 1 . Let’s illustrate this with =3 + 2 ++= = 4 − 11112 2 2 2115 2 1 And, just as naturally, then, we can look at repunits base 3…and so on! Now let’s look at the number 11 – what do we know about it: 8 It’s odd, positive, natural, prime, deficient, and R2 base 10. It’s also, a whole number, and integer, a rational number, and a real number. We know quite a bit about 11 now. To review some interesting facts about numbers, check out the website: Number Gossip It tells you things about numbers up to 99, 999. It’s a fun website. Popper 2, Question 4 Which of the following is NOT TRUE of 111 (base 10) A. It is a repunit B. It is composite C. It is deficient D. It is perfect Now for a note about the homework: There’s a problem with a 3x3 matrix for you to fill in showing how putting various numbers in different places mean different things. If you put a number, like 11 in the center box. You can then begin playing with where to put another number like 2. If you put it out front of 11 like 2(11) you can get 22. If you put it as 211, you’re out there past 200. If you put it as a superscript in the upper right, 11 2 , you’ve got 121. If you put it in the lower right, 11 2 , then you’re talking about the number 3 base 10.