WORKSHEET MATH 215, FALL 15, WHYTE. We begin our course with the natural numbers:

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1 WORKSHEET MATH 215, FALL 15, WHYTE We begin our course with the natural numbers: N = {1, 2, 3,...} which are a subset of the integers: Z = {..., 2, 1, 0, 1, 2, 3,... } We will assume familiarity with their most basic properties: sums and products, commutativity, ordering ( for any two a and b, exactly one of a < b, a > b, or a = b is true ) etc. The first thing we wish to study is divisibility: Definition 0.1. Let a and b be two integers. We say that a divides b ( alternatively, b is a multiple of a) if there is an integer q so that b = aq. We will use the notation a b as shorthand for this property. Proposition 0.2. Let a, b, and c be integers. If a b and a c then a (b + c). Proposition 0.3. Let a, b, and c be integers. If a b and a c then a (b c). Proposition 0.4. Let a, b, and c be integers. If a b and a c then a bc. Proposition 0.5. Let a, b, c, s and t be integers. a (sb + tc). If a b and a c then Proposition 0.6. Let a and b be natural numbers. If a b then a b. Proposition 0.7. If a is an integer then a 0. Proposition 0.8. If a is an integer such that 0 a then a = 0. Proposition 0.9. Let n, a,b, and c be integers. If n (a b) and n (b c) then n (a c). Question What are the converses and contrapositives of the above propositions? Which contrapositives are true? 1

2 2 WORKSHEET MATH 215, FALL 15, WHYTE Definition Let a and b be integers and let n be a natural number. We say a is congruent to b modulo n if n (a b). We use the notation a b mod n for this property. Problem Decide whether each of the following statements is true and justify your answers: 2 93 mod mod mod mod 2 Proposition If a is an integer and n a natural number then a a mod n. Proposition Let a and b be integers and n a natural number. If a b mod n then b a mod n. Proposition Let a, b, and c be integers and n a natural number. If a b mod n and b c mod n then a c mod n. Proposition Let a, b, and c be integers and n a natural number. If a b mod n then a + c b + c mod n. Proposition Let a, b, and c be integers and n a natural number. If a b mod n then ac bc mod n. Proposition Let a, b, and c be integers and n a natural number. If a b mod n and b c mod n then a c mod n. Proposition Let a and b be integers and n a natural number. If a b mod n then a 2 b 2 mod n. Proposition Let a and b be integers and n a natural number. If a b mod n then a 3 b 3 mod n. The last two propositions suggest the following: Conjecture Let a and b be integers and n and m natural numbers. If a b mod n then a m b m mod n. Do you believe this conjecture? If so can you give an explanation for why it is true?

3 WORKSHEET MATH 215, FALL 15, WHYTE 3 The conjecture 0.21 is a natural example of proof by induction. To make such arguments completely rigorous we need another axiom about the natural numbers: Axiom The well-ordering principle for N: If S be any non-empty subset of N then S contains a smallest element. The proof of conjecture 0.21 is a typical application. Start by letting S be the set of natural numbers m which make the statement false ( you ve proven already that {1, 2, 3} are not in S). The conjecture is then equivalent to the claim that S is empty (why?). If it isn t then it has a smallest element (why?). Get a contradiction from this. Problem Give a proof of conjecture 0.21 using well ordering. The well ordering principle is often used more generally than for N. Definition A set S of integers is bounded below if there is an integer n so that n s for all elements s of S. Proposition If S is a subset of N then S is bounded below. Proposition If S is a subset of Z which is bounded below then there is a natural number k so that S + k N. Proposition The well ordering principle for Z Let S be a nonempty subset of Z. If S is bounded below then S has a smallest element. Another basic result about arithmetic follows from the well ordering principle: Proposition Let a be an integer and b a natural number. There is an integers q and r with 0 r < b such that: a = bq + r Here q is called the quotient and r is called the remainder. Here s a hint: think about all possible q and r 0 that make the equation hold (without the assumption r < b) and then use well ordering to find the smallest such r. This proposition is often called the division algorithm because it is tells you exactly what one gets from old fashioned long division of natural numbers - a quotient and a remainder. However there s a subtlety here - the proposition says that q and r exist, but not that they are unique - in other words the division problem might have more than one right answer. Obviously that s not what we expect. Here s how that is phrased precisely (make sure you understand why, then prove it): Proposition Let b a natural number and q 1, q 2, r 1, r 2 integers with 0 r 1 < b and 0 r 2 < b such that q 1 b + r 1 = q 2 b + r 2 then q 1 = q 2 and r 1 = r 2.

4 4 WORKSHEET MATH 215, FALL 15, WHYTE Definition A natural number n > 1 is prime numbers m with m n are m = 1 and m = n if the only natural Proposition If n > 1 is a natural number then there is a prime number p such that p n. Proposition Every natural number n > 1 can be written as a product of primes: n = p k 1 1 pkm m where p 1, p 2,..., p m are prime numbers and k i are natural numbers. This is the prime factorization theorem. It usually also comes with a uniqueness statement. Can you figure out what this should say? It tuns out that the uniqueness is more subtle than it appears, so we will need to develop some other ideas before tackling it. We start with another definition: Definition Given two integers a and b a greatest common divisor is an integer d satisfying: d a and d b For any c with c a and c b, c d (The first part says d is a common divisor of a and b, and the second part says it is the largest one.) Proposition Show that unless a and b are both zero they have a greatest common divisor. Show that there is no greatest common divisor for a = b = 0. Problem Let n be a natural number. divisor of n and 0. Find the greatest common Proposition Let p be a prime number. Show that for any integer a the greatest common divisor of p and a is either p ( if p a) or is 1. Proposition Let a and b be natural numbers. Write a = bq + r using the division algorithm. Show that the greatest common divisor of a and b is the same as the greatest common divisor of r and b. This gives a practical way to compute greatest common divisors: take the larger of a and b and replace it with its remainder when divided by the other, and repeat until one of the numbers is zero. Problem What is the greatest common divisor of 120 and 168? of 59 and 1016? Question Can you show that this process always works?

5 WORKSHEET MATH 215, FALL 15, WHYTE 5 Here is a surprisingly useful way to think about greatest common divisors: Proposition Given a and b, not both zero, let S be the set {ax + by : x, y Z}. Show that if c is a common divisor of a and b then every element of S is a multiple of c ( hint: look back at the first few propositions in the worksheet). Show that S contains a positive integer, so by well ordering has a smallest positive element. Call this smallest positive element D. Show that D a and D b ( hint: think about what the remainders would be if not, and use what it means for D to be in S ) Conclude that D is the greatest common divisor of a and b. Two important facts that follow from this perspective : Proposition Let a and b be integers, not both zero, and let d be their greatest common divisor. There are integers x and y so that ax + by = d. Further, if c is any integer with c a and c b then c d. Using all of this, prove: Proposition Let p be a prime number and a and b integers. If p ab then p a or p b. Problem Prove that the prime factorization of a natural number is unique.