exam 1 (mathematical thought)

definition: divisibility

let d and n be integers. then n is divisible by d, or d divides n, if there is an integer k such that n=dk •notation: d|n "d divides n" - an assertion if its true or false

example: the product of two even integers is divisible by 4

let m and n be the two even integer. then there are integers j and k such that mn=4(jk). m=2j, n=2k... mn=2j(2k)... mn=4(jk)

theorem: the collection of all prime numbers isnt finite proof: ?

let {2,3,5,7} be the finite collection of prime numbers. multiply all the numbers on the list, and call the result M. consider the number M+1. ether its prime or it is NOT

example: show that 1000x999x998x997 is even

1000x999x998x997= 2(500)(999x998x997). there is an integer (500x999x998x997) which multiplied by 2 yields the given integer, thus the given integer is even

divisibility

12 is divisible by 3 - - 3 divides 12 (3|12) 35 is divisible by 7 - - 7 divides 35 (7|35)

case 2

M+1 is not prime. then, it must be divisible by some prime number. lets call it p. p is not in the list. suppose it were, the p|M by the lemma. this is impossible. thus p is NOT on the list

case 1

M+1 is prime. it cannot be in the finite list since it is much bigger than all those primes

natural numbers

The numbers used for counting (1,2,3...)

definition: natural number

a natural number n is even if there is a natural number k such that n=2k

definition: prime numbers

a prime number is a natural number which is larger than 1 and only divisible by 1 and itself *ex: 2,3,5,7,11...

sieve of eratosthenes

a way in which to find the prime numbers than m *ex: m=18 {2,3,5,7,11,13,17}

definition: integer (even)

an integer n is even when there is an integer k such that n=2k *ex: -6 is even, since -6=2(-3)

definition: integer (odd)

an integer n is odd when there is an integer k such that n=2k+1

FALSE STATEMENT let d, m, and n be integers such that d|m and m|n. then m|d+n

d=2 m=6 n=12 d|m-- 2|6 m|n-- 6|12 m|d+n-- 6|14 then, d|m and m|n but, m does not divide d+n

fact

no natural number larger than 1 divides 1

lemma: suppose d is a natural number bigger than 1, and n is an integer such that d|n. then d does not divide n+1 proof: ?

suppose d did divide n+1. then by the proposition, d would divide (n+1)-n=1. this contradicts the fact. hence, d cannot divide n+1

Proposition A

suppose d|m and d|n. then d|m+n

Proposition B:

suppose d|m and d|n. then d|m-n

definition: GCD

the greatest common denominator of two integers is the largest natural number that divides both of the integers • we use (m,n) to denote the greatest common divisor of integers m and n two integers are said to be relatively prime if their greatest common divisor is 1

integers

the set of whole numbers and their opposites (...-2, -1, 0, 1, 2...)

theorem: integer (even)

the sum of two even integers is an even integer

case 1+2 CONCLUSION

thus in either case, the finite list omits some prime number. the argument applies in EVERY finite list of primes. some finite lists contains all prime numbers

proof by contradiction

to show a statement is true, assume it is false and arrive at something which contradicts a known truth

proposition: for any integers m and n, (m,n)=(m-n,n) proof: ?

we show that all divisors of m and n are also divisors of m-n, and that all divisors of m-n and n are also divisors of m. * suppose d|m and d|n. by proposition B, d|m-n. then all numbers that divide m and n divide m-n and n *suppose d|m-n and d|n. then by proposition A, d|m, since (m-n)+n=m some common divisors of m-n and n are common divisors of m and n . since the sets of common divisors coincide, the greatest common divisors are the same

is zero even?

yes, 0 is even, since 0=2(0)

proposition: the natural number is 10 an even number proof: ?

•5 is a natural number, and 10=2(5), so 10 is even

theorem: the sum of two even natural numbers is even proof: ?

•let m and n be the two even numbers. since m is even, there is a natural number j such that m=2j. similarly, since n is even, there is a natural number k such that n=2k. then m+n= 2j+2k. m+n= 2(j+k). since j+k is a natural number, this shows that m+n is even

theorem: the sum of two odd integers is even proof: ?

•let m and n be the two odd integers. since m is odd, there is an integer j such that m=2j+1. similarly, since n is odd, there is an integer k such that n=2k+1. then m+n= 2j+1+2k+1. m+n=2j+2k+2. m+n=2(j+k+1)

proposition: let d, m and n be integers that d|m and m|n. then d|n proof: ?

•since d|m there is an integer j such that m=dj and since m|n there is an integer k such that n=mk. then n=(dj)k. d(jk) so, d|n

proposition: let m and n be integers that 3|m and 3|n. then 3|m+n proof: ?

•since m is divisible by 3, there is an integer j such that m=3j. similarly, since n is divisible by 3, there is an integer k such that n=3k. then 3|m+n = m+n=3j+3k. 3(j+k)

proposition: let m and n be integers that 8|m and 8|n. then 8|m+n proof: ?

•since m is divisible by 8, there is an integer j such that m=8j. similarly, since n is divisible by 8, there is an integer k such that n=8k. then 8|m+n = m+n=8j+8k. 8(j+k)

theorem: let m and n be integers that d|m and d|n. then d|m+n proof:?

•since m is divisible by d, there is an integer j such that m=dj. similarly, since n is divisible by d, there is an integer k such that n=dk. then d|m+n = m+n=dj+dk. d(j+k)