Download Math 261 Spring 2014 Quiz 2 Name: Directions: Complete all of the

Math 261
Spring 2014
Quiz 2
Name:
Directions: Complete all of the following to the best of your ability. If you do not understand a question, please
let me know; I may be able to assist you. SHOW ALL WORK! You will be graded primarily on the method you
use, not your final answer. GOOD LUCK!
1. (Definition, 1 Point) Define composite number.
A natural number n is composite if n is the product of values less than n. In symbols, n = a · b where
either a < n and b < n.
2. (Definition, 1 Point) Finish the statement of the existence part The Fundamental Theorem of
Arithmetic (FTOA). Every natural number n greater than 1 is either prime or . . .
has a prime factorization, i.e, n can be written as
n = pr11 · pr22 · prt t
where p1 , p2 , . . . , pt are prime numbers and r1 , r2 , . . . , rt are positive exponents.
3. (Quick Proof, 3 Points) Prove the following:
Use FTOA to determine the number of zeroes at the end of 15! (Recall: n! = 1 · 2 · · · (n − 1) · n).
Proof: By the FTOA, every factor in 15! can be written as a product of primes. In particular, we
are interested in the number of times the primes 2 and 5 appear since a zero at the end of a number
corresponds to a factor of 10 = 2 · 5.
Note that every other factor in 15! contributes at least one power of 2, with 4 = 22 , 8 = 23 , and
12 = 22 · 3 contributing more. So there are 2 · 22 · 2 · 23 · 2 · 22 · 2 = 211 powers of 2 in the prime
factorization of 15!.
Only every fifth factor in 15! contributes a power of 5, and these come from 5, 10 = 5 · 2, and 15 = 5 · 3.
So there are 5 · 5 · 5 = 53 powers of 5 in the prime factorization of 15!.
Each pair of 2 and 5 (multiplied together) contributes one more zero at the end of 15!. Since there are
only three such pairs (because there are only three factors of 5), there must be THREE zeroes at the
end of 15!.
4. (Medium Proof, 5 Points) Use FTOA to prove there are infinitely many prime numbers by incorporating the following steps into a proof:
(a.) Suppose that there are only finitely many primes, say {p1 , p2 , . . . , pn }.
(b.) Define Q = p1 · p2 · · · pn and Q + 1.
(c.) Use FTOA and arrive at a contradiction.
Proof: Assume, looking for a contradiction, that there are only finitely many primes, say {p1 , p2 , . . . , pn }.
We can form two numbers,
Q = p1 · p2 · · · pn and Q + 1 = p1 · p2 · · · pn + 1.
By definition, Q is divisible by every prime. By FTOA, Q+1 is either prime or has a prime factorization.
If Q + 1 is prime, then we are done since Q + 1 = p1 · p2 · · · pn + 1 is bigger than any prime in the set
{p1 , p2 , . . . , pn }. We have reached a contradiction since we assumed the set {p1 , p2 , . . . , pn } contains
all primes, but Q + 1 is a prime not in this set.
If Q + 1 is not prime, then by the FTOA it must have a prime factorization. Since {p1 , p2 , . . . , pn } is
a complete list of primes,
Q + 1 = pr11 · pr22 · · · prt t
for some positive exponents r1 , r2 , . . . , rt . This too leads to a contradiction. We know that the gcd
of these two numbers is (Q, Q + 1) = 1, so having prime factors in common between Q and Q + 1 is
impossible. We could also see that
1 = Q + 1 − Q = pr11 · pr22 · · · prt t − p1 · p2 · · · pn = p1 · p2 · · · pn · (p1r1 −1 · p2r2 −1 · · · prt t −1 − 1).
This says that p1 · p2 · · · pn divides 1, which is clearly impossible.
Our contradiction means that the assumption that here are only finitely many primes cannot be true.
So there must be an infinite number of primes.