Find the GCD of 2322 and 654. Let a = 2322, b = 654; 2322 = 654*3 + 360 so gcd(2322, 654) = gcd(654, 360);
654 = 360*1 + 294 so gcd(654, 360) = gcd(360, 294); 360 = 294*1 + 66 so gcd(360, 294) = gcd(294, 66); 294 =
66*4 + 30 so gcd(294, 66) = gcd(66, 30); 66 = 30*2 + 6 so gcd(66, 30) = gcd(30, 6); 30 = 6*5 so gcd(30, 6) = 6
A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0,
12, 24. The least common multiple (LCM) of two numbers is the smallest number (not zero)that is a multiple of
both. Multiple of 3: 0,3,6,9,12,15,18,21,24. Multiples of 4: 0,4,8,12,16,20,24. The LCM of 3 and 4 is 12. LCM of
3, 9, 21 Solution: List the prime factors of each. 3: 3, 9: 3 × 3, 21: 3 × 7 Multiply each factor the greatest number of
times it occurs in any of the numbers. 9 has two 3s, and 21 has one 7, so we multiply 3 two times, and 7 once. This
gives us 63, the smallest number that can be divided evenly by 3, 9, and 21. We check our work by verifying that 63
can be divided evenly by 3, 9, and 21. LCM of 12, 80 Solution: List the prime factors of each. 12: 2 × 2 × 3; 80: 2 ×
2 × 2 × 2 × 5 = 80 Multiply each factor the greatest number of times it occurs in either number. 12 has one 3, and 80
has four 2's and one 5, so we multiply 2 four times, 3 once, and five once. This gives us 240, the smallest number
that can be divided by both 12 and 80. We check our work by verifying that 240 can be divided by both 12 and 80.
LCM 135, 90 Solution: 135: 3*3*3*5; 90: 2*3*3*5 127*2*5=270.
Fundamental Theorem of Arithmetic.-Any integer N can be represented as a product of primes. Such a
representation is unique up to the order of prime factors. Since, by definition, a number is composite if it has factors
other than 1 and itself, and these factors are bound to be smaller than the number, we can keep extracting the factors
until only prime factors remain. This shows existence of the representation: N = pqr..., where all p, q, r,... are prime.
To prove uniqueness, assume there are two representations: N = pqr... = uvw... We see that p divides uvw... By
Corollary, it divides one of the factors u,v,w,... Cancel them out. We can go on chipping away on the factors left and
right until no factors remain. Representation of a number as the product of primes is called prime number
decomposition. The Fundamental Theorem of Arithmetic asserts that each integer has a unique prime number
decomposition.
Prime Factorization.-The recognition of the fact that whole numbers bigger than 1 can be represented just one way
as the product of primes is attributed to Euclid, who lived from 325 BC until he died in Alexandria, Egypt in 265
BC. By expressing numbers as products of prime factors, it is easy to find their Greatest Common Divisor, or their
Least Common Multiple. First, you should know what a prime number is. It's a number that can't be expressed as
the product (that means by multiplying together) smaller numbers. An example is 5. 5 can't be expressed as the
product of any smaller numbers. Its only factors are 1 and 5. The Sieve of Eratosthanes can be used to find prime
numbers. Using this method, we see that the first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,
41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. A good way to find the prime factorization of a number is to
see if it can be divided by these small primes. If it can, then "take out" those factors, and try dividing the quotient
by small primes. Each of these successful divisions yields one prime factor. When you're left with a prime number
at the end, then you're done. I'll illustrate the method with an example. Find the prime factorization of 24: 24 ÷ 2 =
12, 12 ÷ 2 = 6 6 ÷ 2 = 3 and 3 is a prime number. As you can see, I started by dividing by the smallest prime
number, and dividing the result as many times as I could, and then at the end I was left with a prime number. So 24
is 2 × 2 × 2 × 3. You might also notice in this example, that the first quotient, 12, was written twice, as was each
quotient that follows it. If you're finding the prime factorization of a few different numbers, you might find it easier
to just write each answer once, on the line below. That looks like this: Find the prime factorization of 24: 24 ÷ 2;
12 ÷ 2 6 ÷ 2
3 Now the last number written on each line is a prime number, so the method is not only easier
to write, but it is clearer to see, as well. Again, the prime factorization of 24 is 2 × 2 × 2 × 3. This can be written as
23 × 3. The little "3" is called an exponent, and means that the 2 appears three times as a factor of 24. You say this
"two to the power three times three". Now let's use the method with a different number. Find the prime factorization
of 180: 180 ÷ 2 90 ÷ 2
45 ÷ 3 15 ÷ 3 5 So 180 = 22 × 32 × 5. One more example -- Find the prime
factorization of 77: 77 ÷ 7 11 So 77 is 7 × 11
Problem 2) Show that the Nth root of any product of two distinct primes “p” and “q” is irrational for any positive
integer N >= 2. Generalize this statement. Solution: By contradiction: Assume it is rational and find a contradiction.
N root of p*q = a / b for some positive integers a, b which are relatively prime. p*q = aN / bN Therefore aN
= p*q*bN So “p” is a factor of p*q*bN, so it is a factor of “a” by FTA since “p” is a prime. So pN is a factor of
aN, so aN = pN*c for some positive integer “c”. pN*c = p*q*bN Therefore pN-1 * c = q*bN, N-1 >= 1 so “p”
divides p*bN. But since p!=q and “p” is prime, so “p” divides bN. Similarly, “p” divides “b”, so contradiction to a,
b begin relatively prime. So assumption is false and statement is true. A generalization: N root of p1, p2,---,pK is
irrational for distinct primes p1,---,pK and N >= 2 any integer.
Problem 3) Use the prime factorization theorem for positive integers to give formulas for the GCD and LCM of two
pos. integers M and N. Solution: m = P1^ α1 P2^ α2---Pk^ αk N = q1^ β1 q2^ β2----qL^ βL By FTA
for α’s and β’s pos. integers. Let r1, r2, ---, rT be the list of primes in increasing order from the p’s and q’s.
Then, m = r1^ α1 -----rT^ αT ; N = r1^ β1 ---- rT^ βT… α’s and β’s are non-negative integers. GCD(m,
N) = r1^ min(α1, β1)---rT^ min(αT, βT)
LMC(M, N) = r1^ max(α1, β1)---rT^ max(αT, βT)
Euclidean Algorithm: this is the easiest way to get the gcd of two pos. integers and it uses successive divisions. This
appeared “first” in Euclides’s Elements. GCD(m, n), m=72, N=27. 72/27=2(reminder 18), 27/18= 1 (reminder 9),
18/9=2 (reminder 0). You stop when reaching 0, the previous reminder is GCD=9.
Problem 4) Let r = M / N be a positive rational solution to the polynomial equation P(x) = 0 with integer
coefficients. Explain why N is a factor of the leading coefficient of P(x) while M is a factor of the constant term of
P(x). Solution: P(r) = P(m / N) = 0 Let P(x) = Ck x^K + Ck-1 x^K-1 + --- + C1 * x + C0, Ck != 0, all the C’s are
integers. So 0 = Ck * (m / N)^k + Ck-1 (m / N)^k-1 +---+ C1 * (m / N) + C0. Multiply thru by N^k giving no
fractions. 0=Ck * m^K + Ck-1 * m^K-1 * N + Ck-2 * m^K-2 * N^2 +----+ C1 * m * N^K-1 + C0 * N^K, K is a
positive integer. –C0 * N^k = Ck * m^k + Ck-1 * m^k-1 * N + --- + C1 * m * N^k-1. a = Ck * m^k; b = Ck-1 *
m^k-1 * N + --- + C1 * m * N^k-1
N divides –C0 * N^k, N divides RHS, and N divides Ck-1 * m^k-1 * N + --- + C1 * m * N^k-1.
N divides a + b, N divides b. So N divides Ck * m^k. So N divides Ck. If M, N are relatively prime using FTA. So
N is a factor of C0.
Problem 6) Solve exactly the polynomial equation –60x^3 + 16x^2 + 17x – 3 = 0, and then use your result to factor
this polynomial completely. Sol: graphically, it has 3 real x’s intercepts. You think one of these solutions is 1/6.
You verify this by checking p(1/6) = 0; So P(x) = (x – 1/6) (quadratic poly) P9x) by
(x – 1/6) the other real solutions are estimated at 3/5 and –1/2. P(x) = -60(x-1/6)(x-3/5)(x+1/2)
(-60x + 10)(x-3/5)(x+1/2)= -60x^2 + 36x + 10x –6 = (-60x^2 + 46x – 6)(x + ½) =
-60x^3 + 46x^2 – 6x – 30x^2 + 23x –3 = -60x^3 + 16x^2 + 17x –3 = 0 http://1728.com/indexalg.htm
Problem 10) a) Explain in full detail why Zn is not a field if n is a composite positive integer. Sol: if n is composite
then n = product of two numbers(no necessarily prime). Say α and β such that n= α*β.Note that α andβ must
be integers and less than n and α & β !=0 and to be elements of Zn.
[α]n * [β]n = [α*β] = [n]n = [0];
b) Explain in full detail why Zp for p a prime has no zero divisors. Sol: Condition: has not zero divisions α and β are
multiples of P. x*y = 0; P(x, y) = 0
Problem 11) Find the following complex numbers in polar form:
a) (3 – 2i)(5 + 4i) = (-2i)(4i)= (-2)(4)(I^2)=> 15 + 12i – 10i + 8 = 23 + 2i; z= x + iy =23 + 2i
z = r cos Θ + r sin Θ * i = re^iΘ; z = r cos Θ = r (cos Θ + i sin Θ)
r =|z|= square root 23^2 + 2^2=23
tg y/x = 2/23; tg-1 23/2 = 5; z=23 + 2i = 23 tg 11.5 = 23.1 e^5i
b) 1+ 6i / -8i + 3i = (1 + 6i / -8 + 3i)(-8 – 3i / -8 – 3i) = (1 + 6i)(-8 –3i)/73 = 10 – 51i / 73
z = (10 / 73) - (5i/73) = square root of (10/73)^2 + (51/73)^2 = .71274; Θ = tg-1 =-51/10
=-79;(1+6i/-8+2i)=.7e^-79i; z=(-5+qi); (7-2i)=(c,d) Θ= 160;w=e^iΘ(z–(c+di))+(c+di);w=e^160i((-5 + qi)(7*2i)+(7-2i)=14.5–16.4i= 21.9e^311i