Definition For a set F, a polynomial over F with variable x is of the form

Size: px
Start display at page:

Download "Definition For a set F, a polynomial over F with variable x is of the form"

Transcription

1 *7. Polynomials Definition For a set F, a polynomial over F with variable x is of the form a n x n + a n 1 x n 1 + a n 2 x n a 1 x + a 0, where a n, a n 1,..., a 1, a 0 F. The a i, 0 i n are the coefficients of the polynomial. If x n is the largest power of x appearing in the polynomial then n is the degree of the polynomial, a n x n is the leading term and a n is the leading coefficient. The collection of all polynomials with one variable x and with coefficients from F will be denoted by F [x]. (Note the square brackets.) Note that 0 F [x], being x 2 + 0x + 0, but it is not said to have a degree, though some books give it degree 1 or even. Part of the aim of this section is to show how similar are the properties of F [x] and Z. Examples 3x 2 + 5x 1 Z [x], 3 7 x x2 + x Q [x], x 2 π R [x]. We could also look at polynomials in Z m [x] or C [x]. If we can add and multiply numbers in the set F then we can add and multiply the polynomials in F [x]. Examples (i) In Z [x] the sum of 3x 2 + 5x 1 and 5x 3 3x 2 + 2x + 1 is ( 3x 2 + 5x 1 ) + ( 5x 3 3x 2 + 2x + 1 ) = 5x 3 + 7x. (ii) In Z [x] the difference of x and 9x + 7 is ( x ) ( 9x + 7 ) = x 2 9x 6. Notice how subtraction is easier than for integers. For example is complicated by the fact that we can subtract 7 from 1 and put 6 in the units place, we have to instead borrow 10 from the next column. 1

2 (iii) In Z [x] the product of x 2 + 2x + 3 and x is ( x 2 + 2x + 3 ) ( x ) = 3x x 3 + 8x +x 4 + 4x 2 = x 4 + 2x 3 + 7x 2 + 8x + 12 We can do all the above again in more interesting sets. (i) Addition in Z 3 [x]. The sum of 2x 3 + 2x 2 + x + 1 and x 3 + 2x is using 3 0 mod 3 and 4 1 mod 3. 2x 3 + 2x 2 + x ( x 3 + 2x ) = x 2 + x, is (ii) Subtraction in Z 5 [x]. The difference of 4x 3 +3x 2 +x+3 and 2x 3 +3x+4 ( 4x 3 + 3x 2 + x + 3 ) ( 2x 3 + 3x + 4 ) = 2x 3 + 3x 2 + 3x + 4 using 2 3 mod 5 and 1 4 mod 5. (iii) Multiplication in Z 2 [x]. The product of x 3 + x + 1 and x 2 + x + 1 is ( x 3 + x + 1 ) ( x 2 + x + 1 ) = x 5 + x 4 + x 3 using 2 0 mod 2. + x 3 + x 2 + x + x 2 + x + 1 = x 5 + x Note that in Z [x], Q [x], R [x], C [x] or Z p [x], with p prime, we have deg fg = deg f + deg g. This may not hold in Z m [x] with m composite. Example In Z 6 [x] multiply f (x) = 2x and g (x) = 3x to get ( 2x ) ( 3x ) = 6x 6 + 3x 4 + 2x = 3x 4 + 2x Hence, in this example, deg fg < deg f + deg g. 2

3 Subtle point. Recall that two functions f and g are equal on a set X if, and only if, f (x) = g (x) for all x X. Consider f (x) = x 4 + 2x and g (x) = (1 + x) 2 over Z 3. There are only three elements in Z 3 and for these, f (0) = 1 = g (0), f (1) = 1 = g (1), f (2) = 0 = g (2). So as functions over Z 3 these are equal though as polynomials they are different. Question If we can multiply polynomials can we factor them? Assumption α F, α 0, α 1 F. So, from now on F will be one of Q, R, C or Z p, with p prime because, to look at factorization, or dividing, it is best to restrict to sets F in which we can find inverses. (We can talk of dividing in Q, R or C but not in Z p. Instead we have to remember that to divide by something is to multiply by it s inverse.) (Aside: In Z 6 does [3] 6 have an inverse? i.e. does there exist an element with [3] 6 [x] 6 = [1] 6. If there did, we could multiply both sides by [2] 6 to get [6] 6 [x] 6 = [2] 6, i.e. [0] 6 = [2] 6, contradiction. So not every element in Z 6 has an inverse. The same argument works in general for Z m with m composite.) The first step to answering this question about factoring is to look at long division for polynomials. Example In Q [x] divide x 2 x 1 into x 4 + 2x 3 + 3x 2 + 4x + 5. Solution x 2 + 3x + 7 x 2 x 1 ) x 4 + 2x 3 + 3x 2 + 4x + 5 x 4 + x 3 + x 2 3x 3 + 4x 2 + 4x 3x 3 + 3x 2 + 3x 7x 2 + 7x + 5 7x 2 + 7x x + 12 Hence x 4 + 2x 3 + 3x 2 + 4x + 5 = (x 2 + 3x + 7) (x 2 x 1) + (14x + 12). 3

4 Always, always check by multiplying out. You should never end with an incorrect answer. Thus given f and g Q [x] we have found q, r Q [x] such that f = qg+r and deg r < deg g. This is very reminiscent of the division theorem for integers and we have exactly the same result for polynomials. Example In Z 3 [x] divide x 2 + x + 1 into x 4 + x x 2 + 2x + 1 x 2 + x + 1 ) x 4 + x x 4 + x 3 + x 2 2x x 3 + 2x 2 + 2x x 2 + x + 1 x 2 + x (Often using 2 = 1, 1 = 2 modulo 3.) Hence x 4 + x = (x 2 + 2x + 1) (x 2 + x + 1) and so x 2 + x + 1 divides x 4 + x Definition If f, g F [x] we say that g (x) divides f (x), and write g f, if there exists q (x) F [x] such that f (x) = g (x) q (x). We also say that f if a multiple of g. Note that if g f then either g 0 (we say it is identically zero) or deg g deg f. Careful If m, n Z, then m n and n m imply m = ±n. If we then demand that both m and n are positive we get m = n. But if we have polynomials f, g F [x] then f g implies deg f deg g. Similarly, g f implies deg g deg f. Hence deg f = deg g. But f g also means f = gu for some polynomial u. Then deg f = deg g means deg u = 0, and so u is a non-zero constant. Hence f g and g f imply only that f = cg for some constant c. Definition If f F [x] can be factored f (x) = g (x) h (x), with g, h F [x] and 1 deg g, deg h < deg f, then we say f is reducible. If we cannot factor f in this way we say it is irreducible. Compare this to the definition of prime numbers. It can be hard to check if a given polynomial is irreducible. And it can depend on the set F, see below. 4

5 Theorem Division Theorem for Polynomials. Let F be as above. Let f (x), g (x) F [x] with g (x) 0. Then there exist unique polynomials q (x), r (x) F [x] such that f (x) = q (x) g (x) + r (x) and either r (x) 0 or r is non-zero and has a lower degree than g. Proof Not given in course, but see appendix. An important deduction connects the roots of a polynomial with factorization. Corollary Let F be as above. Let p (x) F [x]. Then p (a) = 0 if, and only if, (x a) p (x). Proof ( ) Assume p (a) = 0. Apply the Theorem with f = p and g = x a to find q, r F [x] with p (x) = q (x) (x a) + r (x), and either r 0 or deg r < deg g. If r 0 we are finished. Otherwise deg r < deg g = 1 and so deg r = 0, i.e. r is a constant, c say. Thus p (x) = q (x) (x a) + c. Put x = a to see that c = 0. So again x a divides p (x). ( ) Assume (x a) p (x). But this means p (x) = q (x) (x a) for some q (x) F [x]. Putting x = a and we see that p (a) = 0. Since a polynomial in F [x] of degree n can be the product of at most n linear factors x a, it can have at most n roots in F. But will it have exactly n roots? Example x 2 2 Q [x] yet it has no roots in Q. Its roots are ± 2 R. Further x R [x] but it has no roots in R. Its roots are ±i C. Something different happens for C. Theorem Fundamental Theorem of Algebra. If f C [x] has deg f 1 then f has a root in C. Thus f factorizes completely into n linear factors. Proof Not given in this course. So a polynomial in C [x] of degree n has exactly n roots in C. This means that the only irreducible polynomials in C [x] are the linear polynomials, ax + b. The Corollary can help us factorize polynomials; to find linear factors it suffices to find roots. 5

6 Examples (i) Factorize p (x) = x 4 + x 3 2x 2 6x 4 over R. Solution Look at p (a) for small a. So p (0) = 4, p (1) = 8, p ( 1) = 0. Thus x + 1 divides p (x). x 3 2x 4 x + 1 ) x 4 + x 3 2x 2 6x 4 x 4 + x 3 2x 2 6x 4 2x 2 2x 4x 4 4x 4 0 x 4 + x 3 2x 2 6x 4 = (x + 1) ( x 3 2x 4 ). Writing q (x) = x 3 2x 4 we see that q (2) = 0 so x 2 divides q (x). x 2 + 2x + 2 x 2 ) x 3 2x 4 x 3 2x 2 2x 2 2x 4 2x 2 4x 2x 4 2x 4 0 x 3 2x 4 = (x 2) ( x 2 + 2x + 2 ). Does x 2 +2x+2 factorize over R? Note that x 2 +2x+2 = (x + 1) and so can never be zero for real x. Thus x 2 + 2x + 2 has no linear factors. Hence we finish with x 4 + x 3 2x 2 6x 4 = (x + 1) (x 2) ( x 2 + 2x + 2 ). (ii) Factorize p (x) = x 4 + x 3 2x 2 6x 4 over Z 5. The polynomial is the same as in (ii) so we still have But now, x 4 + x 3 2x 2 6x 4 = (x + 1) (x 2) ( x 2 + 2x + 2 ). x x 2 + 2x + 2 mod

7 Thus modulo 5, x 2 +2x+2 has roots at 1 and 2. So, in Z 5 [x], x 2 +2x+2 = (x 1) (x 2) = (x + 4) (x + 3). (Check this by multiplying out, modulo 5.) Hence, in Z 5 [x], x 4 + x 3 2x 2 6x 4 = (x + 1) (x 2) (x + 4) (x + 3) = (x + 1) (x + 3) 2 (x + 4). Aside So here we have seen that x 2 +2x+2 is irreducible in R [x] yet reducible in Z 5 [x] And of course it is reducible in C [x] : x 2 + 2x + 2 = (x + 1) = (x + 1) 2 i 2 = ((x + 1) + i) ((x + 1) i), difference of squares = (x + 1 i) (x i). (iii) Factorize p (x) = x 4 + x 3 + x 2 + 3x + 4 over Z 5. Solution We need only check x = 0, 1, 2, 3 2 or 4 1 mod 5. In turn, p (0) = 4, p (1) = 0, p (2) = 3 while p ( 2) = 0 and p ( 1) = 2, all calculation modulo 5. Thus x 1 and x + 2 divide p (x). So for some q (x) we have p (x) = (x 1) (x + 2) q (x) = (x 2 + x 2) q (x), i.e. x x 2 + x 2 ) x 4 + x 3 + x 2 + 3x + 4 x 4 + x 3 2x 2 3x 2 + 3x + 4 3x 2 + 3x x 4 + x 3 + x 2 + 3x + 4 = (x 1) (x + 2) ( x ). You should check that x has no zeros. If p (x) has repeated roots then x could have 1 or 2 as zeros, just as p (x) did. But x x mod Hence x 4 + x 3 + x 2 + 3x + 4 = (x 1) (x + 2) ( x ) = (x + 4) (x + 2) ( x ) in Z 5 [x]. 7

8 The division Theorem above is very similar to a result in integers. Other ideas can be generalized to polynomials such as Definition Let F be one of Q, R, C or Z p, with p prime. Let f, g F [x] be polynomials not both identically zero. Then a greatest common divisor is a polynomial d F [x] such that (i) d (x) divides both f (x) and g (x), (ii) if c (x) divides both f (x) and g (x) then c (x) divides d (x). Note that if the gcd exists it is not unique. For if d satisfies (i) and (ii) and λ F is non-zero then λd also satisfies the two properties. For two polynomials f and g write gcd (f, g) to represent any greatest common divisor. A particular gcd is often picked out by taking λ = c 1 where c is the leading coefficient of d. Then c 1 d has leading coefficient 1, and is called a monic polynomial. We need to justify that such a gcd exists. argument as used for the gcd of integers. We do not use the same Theorem Let F be as above. Let f, g F [x] be polynomials not both identically zero. Then the gcd (f, g) exists. Proof Not given in course but see the appendix. In any given problem we construct a gcd using repeated use of the Division Theorem. And if we formalize this method we get Theorem The Euclidean Algorithm for Polynomials. Let F be as above. Let f, g F [x] be polynomials. If g divides f then g is a gcd of f and g. Otherwise apply the Division Theorem to obtain a series of equations f = gq 1 + r 1, 0 < deg r 1 < deg g, g = r 1 q 2 + r 2, 0 < deg r 2 < deg r 1, r 1 = r 2 q 3 + r 3, 0 < deg r 3 < deg r 2,. r j 2 = r j 1 q j + r j, 0 < deg r j < deg r j 1, r j 1 = r j q j+1. Then r j, the last non-zero remainder, is a greatest common divisor of f and g. Further, by working back up this list we can find p (x), q (x) F [x] such that gcd (f (x), g (x)) = p (x) f (x) + q (x) g (x). Proof Left to student. 8

9 Examples (i) Over Q [x] find a greatest common divisor of f (x) = x 5 + 3x 4 + 5x 3 + 5x 2 + 4x + 2 and g (x) = x 4 + 2x 3 + 4x 2 + 4x + 4 and write the answer as a linear combination of f and g. Solution Applying the division algorithm we get x + 1 x 4 + 2x 3 + 4x 2 + 4x + 4 ) x 5 + 3x 4 + 5x 3 + 5x 2 + 4x + 2 x 5 2x 4 4x 3 4x 2 4x x 4 + x 3 + x x 4 2x 3 4x 2 4x 4 x 3 3x 2 4x 2 x + 1 x 3 3x 2 4x 2 ) x 4 + 2x 3 + 4x 2 + 4x + 4 x 4 3x 3 4x 2 2x x 3 + 2x + 4 x 3 + 3x 2 + 4x + 2 3x 2 + 6x x 1 3 3x 2 + 6x + 6 ) x 3 3x 2 4x 2 x 3 + 2x 2 + 2x x 2 2x 2 x 2 + 2x So we have found a zero remainder. The last non-zero remainder can be taken as the greatest common divisor i.e. 3x 2 + 6x + 6. We can work back and write the gcd as a linear multiple of f and g. So r 2 (x) = g (x) ( x + 1) ( x 3 3x 2 4x 2 ) = g (x) ( x + 1) (f (x) (x + 1) g (x)) = (1 + ( x + 1) (x + 1)) g (x) ( x + 1) f (x) = ( 2 x 2) g (x) + (x 1) f (x). Always, always check your answer by multiplying out: ( 2 x 2 ) ( x 4 + 2x 3 + 4x 2 + 4x + 4 ) + (x 1) ( x 5 + 3x 4 + 5x 3 + 5x 2 + 4x + 2 ) = 3x 2 + 6x + 6 9

10 (ii) Over Z 3 [x] find a greatest common divisor of x 3 + 2x 2 + 2x + 1 and x and express your answer as a linear combination of the original polynomials. Solution Firstly Next x = x ( x 3 + 2x 2 + 2x + 1 ) + x 3 + x 2 + 2x + 2 = (x + 1) ( x 3 + 2x 2 + 2x + 1 ) + 2x x 3 + 2x 2 + 2x + 1 = 2x ( 2x ) + 2x = (2x + 1) ( 2x ). So gcd (x 3 + 2x 2 + 2x + 1, x 4 + 2) = 2x And, quite simply, 2x = ( x ) (x + 1) ( x 3 + 2x 2 + 2x + 1 ). 10

11 Appendix 1) Theorem Division Theorem for Polynomials. (H.P p. 264) Let F be as above. Let f (x), g (x) F [x] with g (x) 0. Then there exist unique polynomials q (x), r (x) F [x] such that f (x) = q (x) g (x) + r (x) and either r (x) 0 or r is non-zero and has a lower degree than g. Proof (Existence) If f (x) 0 we can write 0 = 0g (x)+0, so we can assume that f (x) 0. If g f then f (x) = g (x) q (x) for some q (x) and the result follows with r (x) 0. In particular if deg g = 0, i.e. g (x) c a non-zero constant, we can take q (x) = c 1 f (x). Hence we can assume that g f and n = deg g 1. Consider the set S = {f qg : q F [x]}. Because g f this set does not contain 0 so every element in this set has a degree. This set is non-empty, since it contains f 0g. If we look at D, the set of degrees of the polynomials in S, we have a non-empty set of nonnegative integers so, by the well-ordering principle, it contains a minimum element. Choose a q F [x] such that f qg takes this minimum value and set r = f qg. Since g f we have r (x) 0 and so we have to show that deg r < deg g. Assume for the sake of contradiction that deg r deg g. Let m = deg r and write r (x) = a m x m + a m 1 x m a 1 x + a 0, along with g (x) = b n x n + b n 1 x n b 1 x + b 0. So m n, a m 0 and b n 0. Consider the polynomial r (x) a m b 1 n x m n g (x) = (f (x) q (x) g (x)) a m b 1 n x m n g (x) = f (x) ( q + a m b 1 n x m n) g (x) S. It is in S but the degree of this polynomial is m 1, i.e. it is < deg r, contradicting the choice of r as having minimum degree. Hence our assumption is false and deg r < deg g as required. (Uniqueness) Assume for contradiction that, for some f and g, the division is not unique. For this pair we can find distinct divisions f = qg + r and f = q g + r 11

12 with either r 0 or deg r < deg g and either r 0 or deg r < deg g. Then qg + r = q g + r, or r r = (q q) g. Thus g divides r r. This means either r r 0, i.e. r = r, and thus q = q, contradicting the fact that we started with distinct divisions, or deg g deg (r r ) max (deg r, deg r ) < deg g, again a contradiction. unique. Hence our assumption is false and the division is 2) Theorem Let F be as above. Let f, g F [x] be polynomials not both identically zero. Then the gcd (f, g) exists. Proof Let S = {f (x) K (x) + g (x) L (x) : K (x), L (x) Q [x]} {0}. Let D be the set of degrees of polynomials in S. W.l.o.g. assume f is not identically zero and take K 1 and L 0 to see that f = 1 f + 0 g S. Thus S, and and in turn D, are non-empty. D is a set of nonnegative integers so, by the well-ordering principle it has a least element. Take K (x), L (x) F [x] so that d (x) = f (x) K (x) + g (x) L (x) has this minimum degree. We have to show that (i) and (ii) of the definition of gcd both hold for this d (x). (i) Use the division Theorem to write f (x) = q (x) d (x) + r (x) for some q (x), r (x) F [x] with either r 0 or deg r < deg d. Then r (x) = f (x) q (x) d (x) = f (x) q (x) (f (x) F (x) + g (x) L (x)) = f (x) (1 q (x) K (x)) + ( q (x) L (x)) g (x) which is either 0 or in S. If it is not identically zero then we have r (x) S and deg r < deg d, which contradicts the minimality of deg d. Hence r 0 and f (x) = q (x) d (x), i.e. d f. Similarly you can show that d g. (ii) Assume that c (x) divides both f (x) and g (x), so f (x) = c (x) u (x) and g (x) = c (x) v (x) for some u (x), v (x) F [x]. Then d (x) = f (x) K (x) + g (x) L (x) = c (x) u (x) K (x) + c (x) v (x) L (x) = c (x) (u (x) K (x) + v (x) L (x)) and so c (x) divides d (x). 12

13 3) Recall, in Z we chose an m Z, and looked at the congruence classes [a] = {n Z : m (n a)}. We then defined addition and multiplication on the set of classes, Z m. We have seen above many similarities between Z and F [x] where F is one of Q, R, C or Z p, with p prime. So we can try to copy what we did in constructing the congruence classes of Z. Fix a polynomial f (x) F [x] and look at the classes [g (x)] = {h (x) : f (x) (h (x) g (x))}. Then, using the factor notation from the section on relations (since we could define a relation h (x) g (x) if and only if f (x) (h (x) g (x))), we denote the collection of classes by F [x] / (f (x)). We then define addition and multiplication on this set of classes by [g 1 (x)] + [g 2 (x)] = [g 1 (x) + g 2 (x)] and [g 1 (x)] [g 2 (x)] = [g 1 (x) g 2 (x)]. remember, you should check that these are well-defined operations. Example Let f (x) = x R [x]. Given any polynomial p (x) we can use the division theorem to write p (x) = q (x) (x 2 + 1) + r (x) with either r 0 or deg r (x) < deg (x 2 + 2) = 2. Hence [p (x)] = [r (x)] = [a + bx] with a, b R. So all the classes in R [x] / (x 2 + 1) can be labelled simply as [a + bx], a, b R. For addition and multiplication we have [a + bx] + [c + dx] = [(a + c) + (b + d) x] [a + bx] [c + dx] = [ ac + bcx + adx + bdx 2] = [ (ac bd) + (bc + ad) x + db ( x 2 1 )] Note that = [(ac bd) + (bc + ad) x]. (a + bi) (c + di) = (ac bd) + (bc + ad) i. So the map R [x] / (x 2 + 1) C, [a + bx] a+bi is a map that preserves addition and multiplication. Or we can say that R [x] / (x 2 + 1) is a way of constructing the complex numbers from the reals. Example Take x 2 +x+2 Z 3 [x]. Again the classes are of the form [a + bx], a, b Z 3. There are 9 classes, 3 choices for a and 3 for b. They are [0], [1], [2], [x], [1 + x], [2 + x], [2x], [1 + 2x] and [2 + 2x]. 13

14 For the addition and multiplication we can write out the tables table[1].pdf For the multiplication table we need only look at non-zero classes. [1] [2] [x] [1 + x] [2 + x] [2x] [1 + 2x] [2 + 2x] [1] [1] [2] [x] [1 + x] [2 + x] [2x] [1 + 2x] [2 + 2x] [2] [2] [1] [2x] [2 + 2x] [1 + 2x] [x] [2 + x] [1 + x] [x] [x] [2x] [1 + 2x] [1] [1 + x] [2 + x] [2 + 2x] [2] [1 + x] [1 + x] [2 + 2x] [1] [2 + x] [2x] [2] [x] [1 + 2x] [2 + x] [2 + x] [1 + 2x] [1 + x] [2x] [2] [2 + 2x] [1] [x] [2x] [2x] [x] [2 + x] [2] [2 + 2x] [1 + 2x] [1 + x] [1] [1 + 2x] [1 + 2x] [2 + x] [2 + 2x] [x] [1] [1 + x] [2] [2x] [2 + 2x] [2 + 2x] [1 + x] [2] [1 + 2x] [x] [1] [2x] [2 + x] Note that every row (and column) is simply a permutation of the classes, none is missed and none replicated. We see no zero classes, i.e. we have no 14

15 divisors of zero. And all classes have inverses, namely [1] 1 = [1], [2] 1 = [1], [x] 1 = [1 + x], [1 + x] 1 = [x], [2 + x] 1 = [1 + 2x], [2x] 1 = [2 + 2x], [1 + 2x] 1 = [2 + x] and [2 + 2x] 1 = [2x]. Example Student to take x 3 + x + 1 Z 2 [x]. Draw up the tables for (Z 2 [x] / (x 3 + x + 1), +) and (Z 2 [x] / (x 3 + x + 1) \ {[0]}, ). The different classes are [0], [1], [x], [1 + x], [ x 2], [ 1 + x 2], [ x + x 2] and [ 1 + x + x 2]. [1] [x] [1+x] [x 2 ] [1+x 2 ] [x+x 2 ] [1+x+x 2 ] [1] [1] [x] [1+x] [x 2 ] [1+x 2 ] [x+x 2 ] [1+x+x 2 ] [x] [x] [x 2 ] [x+x 2 ] [1+x] [1] [1+x+x 2 ] [1+x 2 ] [1+x] [1+x] [x+x 2 ] [1+x 2 ] [1+x+x 2 ] [x 2 ] [1] [x] [x 2 ] [x 2 ] [1+x] [1+x+x 2 ] [x+x 2 ] [x] [1+x 2 ] [1] [1+x 2 ] [1+x 2 ] [1] [x 2 ] [x] [1+x+x 2 ] [1+x] [x+x 2 ] [x+x 2 ] [x+x 2 ] [1+x+x 2 ] [1] [1+x 2 ] [1+x] [x] [x 2 ] [1+x+x 2 ] [1+x+x 2 ] [1+x 2 ] [x] [1] [x+x 2 ] [x 2 ] [1+x] Note that [x] 2 = [x 2 ], [x] 3 = [1 + x], [x] 4 = [x + x 2 ], [x] 5 = [1 + x + x 2 ], [x] 6 = [1 + x 2 ] and [x] 7 = [1]. 15

7.2 Applications of Euler s and Fermat s Theorem.

7.2 Applications of Euler s and Fermat s Theorem. 7.2 Applications of Euler s and Fermat s Theorem. i) Solving non-linear congruences. Example Find a solution to x 12 3 mod 11. Solution Any solution of this must satisfy gcd (x, 11) = 1 so Fermat s Little

More information

Quotient Rings and Field Extensions

Quotient Rings and Field Extensions Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

More information

Greatest common divisors

Greatest common divisors Greatest common divisors Robert Friedman Long division Does 5 divide 34? It is easy to see that the answer is no, for many different reasons: 1. 34 does not end in a 0 or 5. 2. Checking directly by hand,

More information

PUTNAM TRAINING NUMBER THEORY. Exercises 1. Show that the sum of two consecutive primes is never twice a prime.

PUTNAM TRAINING NUMBER THEORY. Exercises 1. Show that the sum of two consecutive primes is never twice a prime. PUTNAM TRAINING NUMBER THEORY (Last updated: November 8, 2016) Remark. This is a list of exercises on Number Theory. Miguel A. Lerma Exercises 1. Show that the sum of two consecutive primes is never twice

More information

Mathematics of Cryptography Part I

Mathematics of Cryptography Part I CHAPTER 2 Mathematics of Cryptography Part I (Solution to Odd-Numbered Problems) Review Questions 1. The set of integers is Z. It contains all integral numbers from negative infinity to positive infinity.

More information

Mathematics of Cryptography

Mathematics of Cryptography CHAPTER 2 Mathematics of Cryptography Part I: Modular Arithmetic, Congruence, and Matrices Objectives This chapter is intended to prepare the reader for the next few chapters in cryptography. The chapter

More information

Math 135 Algebra Midterm Solutions

Math 135 Algebra Midterm Solutions Math 135 - Midterm Examination 1 Math 135 Algebra Midterm Solutions Question 1. (a) [4 marks] Consider the following two statements: x y(p (x) = Q(y)), and ( x P (x)) = ( y Q(y)) Give an example of a universe

More information

Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

More information

Polynomial Rings If R is a ring, then R[x], the ring of polynomials in x with coefficients in R, consists of all formal

Polynomial Rings If R is a ring, then R[x], the ring of polynomials in x with coefficients in R, consists of all formal Polynomial Rings 8-17-2009 If R is a ring, then R[x], the ring of polynomials in x with coefficients in R, consists of all formal sums a i x i, where a i = 0 for all but finitely many values of i. If a

More information

Your Name: MA 261 Worksheet Monday, April 21, 2014

Your Name: MA 261 Worksheet Monday, April 21, 2014 Your Name: MA 261 Worksheet Monday, April 21, 2014 1. Show that 2 11,213 1 is not divisible by 11. Solution: To say that 2 11,213 1 is divisible by 11 is equivalent to saying that 2 11,213 1 (mod 11).

More information

Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices. A Biswas, IT, BESU SHIBPUR

Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices. A Biswas, IT, BESU SHIBPUR Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices A Biswas, IT, BESU SHIBPUR McGraw-Hill The McGraw-Hill Companies, Inc., 2000 Set of Integers The set of integers, denoted by Z,

More information

Basic Number Theory 1

Basic Number Theory 1 Basic Number Theory 1 Divisibility Basic number theory uncovers the multiplicative structure of the integers. As such, the most important relation between integers is divisibility: the nonzero integer

More information

MERSENNE PRIMES SARAH MEIKLEJOHN AND STEVEN J. MILLER. 4 if i = 0, s i =

MERSENNE PRIMES SARAH MEIKLEJOHN AND STEVEN J. MILLER. 4 if i = 0, s i = MERSENNE PRIMES SARAH MEIKLEJOHN AND STEVEN J. MILLER ABSTRACT. A Mersenne prime is a prime that can be written as 2 p 1 for some prime p. The first few Mersenne primes are 3, 7 and 31 (corresponding respectively

More information

Lecture 1-3: Abstract algebra and Number theory

Lecture 1-3: Abstract algebra and Number theory Lecture 1-3: Abstract algebra and Number theory Thomas Johansson T. Johansson (Lund University) 1 / 61 a divides b {..., 2, 1, 0, 1, 2,...} = Z. Definition a, b Z. We say that a divides b (written a b)

More information

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

More information

Homework 5 Solutions

Homework 5 Solutions Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which

More information

In Problems 55 through 58, evaluate the given sum. 55. (3j 1) 56.

In Problems 55 through 58, evaluate the given sum. 55. (3j 1) 56. Appendix A2 Factoring Polynomials and Solving Systems of Equations 605 SUMMATION NOTATION In Problems 55 through 58, evaluate the given sum. 4 j 1 10 j 1 55. (3j 1) 56. 5 j 2 j 1 5 j 1 57. ( 1) j 58. 2

More information

MATH 118 HW 1 LINDSAY CROSS, KELLY DOUGAN, KEVIN LIU

MATH 118 HW 1 LINDSAY CROSS, KELLY DOUGAN, KEVIN LIU MATH 118 HW 1 LINDSAY CROSS, KELLY DOUGAN, KEVIN LIU 1.2 Problem 2. Find the greatest common divisor g of the numbers 1819 and 3574, and then find integers x and y to satisfy 1819x + 3587y = g. Use the

More information

Fermat s Little Theorem

Fermat s Little Theorem Fermat s Little Theorem Theorem (Fermat s Little Theorem): Let p be a prime. Then p n p n (1) for any integer n 1. Proof: We distinguish two cases. Case A: Let p n, then, obviously, p n p n, and we are

More information

2215 Solution Problem Sheet 4 1. Irreducibility and Factorization of Polynomials

2215 Solution Problem Sheet 4 1. Irreducibility and Factorization of Polynomials 2215 Solution Problem Sheet 4 1 Irreducibility and Factorization of Polynomials 1. Show that p(x) is irreducible in R[X], R any ring, if and only if p(x a) is irreducible in R[x] for any a R. Solution:

More information

Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that. a = bq + r and 0 r < b.

Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that. a = bq + r and 0 r < b. Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that a = bq + r and 0 r < b. We re dividing a by b: q is the quotient and r is the remainder,

More information

Number Theory Homework.

Number Theory Homework. Number Theory Homework. 1. The greatest common divisor and Bezout s Theorem Definition 1. If a and b are integers, not both zero, then c is a common divisor of a and b iff c a and c b. More generally if

More information

Number Theory Vocabulary (For Middle School Teachers)

Number Theory Vocabulary (For Middle School Teachers) Number Theory Vocabulary (For Middle School Teachers) A Absolute value the absolute value of a real number is its distance from zero on the number line. The absolute value of any real number, a, written

More information

Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

More information

MATH10101: problems for supervision in week 12 SOLUTIONS

MATH10101: problems for supervision in week 12 SOLUTIONS MATH10101: problems for supervision in week 12 SOLUTIONS Q37.. Use the table below to sieve the integers up to 200 for primes. Thus calculate π200. What is the smallest composite number greater than 200

More information

1.2 Properties of Integers

1.2 Properties of Integers 10 CHAPTER 1. PRELIMINARIES 1.2 Properties of Integers 1.2.1 The Well Ordering Principle and the Division Algorithm We now focus on a special set, the integers, denoted Z, as this set plays an important

More information

Basic Properties of Rings

Basic Properties of Rings LECTURE 15 Basic Properties of Rings Theorem 15.1. For any element a in a ring R, the equation a + x 0 R has a unique solution. We know that a + x 0 R has at least one solution u R by Axiom (5) in the

More information

Finding Multiplicative Inverses Modulo n

Finding Multiplicative Inverses Modulo n Fall 2005 Chris Christensen Cryptology Notes Finding Multiplicative Inverses Modulo n It is not necessary to do trial and error to determine the multiplicative inverse of an integer modulo n. If the modulus

More information

Section III.6. Factorization in Polynomial Rings

Section III.6. Factorization in Polynomial Rings III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)

More information

3.3 Solving Simultaneous Pairs of Linear Congruences

3.3 Solving Simultaneous Pairs of Linear Congruences 3.3 Solving Simultaneous Pairs of Linear Congruences Consider the two linear congruences x 2 mod 5 and x 1 mod 3. Integers satisfying the first congruence include... 8, 5, 2, 7, 12, 17, 22, 27, 32,...

More information

3. PRIME POLYNOMIALS

3. PRIME POLYNOMIALS 3. PRIME POLYNOMIALS 3.1. Tests for Primeness A polynomial p(x) F[x] is prime (irreducible) over F if its degree is at least 1 and it cannot be factorised into polynomials of lower degree. Constant polynomials

More information

Solutions to Practice Final 1

Solutions to Practice Final 1 s to Practice Final 1 1. (a) What is φ(0 100 ) where φ is Euler s φ-function? (b) Find an integer x such that 140x 133 (mod 301). Hint: gcd(140, 301) = 7. (a) φ(0 100 ) = φ(4 100 5 100 ) = φ( 00 5 100

More information

Lecture 6 : Divisibility and the Euclidean Algorithm

Lecture 6 : Divisibility and the Euclidean Algorithm Lecture 6 : Divisibility and the Euclidean Algorithm July 24, 2007 1. If a and b are relatively prime integers, show that ab and a + b are also relatively prime. 2. (a) If 2 n + 1 is prime for some integer

More information

Topics in Number Theory

Topics in Number Theory Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall

More information

SOLUTIONS TO PROBLEM SET 6

SOLUTIONS TO PROBLEM SET 6 SOLUTIONS TO PROBLEM SET 6 18.6 SPRING 16 Note the difference of conventions: these solutions adopt that the characteristic polynomial of a matrix A is det A xi while the lectures adopt the convention

More information

Math 139 Problem Set #6 Solution 15 April 2003

Math 139 Problem Set #6 Solution 15 April 2003 Math 139 Problem Set #6 Solution 15 April 2003 A1 (Based on Barr, 41, exercise 3) (a) Let n be an integer greater than 2 Consider the list of numbers n! + 2, n! + 3, n! + 4,, n! + n Explain why all the

More information

IMPORTANT NOTE: All the following statements are equivalent:

IMPORTANT NOTE: All the following statements are equivalent: .5 Congruences For this section, we think of m as a fixed positive integer. Definition 15. We say that a is congruent to b modulo m, and we write if m divides (a b). a b (mod m) IMPORTANT NOTE: All the

More information

Polynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if

Polynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if 1. Polynomials 1.1. Definitions A polynomial in x is an expression obtained by taking powers of x, multiplying them by constants, and adding them. It can be written in the form c 0 x n + c 1 x n 1 + c

More information

Chapter 6. Number Theory. 6.1 The Division Algorithm

Chapter 6. Number Theory. 6.1 The Division Algorithm Chapter 6 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization,

More information

Congruences. Robert Friedman

Congruences. Robert Friedman Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

More information

CONGRUENCES. m is said to be the modulus of the congruence and is defined to be a positive integer.

CONGRUENCES. m is said to be the modulus of the congruence and is defined to be a positive integer. Dr. J.S. Lipowski (Congruences.pdf) CONGRUENCES Let a, b, m Z = {, -3, -2, -1, 0, 1, 2, 3, } with m > 0, then a is said to be congruent to b modulo m, denoted a b (mod m), if m a b (a = b + k m, where

More information

MATH 523 LECTURE NOTES Summer 2008

MATH 523 LECTURE NOTES Summer 2008 MATH 523 LECTURE NOTES Summer 2008 These notes are intended to provide additional background from abstract algebra that is necessary to provide a good context for the study of algebraic coding theory.

More information

a 2 + b 2) ( c 2 + d 2) = (ac bd) 2 + (ad + bc) 2 = (ac + bd) 2 + (ad bc) 2

a 2 + b 2) ( c 2 + d 2) = (ac bd) 2 + (ad + bc) 2 = (ac + bd) 2 + (ad bc) 2 We will be talking about integers. We start with the rational integers. These are the ordinary integers 0, ±1, ±, ±3,.... The text denotes the set of rational integers by J. A lot of people use the letter

More information

Problem Set 3 Solutions

Problem Set 3 Solutions Problem Set 3 Solutions Section 3.1. For each of the following, use a counterexample to prove the statement is false. (a) For each odd natural number n, if n > 3 then 3 divides (n 1). Consider n = 9. Then,

More information

Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2)

Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2) Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2) Problems from 21 211 Prove that a b (mod n) if and only if a and b leave the same remainder when divided by n Proof Suppose a

More information

Section 4-1 Divisibility. Even integers can be represented by the algebraic expression: Odd integers can be represented by the algebraic expression:

Section 4-1 Divisibility. Even integers can be represented by the algebraic expression: Odd integers can be represented by the algebraic expression: Section 4-1 Divisibility Even and Odd Numbers: A number is even if: It has 2 as a factor It is divisible by 2 (when divided by 2 the remainder is 0) A number is odd if: It does not have a factor of 2 When

More information

MODULAR ARITHMETIC STEVEN DALE CUTKOSKY

MODULAR ARITHMETIC STEVEN DALE CUTKOSKY MODULAR ARITHMETIC STEVEN DALE CUTKOSKY 1. Relations, Rational Numbers and Modular Arithmetic 1.1. Relations. Suppose that X is a set. a relation on X is a subset R X X = {(a, b) a, b X}. Write a b if

More information

Summer High School 2009 Aaron Bertram

Summer High School 2009 Aaron Bertram Summer High School 2009 Aaron Bertram 2. Modular Arithmetic and Algebra. Notation: The expression k n means k divides n. Now fix a natural number k > 1. Definition 2.1. Integers a and b are congruent modulo

More information

= 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

= 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

More information

GALOIS GROUPS OF CUBICS AND QUARTICS (NOT IN CHARACTERISTIC 2)

GALOIS GROUPS OF CUBICS AND QUARTICS (NOT IN CHARACTERISTIC 2) GALOIS GROUPS OF CUBICS AND QUARTICS (NOT IN CHARACTERISTIC 2) KEITH CONRAD We will describe a procedure for figuring out the Galois groups of separable irreducible polynomials in degrees 3 and 4 over

More information

BASIC CONCEPTS ON NUMBER THEORY

BASIC CONCEPTS ON NUMBER THEORY BASIC CONCEPTS ON NUMBER THEORY This is a brief introduction to number theory. The concepts to cover will enable us to answer questions like: How many integers between 50 and 150 are divisible by 4? How

More information

Given two integers a, b with a 0. We say that a divides b, written

Given two integers a, b with a 0. We say that a divides b, written 1 Divisibility Given two integers a, b with a 0. We say that a divides b, written a b, if there exists an integer q such that b = qa. When this is true, we say that a is a factor (or divisor) of b, and

More information

Basic Definitions. A group (G, ) is commutative (abelian) if a b = b a a,b G. , +), (Z p -{0}, ) All these examples are abelian groups.

Basic Definitions. A group (G, ) is commutative (abelian) if a b = b a a,b G. , +), (Z p -{0}, ) All these examples are abelian groups. Finite Fields Basic Definitions A group is a set G with a binary operation (i.e., a function from G G into G) such that: 1) The operation is associative, a (b c) = (a b) c a,b,c G. 2) There exists an identity

More information

MTH 382 Number Theory Spring 2003 Practice Problems for the Final

MTH 382 Number Theory Spring 2003 Practice Problems for the Final MTH 382 Number Theory Spring 2003 Practice Problems for the Final (1) Find the quotient and remainder in the Division Algorithm, (a) with divisor 16 and dividend 95, (b) with divisor 16 and dividend -95,

More information

, but this gives r Q, so there are no 2 PT s in a geometric progression.

, but this gives r Q, so there are no 2 PT s in a geometric progression. (1) (Niven 5.3.3) Find all PT s whose terms form an (a) Arithmetic progression Such a triple would be of the form (b d, b, b + d) for b, d Z +. To be Pythagorean triples, we require (b d) + b = (b + d)

More information

MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction. MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

More information

ALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS. 1. Ideals in Commutative Rings In this section all groups and rings will be commutative.

ALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS. 1. Ideals in Commutative Rings In this section all groups and rings will be commutative. ALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS PETE L. CLARK 1. Ideals in Commutative Rings In this section all groups and rings will be commutative. 1.1. Basic definitions and examples. Let R be a (commutative!)

More information

Factoring Polynomials

Factoring Polynomials Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

More information

Final Exam. Math March 2015 Jenny Wilson. Instructions: This exam has 17 questions for a total of 70 points and 4 bonus points.

Final Exam. Math March 2015 Jenny Wilson. Instructions: This exam has 17 questions for a total of 70 points and 4 bonus points. Final Exam Math 110 19 March 2015 Jenny Wilson Name: Instructions: This exam has 17 questions for a total of 70 points and 4 bonus points. You may bring in a basic calculator. You may bring in a one double-sided

More information

Abstract Algebra. Ch 0: Preliminaries. Properties of Integers

Abstract Algebra. Ch 0: Preliminaries. Properties of Integers Abstract Algebra Ch 0: Preliminaries Properties of Integers Note: Solutions to many of the examples are found at the end of this booklet, but try working them out for yourself first. 1. Fundamental Theorem

More information

Math 581. Basic Background Results. Below are some basic results from algebra dealing with fields and polynomial rings which will be handy to

Math 581. Basic Background Results. Below are some basic results from algebra dealing with fields and polynomial rings which will be handy to Math 581 Basic Background Results Below are some basic results from algebra dealing with fields and polynomial rings which will be handy to recall in 581. This should be a reasonably logical ordering,

More information

2 The Euclidean algorithm

2 The Euclidean algorithm 2 The Euclidean algorithm Do you understand the number 5? 6? 7? At some point our level of comfort with individual numbers goes down as the numbers get large For some it may be at 43, for others, 4 In

More information

3.7 Complex Zeros; Fundamental Theorem of Algebra

3.7 Complex Zeros; Fundamental Theorem of Algebra SECTION.7 Complex Zeros; Fundamental Theorem of Algebra 2.7 Complex Zeros; Fundamental Theorem of Algebra PREPARING FOR THIS SECTION Before getting started, review the following: Complex Numbers (Appendix,

More information

Basic Definitions. A group (G, ) is commutative (abelian) if a b = b a a,b G. , +), (Z p -{0}, ) All these examples are abelian groups.

Basic Definitions. A group (G, ) is commutative (abelian) if a b = b a a,b G. , +), (Z p -{0}, ) All these examples are abelian groups. Finite Fields Basic Definitions A group is a set G with a binary operation (i.e., a function from G G into G) such that: 1) The operation is associative, a (b c) = (a b) c a,b,c G. 2) There exists an identity

More information

Unique Factorization

Unique Factorization Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

More information

Exercise Set 3 Solutions Math 2020 Due: March 6, Exercises for practice: Do the following exercises from the text:

Exercise Set 3 Solutions Math 2020 Due: March 6, Exercises for practice: Do the following exercises from the text: Exercises for practice: Do the following exercises from the text: Section NT-1 (page 60): 1.1, 1.2, 1.3, 1.4, 1.6, 1.7, 1.13, 1.14, 1.15, 1.17. Section NT-2 (page 76): 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7,

More information

MODULAR ARITHMETIC (SHORT VERSION)

MODULAR ARITHMETIC (SHORT VERSION) MODULAR ARITHMETIC (SHORT VERSION) KEITH CONRAD. Introduction We will define the notion of congruent integers (with respect to a modulus) and develop some basic ideas of modular arithmetic, which lets

More information

HOMEWORK SOLUTIONS MATH 114

HOMEWORK SOLUTIONS MATH 114 HOMEWORK SOLUTIONS MATH 114 Problem set 10. 1. Find the Galois group of x 4 + 8x + 12 over Q. Solution. The resolvent cubic x 3 48x + 64 does not have rational roots. The discriminant 27 8 4 + 256 12 3

More information

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

Elementary Number Theory We begin with a bit of elementary number theory, which is concerned CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

More information

Definition: Twin Primes.

Definition: Twin Primes. 4.5.. Definition: Twin Primes. If p and p + are both prime numbers, then p and p + are called twin primes Examples: Find all twin primes between 5 and 50. The primes 5 50 are: 7, 9, 3, 9, 3, 37, 4, 43,

More information

Clock Arithmetic and Euclid s Algorithm

Clock Arithmetic and Euclid s Algorithm Clock Arithmetic and Euclid s Algorithm Lecture notes for Access 2006 by Erin Chamberlain. Earlier we discussed Caesar Shifts and other substitution ciphers, and we saw how easy it was to break these ciphers

More information

13. Suppose that a and b are integers, a 4 (mod 13), and. b 3 (mod 19). Find the integer c with 0 c 18 such

13. Suppose that a and b are integers, a 4 (mod 13), and. b 3 (mod 19). Find the integer c with 0 c 18 such 244 4 / Number Theory and Cryptography Remark: Because Z m with the operations of addition and multiplication modulo m satisfies the properties listed, Z m with modular addition is said to be a commutative

More information

8 Primes and Modular Arithmetic

8 Primes and Modular Arithmetic 8 Primes and Modular Arithmetic 8.1 Primes and Factors Over two millennia ago already, people all over the world were considering the properties of numbers. One of the simplest concepts is prime numbers.

More information

The Division Algorithm for Polynomials Handout Monday March 5, 2012

The Division Algorithm for Polynomials Handout Monday March 5, 2012 The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,

More information

minimal polyonomial Example

minimal polyonomial Example Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

More information

3. QUADRATIC CONGRUENCES

3. QUADRATIC CONGRUENCES 3. QUADRATIC CONGRUENCES 3.1. Quadratics Over a Finite Field We re all familiar with the quadratic equation in the context of real or complex numbers. The formula for the solutions to ax + bx + c = 0 (where

More information

MA Number Theory Spring Semester Prof. Meuser MIDTERM EXAM - March 1, 2013

MA Number Theory Spring Semester Prof. Meuser MIDTERM EXAM - March 1, 2013 MA 341 - Number Theory Spring Semester - 13 - Prof. Meuser MIDTERM EXAM - March 1, 2013 1) (16 points) Let a, b N. a) (4 points) Give the definition of gcd(a, b). b) (6 points) Use the definition of gcd(a,

More information

12. POLYNOMIAL CODES

12. POLYNOMIAL CODES 12. POLYNOMIAL CODES 12.1. Error Detection and Error Correction Generally when people talk of codes they re thinking of cryptography where messages are coded for reasons of secrecy. But there's another

More information

Integer Divisibility and Division

Integer Divisibility and Division Integer Divisibility and Division Let a and b be integers. We say that a divides b, if b = ka for some integer k. In symbols, this relationship is written as a b. In this case we also say that a is a divisor

More information

a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)

a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4) ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x

More information

Finite Fields and Error-Correcting Codes

Finite Fields and Error-Correcting Codes Lecture Notes in Mathematics Finite Fields and Error-Correcting Codes Karl-Gustav Andersson (Lund University) (version 1.013-16 September 2015) Translated from Swedish by Sigmundur Gudmundsson Contents

More information

Number Theory. Proof. Suppose otherwise. Then there would be a finite number n of primes, which we may

Number Theory. Proof. Suppose otherwise. Then there would be a finite number n of primes, which we may Number Theory Divisibility and Primes Definition. If a and b are integers and there is some integer c such that a = b c, then we say that b divides a or is a factor or divisor of a and write b a. Definition

More information

A.4 Polynomial Division; Synthetic Division

A.4 Polynomial Division; Synthetic Division SECTION A.4 Polynomial Division; Synthetic Division 977 A.4 Polynomial Division; Synthetic Division OBJECTIVES 1 Divide Polynomials Using Long Division 2 Divide Polynomials Using Synthetic Division 1 Divide

More information

Quadratic Equations and Inequalities

Quadratic Equations and Inequalities MA 134 Lecture Notes August 20, 2012 Introduction The purpose of this lecture is to... Introduction The purpose of this lecture is to... Learn about different types of equations Introduction The purpose

More information

ECE 7670 Lecture gfint Introduction to multiple-error correcting codes

ECE 7670 Lecture gfint Introduction to multiple-error correcting codes ECE 7670 Lecture gfint Introduction to multiple-error correcting codes Objective: To demonstrate important concepts that lead to more general families of codes; to introduce the idea behind Galois fields.

More information

Monomials. Polynomials. Objectives: Students will multiply and divide monomials Students will solve expressions in scientific notation

Monomials. Polynomials. Objectives: Students will multiply and divide monomials Students will solve expressions in scientific notation Students will multiply and divide monomials Students will solve expressions in scientific notation 5.1 Monomials Many times when we analyze data we work with numbers that are very large. To simplify these

More information

9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11. 9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

More information

62 CHAPTER 2. POLYNOMIALS. Roots are the key to a deeper understanding of polynomials. Definition: Any value r F that solves: f(r) = 0

62 CHAPTER 2. POLYNOMIALS. Roots are the key to a deeper understanding of polynomials. Definition: Any value r F that solves: f(r) = 0 6 CHAPTER. POLYNOMIALS.3 Roots Roots are the key to a deeper understanding of polynomials. Definition: Any value r F that solves: f(r) = 0 is called a root of the polynomial f(x) F [x]. Examples: (a) Every

More information

STUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS

STUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS STUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS The intermediate algebra skills illustrated here will be used extensively and regularly throughout the semester Thus, mastering these skills is an

More information

The Euclidean Algorithm and Multiplicative Inverses

The Euclidean Algorithm and Multiplicative Inverses 1 The Euclidean Algorithm and Multiplicative Inverses Lecture notes for Access 2011 The Euclidean Algorithm is a set of instructions for finding the greatest common divisor of any two positive integers.

More information

Math 430 Problem Set 1 Solutions

Math 430 Problem Set 1 Solutions Math 430 Problem Set 1 Solutions Due January 22, 2016 1.2. If A = {a, b, c}, B = {1, 2, 3}, C = {x}, and D =, list all of the elements of each of the following sets. (a) A B (b) B A {(a, 1), (a, 2), (a,

More information

FIELDS AND GALOIS THEORY

FIELDS AND GALOIS THEORY FIELDS AND GALOIS THEORY GEORGE GILBERT 1. Rings and Polynomials Given a polynomial of degree n, there exist at most n roots in the field. Given a polynomial that factors completely, x r 1 ) x r 2 )...

More information

Sect 3.3 Zeros of Polynomials Functions

Sect 3.3 Zeros of Polynomials Functions 101 Sect 3.3 Zeros of Polynomials Functions Objective 1: Using the Factor Theorem. Recall that a zero of a polynomial P(x) is a number c such that P(c) = 0. If the remainder of P(x) (x c) is zero, then

More information

Modular Arithmetic continued

Modular Arithmetic continued Modular Arithmetic continued Lecture notes for Access 2011 by Erin Chamberlain and Nick Korevaar Number theory refresher 1 Here are some words which will occur in our discussion today. Definition 1. An

More information

Consider = 19, and consider the remainders when these numbers are divided by 7.

Consider = 19, and consider the remainders when these numbers are divided by 7. Congruences Note: I have attempted to restore as much of the fonts as I could, unfortunately I do not have the original document, so there could be some minor mistakes. It all started a long time ago when

More information

EUCLID S ALGORITHM AND THE FUNDAMENTAL THEOREM OF ARITHMETIC. after N. Vasiliev and V. Gutenmacher (Kvant, 1972)

EUCLID S ALGORITHM AND THE FUNDAMENTAL THEOREM OF ARITHMETIC. after N. Vasiliev and V. Gutenmacher (Kvant, 1972) Intro to Math Reasoning Grinshpan EUCLID S ALGORITHM AND THE FUNDAMENTAL THEOREM OF ARITHMETIC. after N. Vasiliev and V. Gutenmacher (Kvant, 1972) We all know that every composite natural number is a product

More information

23 Elliptic curves mod p

23 Elliptic curves mod p 52 MA6011 23 Elliptic curves mod p Elliptic curves have been studied by mathematicians for a long time. Starting in about 1985 such curves were used in cryptography. In the cryptographic applications of

More information

Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.

Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

More information

CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 Solutions CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

More information

THE PROBABILITY OF RELATIVELY PRIME POLYNOMIALS IN Z p k[x]

THE PROBABILITY OF RELATIVELY PRIME POLYNOMIALS IN Z p k[x] THE PROBABILITY OF RELATIVELY PRIME POLYNOMIALS IN Z p [x] THOMAS R. HAGEDORN AND JEFFREY HATLEY Abstract. Let P Rm, n denote the probability that two randomly chosen monic polynomials f, g R[x] of degrees

More information