By Lindsay N. Childs
This booklet is written as an creation to raised algebra for college kids with a heritage of a yr of calculus. the 1st variation of this ebook emerged from a collection of notes written within the 1970sfor a sophomore-junior point path on the college at Albany entitled "Classical Algebra." the target of the path, and the e-book, is to provide scholars sufficient event within the algebraic conception of the integers and polynomials to appre ciate the fundamental techniques of summary algebra. the most theoretical thread is to enhance algebraic houses of the hoop of integers: specified factorization into primes, congruences and congruence periods, Fermat's theorem, the chinese language the rest theorem; after which back for the hoop of polynomials. Doing so ends up in the learn of straightforward box extensions, and, particularly, to an exposition of finite fields. common houses of earrings, fields, teams, and homomorphisms of those items are brought and used as wanted within the improvement. simultaneously with the theoretical improvement, the e-book provides a vast number of functions, to cryptography, error-correcting codes, Latin squares, tournaments, innovations of integration, and particularly to elemen tary and computational quantity thought. A pupil who asks, "Why am I studying this?," willfind solutions often inside a bankruptcy or . For a primary direction in algebra, the ebook deals a few benefits. • by way of development the algebra out of numbers and polynomials, the publication takes maximal good thing about the student's earlier event in algebra and mathematics. New ideas come up in a well-known context.
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Additional resources for A Concrete Introduction to Higher Algebra
Long division works in any base as well. In fact, in base 2 it is particularly easy, because in determining the correct digits of the quotient, no 24 2. Induction guesswork is involved. For example: 1010011 100101 ) 101111111 ill 100101 00101011 100101 000110111 100101 100101 100101 O. On the other hand, in base a where a is large it is harder to determine the digits of the quotient. A. L. Stein in 1960 that if the first digit of the divisor is close to the base a, then guessing the digits of the quotient is easier than if the first digit of the quotient is near O.
Notice that 1 divides any integer (so does -1); and any integer divides O. We will often use the notation alb to mean, a divides b, and the notation a %b to mean, a does not divide b. Thus 6112, 6%14. (The notation I can be confusing. ") El. Show that of any three consecutive integers, exactly one is divisible by 3. E2. Show that of any m consecutive integers, exactlyone is divisible by m. E3. Let m > n > 1 be natural numbers. Show that there is some t, n ~ t < m, such that m - n divides t. If a and b are integers, a common divisor of a and b is an integer e such that e divides a and e divides b.
It is not obvious how to do this! Here is how. Do the Euclidean algorithm: 1876 = 365·5 + 51, 365 = 51. 7 + 8, 51 = 8·6 + 3, 8=3'2+2, 3 = 2 ·1 + 1; so 1 is the greatest common divisor. Now solve for the remainders (the remainders are in boldface to distinguish them from the quotients) 1=3-2'1, 2=8-3'2, 3 = 51 - 8 '6, 8 = 365 - 51 ' 7, 51 = 1876 - 365· 5, and successively substitute the remainders into the equation 1 = 3 - 2· 1, starting with 2: 1=3-2 = 3 - (8 - 3 . 2) = 3 · 3 - 8 = 3(51 - 8· 6) - 8 = 3· 51 - 8 ·19 = 3· 51 - 19(365 - 51· 7) = 136· 51 - 19·365 = 136(1876 - 5 · 365) - 19· 365 = 136·1876 - 699·365.