Download e-book for kindle: Abstract Algebra : An Inquiry Based Approach by Hodge, Jonathan K.; Schlicker, Steven; Sundstrom, Ted
By Hodge, Jonathan K.; Schlicker, Steven; Sundstrom, Ted
""This booklet arose from the authors' method of instructing summary algebra. They position an emphasis on lively studying and on constructing scholars' instinct via their research of examples. ... The textual content is equipped in this kind of manner that it really is attainable firstly both earrings or groups.""
-Florentina Chirteş, Zentralblatt MATH 1295
""This e-book arose from the authors' method of educating summary algebra. They position an emphasis on energetic studying and on constructing scholars' instinct via their research of examples. ... The textual content is geared up in one of these approach that it really is attainable first of all both earrings or groups.""
-Florentina Chirteş, Zentralblatt MATH 1295
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Extra resources for Abstract Algebra : An Inquiry Based Approach
8, you may have noticed that since the days of the week follow a 7-day cycle, the difference between any two numbers on this list is divisible by 7. You may have also noticed that all of the numbers on the list have the same remainder (as specified by the Division Algorithm) when divided by 7. These two observations are important and useful; the first forms the basis of our definition of congruence, and the second is a consequence of this definition. 9. Let n be a natural number, and let a and b be integers.
For convenience, use q1 , q2 , r1 , r2 to denote the resulting quotients and remainders. (b) If you haven’t already done so, write your equations from part (a) so that they are in the form a = . . and b = . .. Then use subtraction to obtain a new equation of the form a − b = . .. (c) Now assume that n | (a − b). Use your equation from part (b) to argue that n | (r1 − r2 ) as well. (d) Use the result you proved in part (c) to deduce that r1 = r2 . (Hint: Both r1 and r2 satisfy a certain inequality.
In particular, d = gcd(a, b) if • d ∤ a or d ∤ b; or • there exists some integer k > d such that k | a and k | b. 3, and its negation, may seem quite natural, the inequalities involved in them can be somewhat troublesome, especially when it comes to writing proofs. Later on in this investigation, we will see how these inequalities can be replaced with conditions that involve divisibility instead. 3 in a way that is both easier to work with in proofs and easier to generalize to other number systems.
Abstract Algebra : An Inquiry Based Approach by Hodge, Jonathan K.; Schlicker, Steven; Sundstrom, Ted