Quiz 4 answers PDF

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Math 311, Quiz 2 Section: Name: Instructions: Clearly answer each of the questions below. Remember to check the back side. Use full sentences and proper grammar. Show your work and any formulas you employ. Simplify all answers as far as possible. Box your answers, when needed. 1. What is a congruence class? Answer:A congruence class [a]n is the set of all integers with the same remainder as a when divided by n. Said slightly differently with symbols, [a]n = {x ∈ Z : x%n = a%n}

2. What is a zero-divisor for a congruence class? Given an example. Answer: A zero-divisor of a non-zero congruence class is any non-zero congruence class such that the multiple of the two is the zero congruence class.

3. Use an algebraic argument to show that the simultaneous existence of a zero divisor and an inverse of a congruence class is impossible. Answer: We will argue by contradiction. Given a congruence class [a]n , assume both a zero-divisor (call it [z]n ) and an inverse (call it [i]n ) exist. By definition [z]n 6= [0]n . Now, starting from our zero-division, we do some algebra. [z ]n = [z]n [1]n = [z]n ([a]n [i]n ) = ([z ]n [a]n )[i]n = [0]n [i]n = [0]n In this manipulation, we have used the special properties of [0]n and [1]n , and the associativity of multiplication. From it, we derive the contradiction that [z ]n simultaneously is and is not [0]n . Thus, our assumption must be false, and no congruence class can have both an inverse and a zero divisor.

4. Calculate [5]14 [9]14 − [8]14 and reduce your answer to standard form. Answer: [5]14 [9]14 = [45]14 = [3 × 14 + 3]14 = [3]14 [3]14 − [8]14 = [−5]14 = [14 − 5]14 = [9]14

5. Prove

√

3 does not exist as a ratio of two integers.

√ 3. Then with some algebra, 2 2 we can show that a = 3b . Now, according to the fundamental theorem of arithmetic, both a and b have unique prime factorizations. Let

Answer: Suppose there are two positive integers a and b such that a/b =

a = 2x2 × 3x3 × 5x5 . . .

and

b = 2y 2 × 3y 3 × 5y 5 . . . .

Then a2 = 22x2 × 32x3 × 52x5 . . .

and

3b2 = 22y2 × 32y3 +1 × 52y5 . . . .

Since a2 = 3b2 , these must be the same prime factorizations, because the fundamental theory of arithmetic says prime factorizations are unique. But the factorization of a2 has an even number of 3’s, while the factorization of 3b2 has an odd number of 3’s. This √ is impossible, no matter what the other prime factors are, so there can not be a ratio of integers a/b = 3.

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