A Transition to Advanced Mathematics
A Transition to Advanced Mathematics
8th Edition
ISBN: 9781285463261
Author: Douglas Smith, Maurice Eggen, Richard St. Andre
Publisher: Cengage Learning
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Question
Chapter 3.4, Problem 1E

(a)

To determine

To find the value 6+6  in  7 .

(a)

Expert Solution
Check Mark

Answer to Problem 1E

  6+6 =5 in  7 .

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is 7 .

Formula used: Definition congruence modulo m.

For x,y , x is congruentto y if m divides xy and it is written as

  xy(mod m)

Calculation:

  6+6 =125(mod7)

  6+6=5 in  7

(b)

To determine

To find the value 54in 7 .

(b)

Expert Solution
Check Mark

Answer to Problem 1E

  54in 7  is  6 .

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is 7 .

Formula used: Definition of congruence modulo m.

For x,y , x is congruent to y if m divides xy and it is written as

  xy(mod m)

Calculation:

  54=206(mod  7)

  54=6in  7

c.

To determine

To find 33+52  in 7 .

c.

Expert Solution
Check Mark

Answer to Problem 1E

  33+52 =5 in 7

Explanation of Solution

Given information: Equivalence class under congruence modulo 7is 7 .

Formula used: Definition of congruence modulo m.

For x,y , x is congruentto y if m divides xy and it is written as

  xy(mod m)

Calculation:

  33+52 =9+102(mod 7)+3(mod7)5(mod7)

   33+52 =5in 7

d.

To determine

To find 24+35  in  7 .

d.

Expert Solution
Check Mark

Answer to Problem 1E

  24+35  in  7  is  2

Explanation of Solution

Given information: Equivalence class under congruence modulo 7is 7 .

Formula used: Definition of congruence modulom.

For x,y , x is congruentto y if m divides xy and it is written as

  xy(mod m)

Calculation:

  24+35 =8+151(mod 7)+1(mod7)2(mod7)

   24+35 =2in 7

e.

To determine

To find the value 51+32 in 7 .

e.

Expert Solution
Check Mark

Answer to Problem 1E

  51+32 =4in 7

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is 7 .

Formula used: Definition of congruence modulo m.

For x,y , x is congruent to y if m divides xy and it is written as

  xy(mod m)

Calculation:

  51+32 =5+6=114(mod7)

   51+32 =4in 7

f.

To determine

To find 03+24  in 7  .

f.

Expert Solution
Check Mark

Answer to Problem 1E

  03+24  in 7  is  1 .

Explanation of Solution

Given information: Equivalence class under congruence modulo 7is 7 .

Formula used: Definition of congruence modulo m.

For x,y , x is congruent to y if m divides xy and it is written as

  xy(mod m)

Calculation:

  03+24 =0+8=81(mod7)

   03+24 =1in 7

h.

To determine

To find (4+5)(5+6)

  in 7 .

h.

Expert Solution
Check Mark

Answer to Problem 1E

  (4+5)(5+6)

  in 7 is 1

Explanation of Solution

Given information: Equivalence class under congruence modulo7 is 7 .

Formula used: Definition of congruence modulom.

For x,y , x is congruent to y if m divides xy and it is written as

  xy(mod m)

Calculation:

  (4+5)(5+6)=9112(mod 7)4(mod7)=81(mod7)

(4+5)(5+6)

  in 7 is 1

i.

To determine

To find 225  in  7 .

i.

Expert Solution
Check Mark

Answer to Problem 1E

  225=2  in  7

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is 7 .

Formula used: Definition of congruence modulom.

For x,y , x is congruent to y ifm divides xy and it is written as

  xy(mod m)

Calculation:

  225=(25)5=(23)82=(8)82{1(mod  7)}82=(1)82=2  

225=2  in  7

j.

To determine

To find 523

  in  7 .

j.

Expert Solution
Check Mark

Answer to Problem 1E

  523

  in  7 is 3.

Explanation of Solution

Given information: Equivalence class under congruence modulo7 is 7 .

Formula used: Definition of congruence modulo m.

For x,y , x is congruentto yif m divides xy and it is written as

  xy(mod m)

Calculation:

  523=51855=(56)35513(mod7)551545(mod7)(25)25(mod7)425(mod7)2(mod7)5=10(mod7)3(mod7)

523

  in  7 is 3

k.

To determine

To find 444

  in  7 .

k.

Expert Solution
Check Mark

Answer to Problem 1E

  444

  in  7 is 2

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is 7 .

Formula used: Definitionof congruence modulom.

For x,y ,x is congruent to y if m divides xy and it is written as

  xy(mod m)

Calculation:

  444=(43)1442=(64)1416{1(mod 7)}8{2(mod 7)}=(1)82=2  

444

  in  7 is 2

l.

To determine

To find 226

  in  7 .

l.

Expert Solution
Check Mark

Answer to Problem 1E

  226

  in  7 is 4.

Explanation of Solution

Given information: Equivalence class under congruence modulo 7 is 7 .

Formula used: Definition of congruence modulo m.

For x,y , x is congruent to y if m divides xy and it is written as

  xy(mod m)

Calculation:

  226=2252=(25)52{4(mod7)}52=2mod(7)

226

  in  7 is 4.

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Chapter 3 Solutions

A Transition to Advanced Mathematics

Ch. 3.1 - Prove that if G is a group and H is a subgroup of...Ch. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.2 - (a)Show that any two groups of order 2 are...Ch. 3.2 - (a)Show that the function h: defined by h(x)=3x is...Ch. 3.2 - Let R be the equivalence relation on ({0}) given...Ch. 3.2 - Let (R,+,) be an integral domain. Prove that 0 has...Ch. 3.2 - Complete the proof of Theorem 6.5.5. That is,...Ch. 3.2 - Prob. 6ECh. 3.2 - Assign a grade of A (correct), C (partially...Ch. 3.2 - Prob. 8ECh. 3.2 - Prob. 9ECh. 3.2 - Use the method of proof of Cayley's Theorem to...Ch. 3.2 - Prob. 11ECh. 3.2 - Assign a grade of A (correct), C (partially...Ch. 3.2 - Prob. 13ECh. 3.2 - Define on by setting (a,b)(c,d)=(acbd,ad+bc)....Ch. 3.2 - Prob. 15ECh. 3.2 - Let f:(A,)(B,*) and g:(B,*)(C,X) be OP maps. Prove...Ch. 3.2 - Prob. 17ECh. 3.2 - Let Conj: be the conjugate mapping for complex...Ch. 3.2 - Prove the remaining parts of Theorem 6.4.1.Ch. 3.3 - Let 3={3k:k}. Apply the Subring Test (Exercise...Ch. 3.3 - Use these exercises to check your understanding....Ch. 3.3 - Use these exercises to check your understanding....Ch. 3.3 - Use these exercises to check your understanding....Ch. 3.3 - Use these exercises to check your understanding....Ch. 3.3 - Prob. 6ECh. 3.3 - Use the definition of “divides” to explain (a) why...Ch. 3.3 - Prob. 8ECh. 3.3 - Prob. 9ECh. 3.3 - Complete the proof that for every m,(m+,) is a...Ch. 3.3 - Define addition and multiplication on the set ...Ch. 3.3 - Prob. 12ECh. 3.3 - Let (R,+,) be a ring and a,b,R. Prove that b+(a)...Ch. 3.3 - Prove the remaining parts of Theorem 6.5.3: For...Ch. 3.3 - We define a subring of a ring in the same way we...Ch. 3.4 - Prob. 1ECh. 3.4 - Prob. 2ECh. 3.4 - If possible, give an example of a set A such that...Ch. 3.4 - Let A. Prove that if sup(A) exists, then...Ch. 3.4 - Let A and B be subsets of . Prove that if sup(A)...Ch. 3.4 - a.Give an example of sets A and B of real numbers...Ch. 3.4 - a.Give an example of sets A and B of real numbers...Ch. 3.4 - An alternate version of the Archimedean Principle...Ch. 3.4 - Prob. 9ECh. 3.4 - Prob. 10ECh. 3.4 - Prob. 11ECh. 3.4 - Prob. 12ECh. 3.5 - Prob. 1ECh. 3.5 - Prob. 2ECh. 3.5 - Let A be a subset of . Prove that the set of all...Ch. 3.5 - Prob. 4ECh. 3.5 - Let be an associative operation on nonempty set A...Ch. 3.5 - Suppose that (A,*) is an algebraic system and * is...Ch. 3.5 - Let (A,o) be an algebra structure. An element lA...Ch. 3.5 - Let G be a group. Prove that if a2=e for all aG,...Ch. 3.5 - Give an example of an algebraic structure of order...Ch. 3.5 - Prove that an ordered field F is complete iff...Ch. 3.5 - Prove that every irrational number is "missing"...Ch. 3.5 - Find two upper bounds (if any exits) for each of...Ch. 3.5 - Prob. 13ECh. 3.5 - Prob. 14ECh. 3.5 - Prob. 15ECh. 3.5 - Let A and B be subsets of . Prove that if A is...Ch. 3.5 - Prob. 17ECh. 3.5 - Prob. 18ECh. 3.5 - Give an example of a set A for which both A and Ac...Ch. 3.5 - Prob. 20ECh. 3.5 - Prob. 21ECh. 3.5 - Prob. 22E
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