Question

1. Compute additive and multiplicative inverses of 7 and 9 in Z11 (mod 11).

2. Find out whether or not 4 and 7 have multiplicative inverse in Z14 (mod 14).

3. Let S be the set of even integers under the operations of addition and multiplication. Is S a ring? Is it commutative? Is it a field? Justify your answer.

4. Compute the multiplicative inverse of 9 under modulo 31 using the extended Euclid’s algorithm.

Answer 1

Answer 1

Multiplicative inverse of integer a is x such that

For a=7 and n=11, multiplicative inverse will be 8

For a=9 and n=11, multiplicative inverse will be 5

Additive inverse of integer a is x such that

For a=7 and n=11, additive inverse will be 4

For a=9 and n=11, additive inverse will be 2

Answer 2

Multiplicative inverse of integer a is x such that

For a=4 and n=14,  

For any value of x from 1 to 13, there is no value of y which can satisfy this equation so there is no multiplicative inverse.

For a=7 and n=14,  

For any value of x from 1 to 13, there is no value of y which can satisfy this equation so there is no multiplicative inverse for Z14.

Answer 3

S is a set of even integers with operations addition and multiplication.

Addition of two even number is even.

Addition of two even integer will satisfy relation below.

Addition of three even integer will satisfy relation below as well.

There exists a element 0 in set S (0 is even integer) such that

There exists a element y in set S (y = -a which is also even integer if a is even integer) such that

Multiplication of two even number is even.

Multiplication of two even integer will satisfy relation below.

Multiplication of three even integer will satisfy relation below as well.

There does not exists a element 1 in set S (1 is not even integer) such that

There does not exists a element y in set S (y = 1/a which is not integer) such that

Multiplication of three even integer will satisfy relation below as well.

So as we can see that set S is Ring but not Field as R satisfies below.

• (R, addition) is an abelian group ( i.e commutative group)
• (R, multiplication) is a semigroup
• For any three elements a, b, c R the left distributive law a.(b+c) =a.b + a.c.
• Right distributive property (b + c).a =b.a + c.a is satisfied.

Answer 4

Multiplicative inverse of 9 under modulo 31 using the extended Euclid’s algorithm

a = 9, n=31

Step 0 : 31 = 3(9) + 4, p₀ = 0, q₀ = 3

Step 1: 9 = 2(4) + 1, p₁ = 1, q₁ = 2

Step 2: 4 = 4(1) + 0, p₂ = p₀ - p₁ q₀ (mod n) = 0 - 1(3) mod 31 = -3 mod 31 = 28

p₃ = p₁ - p₂ q₁ (mod n) = 1 - 28(2) mod 31 = -55 mod 31 = 7

So, 9(7) = 63 = 1+ 31(2) 1 (mod 31) . Answer is 7