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The set {1, 2, 4, 7, 8,11,13,14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectively
- a)3 and 13
- b)2 and 11
- c)4 and 13
- d)8 and 14
Correct answer is option 'C'. Can you explain this answer?
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Solution:
To check whether the given set is a group under multiplication modulo 15, we need to verify the following properties:
1. Closure property: For any a, b ∈ S, a × b ∈ S
2. Associativity property: For any a, b, c ∈ S, (a × b) × c = a × (b × c)
3. Identity property: There exists an element e ∈ S such that for any a ∈ S, a × e = e × a = a
4. Inverse property: For any a ∈ S, there exists an element a^-1 ∈ S such that a × a^-1 = a^-1 × a = e
Checking Closure property:
1 × 2 = 2
2 × 4 = 8
8 × 7 = 1
1 × 11 = 11
11 × 13 = 8
8 × 14 = 7
All the elements obtained are in the given set. Hence, the closure property holds.
Checking Associativity property:
Let's take a = 2, b = 4, and c = 7
(2 × 4) × 7 = 8 × 7 = 1
2 × (4 × 7) = 2 × 13 = 11
Since (2 × 4) × 7 ≠ 2 × (4 × 7), the associativity property does not hold.
Hence, the given set is not a group under multiplication modulo 15.
Finding Inverses:
To find the inverse of an element 'a' in a group, we need to find an element 'a^-1' such that a × a^-1 = a^-1 × a = e, where e is the identity element.
Let's find the inverse of 4:
4 × 4 = 16 ≡ 1 (mod 15)
Hence, the inverse of 4 is 4^-1 ≡ 4 (mod 15)
Let's find the inverse of 7:
7 × 13 = 91 ≡ 1 (mod 15)
Hence, the inverse of 7 is 7^-1 ≡ 13 (mod 15)
Therefore, the correct option is (C) 4 and 13.
To check whether the given set is a group under multiplication modulo 15, we need to verify the following properties:
1. Closure property: For any a, b ∈ S, a × b ∈ S
2. Associativity property: For any a, b, c ∈ S, (a × b) × c = a × (b × c)
3. Identity property: There exists an element e ∈ S such that for any a ∈ S, a × e = e × a = a
4. Inverse property: For any a ∈ S, there exists an element a^-1 ∈ S such that a × a^-1 = a^-1 × a = e
Checking Closure property:
1 × 2 = 2
2 × 4 = 8
8 × 7 = 1
1 × 11 = 11
11 × 13 = 8
8 × 14 = 7
All the elements obtained are in the given set. Hence, the closure property holds.
Checking Associativity property:
Let's take a = 2, b = 4, and c = 7
(2 × 4) × 7 = 8 × 7 = 1
2 × (4 × 7) = 2 × 13 = 11
Since (2 × 4) × 7 ≠ 2 × (4 × 7), the associativity property does not hold.
Hence, the given set is not a group under multiplication modulo 15.
Finding Inverses:
To find the inverse of an element 'a' in a group, we need to find an element 'a^-1' such that a × a^-1 = a^-1 × a = e, where e is the identity element.
Let's find the inverse of 4:
4 × 4 = 16 ≡ 1 (mod 15)
Hence, the inverse of 4 is 4^-1 ≡ 4 (mod 15)
Let's find the inverse of 7:
7 × 13 = 91 ≡ 1 (mod 15)
Hence, the inverse of 7 is 7^-1 ≡ 13 (mod 15)
Therefore, the correct option is (C) 4 and 13.
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The set {1, 2, 4, 7, 8,11,13,14} is a group under multiplication modulo 15. The inverses of 4 and 7 are respectivelya)3 and 13b)2 and 11c)4 and 13d)8 and 14Correct answer is option 'C'. Can you explain this answer?
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