week04tutsols

# week04tutsols - THE UNIVERSITY OF SYDNEY Math2968...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY Math2968 Algebra (Advanced) Semester 2 Tutorial Solutions Week 4 2008 1. (for general discussion) Determine which of the following form semigroups, groups or abelian groups: (a) the positive real numbers, under multiplication (b) the positive real numbers, under division (c) the integers that are divisible by 10, under addition (d) the nonzero rational numbers, where a b is the ‘product’ of a and b (e) { a + b √ 2 | a,b ∈ Q , not both zero } , under multiplication (f) the integers, with 3 a + 4 b as the ‘product’ of a and b (g) elements of Z 6 , with 3 a + 4 b as the ‘product’ of a and b (h) 2 × 2 matrices with real entries, under matrix addition (i) 2 × 2 matrices with real entries, under matrix multiplication (j) 2 × 2 singular matrices with real entries, under matrix addition (k) 2 × 2 singular matrices with real entries, under matrix multiplication (l) 2 × 2 nonsingular matrices with real entries, under matrix multiplication (m)geometric vectors, under vector addition (n) geometric vectors, under cross product (o) a single orangatang, where the ‘product’ of the orangatang with itself is itself Solution (a) The positive reals from an abelian group under multiplication, noting that products of positive reals are positive and the inverse of a positive real is positive. (b) Division is not associative, so we don’t even get a semigroup. For example, (1 ÷ 1) ÷ 2 = 1 2 negationslash = 2 = 1 ÷ (1 ÷ 2) . (c) The integers divisible by 10 form an abelian group under addition, with 0 as the identity element and negation as group inversion. (d) The given ‘product’ is not even a binary operation on nonzero rationals. For example, 2 1 2 = √ 2 negationslash∈ Q \{ } . Also, when the ‘product’ is defined, it need not be associative, so we don’t even get a partial semigroup. For example, (2 1 ) 2 = 4 negationslash = 2 = 2 (1 2 ) . (e) The set { a + b √ 2 | a,b ∈ Q , not both zero } forms an abelian group. The tricky bit to check is the existence of multiplicative inverses, which involves the usual technique of rationalising the denominator: ( a + b √ 2) − 1 = a a 2 − 2 b 2 + − b a 2 − 2 b 2 √ 2 . (f) The integers with ‘product’ a ∗ b = 3 a + 4 b do not even form a semigroup. For example, (1 ∗ 0) ∗ 0 = 9 negationslash = 3 = 1 ∗ (0 ∗ 0) . (g) The elements of Z 6 with ‘product’ a ∗ b = 3 a +4 b form a semigroup, but not a group. Associativity follows from the arithmetic in Z 6 : ( a ∗ b ) ∗ c = (3 a + 4 b ) ∗ c = 9 a + 12 b + 4 c = 3 a + 4 c = 3 a + 12 b + 16 c = a ∗ (3 b + 4 c ) = a ∗ ( b ∗ c ) ....
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## This note was uploaded on 09/12/2009 for the course MATH 2968 taught by Professor Easdown during the One '09 term at University of Sydney.

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week04tutsols - THE UNIVERSITY OF SYDNEY Math2968...

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