Math Challenges

Submissions for Problem #18

Problem #18

Which of the following are groups? Explain

\[ \displaylines{\text{i)}\left(\mathbb{R}_{+},+\right)\\ \text{ii)}\left(2\mathbb{Z},+\right)\\ \text{iii)}\left(\left\lbrace-2,-1,1,2\right\rbrace,\cdot\right)\\ \text{iv)}\left(\left\lbrace-1,1\right\rbrace,\cdot\right)\\ \text{v)}\left(\mathbb{Q}-\left\lbrace0\right\rbrace,\cdot\right)\\ \text{vi)}\left(\mathbb{R},\cdot\right)} \]

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fish-face

Solution: \( \)

Explanation:
i) no, no inverses ii) yes iii) no, not closed iv) yes v) yes. Excluding 0 necessary for inverses. vi) no, no inverse of 0.

zapwai

Solution: \( \displaylines{\text{1) The identity zero is not in the group, and neither are inverses. No, not a group.}\\ \text{2) Both the identity zero, and inverses (-2K for any given element 2K in the set) are in the set. It is also closed under addition, so + is a binary operation on the set. Yes, this is a group.}\\ \text{3) 2x2 = 4, which is not in the set. So multiplication is not a binary operation on this set, as it produces something outside of the set. (Also inverses do not exist for the elements -2 and 2.) No, not a group.}\\ \text{4) This is closed under multiplication, 1 is the identity, and the inverse of each element is itself. Yes, it's a group.}\\ \text{5) The product of two rationals is again a rational, so it is closed under the operation. The identity is 1, which is there. Also inverses are simply q/p for a given p/q in the set. The only issue would have been if p/q = 0, which does not have an inverse. So yes, this is a group.}\\ \text{6) The reals are closed under multiplication, and the identity is 1. For all reals (except r = 0) 1/r is in the group. Since the set does not exclude 0, }0^{-1}\text{ is not in the set, and the set is not closed under inverses. No, not a group.}} \)

Explanation:
A set G is a group if i) it has a binary operation * (under which the set G is "closed", meaning if a and b are in G, then so is a*b) ii) the operation is associative ((a*b)*c = a*(b*c)) iii) the identity for the operation is in set G (a*e = e*a = a) iv) inverses exist in G for every element of G (given a in G, there exists b in G such that a*b = b*a = e) In our case multiplication and addition is associative so I will skip that step.