course Mth 151 ‡ç£›àyÜWxàÜåªv¶î¦–ûêÖ¦‡–×Uassignment #022
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10:43:57 4.5.9 {-1,0,1} group on multiplication?
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RESPONSE --> no - there is no inverses in this ""group,"" so it is not a group.
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10:44:31 ** There are four criteria for the group: closure, identity, inverse property, and associativity. The lack of any one of these properties means that the set and operation do not form a group. The set is closed on multiplication. The identity is the element that when multiplied by other elements does not change them. The identity for this operation is 1, since 1 * -1 = -1, 1 * 0 = 0 and 1 * 1 = 1. Inverses are pairs of elements that give you 1 when you multiply them. For example -1 * -1 = 1 so -1 is its own inverse. 1 * 1 = 1 so 1 is also its own inverse. However, 0 does not have an inverse because there is nothing you can multiply by 0 to get 1. Since there is an element without an inverse this is not a group. **
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RESPONSE --> ok
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10:45:24 4.5.25 verify (NT)R = N(TR)
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RESPONSE --> (NT)R = N(TR) VR = NP M = M
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10:45:31 ** From the table (NT)R= V R = M and N(TR)= N P = M This verifies the identity. **
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RESPONSE --> ok
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10:46:29 query 4.5.33 inverse of T
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RESPONSE --> The inverse of T is T. It is it's own inverse. If you look at the identity of the group, which is Q, then you would see that T is its own inverse.
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10:46:35 ** T is its own inverse because T T gives you the identity **
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RESPONSE --> ok
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10:47:40 4.5.42. Explain what property is gained when the system of integers is extended to the system of rational numbers.
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RESPONSE --> I don't understand this question. I worked out the problem about the group of integers, but didn't complete the problem about the group of rational numbers.
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10:49:37 ** The set of integers is a group on addition, with identity 0 and every number x having additive inverse -x. It is not a group on multiplication. It contains the identity 1 but does not contain inverses, except for 1 itself. This is because, for example, there is no integer you can multiply by 2 to get the identity 1. If we extend the integers to the rational numbers we do get the inverses. The inverse of 2 is 1/2 since x * 1/2 = 1, the identity. In general the multiplicative inverse of x is 1 / x. However we still don't have a group on multiplication since 0 still doesn't have an inverse, 1 / 0 being undefined on the real numbers. **
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RESPONSE --> I am sure that the explanation you gave is a great one, but I still don't quite understand this problem. Maybe if I saw it worked out. I'll try working it out.
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