Assignment 22

course Mth 151

磛yWxv֦Uassignment #022

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Liberal Arts Mathematics I

07-18-2006

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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.

What we gain is multiplicative inverses. The solution shows this.

If you deconstruct every statement in the given solution, and let me know specifically what you do and do not understand, I'm sure we can make this clear to you.

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Good. See my note on the last problem, and let me know if you have questions.