Proof of (m+n*)* = m* + n* in Modern Abstract Algebra

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In summary, the person has been taking courses to improve their ability to construct proofs. They have received guidance from Tom on a proof by mathematical induction and are now seeking input on another proof they need to do. They present a question from a textbook and provide a detailed proof using Postulate II, rules of addition, and Peano's axioms. They ask for feedback on their proof and express their willingness to go through "picky" steps to ensure it is done correctly. Ivy approves of the proof and offers additional advice.
  • #1
wubie
[SOLVED] Another Proof

Hello,

I have stated in other threads that my ability to construct proofs is somewhat lacking. I have taken it upon myself to take courses in which constructing proofs is a large part of the content of these courses. I don't expect to do well in these courses. But the intent was to correct a deficiency in my mathematical education. So that is good enough for me.

Tom was kind enough to guide me through a proof by mathematical induction. I have another proof to do in addition to the proof by induction. It is not as intensive as the proof by mathematical induction. But my confidence in constructing proofs is lacking. So I would still like input on the proof that I am about to present. For anyone who feels they can spare some time and give me advice on proofs, that would be great. If not, that's ok too.

Here is my question as given in "Schaum's Outlines: Modern Abstract Algebra", page 37, 16 a):

Prove (m+n*)* = m* + n*

Given:

Postulate II: For each n which is an element of N, there exists a unique n* which is an element of N, called the successor of n.

Addition on N is defined by

i) n+1 = n*, for every n which is an element of N.

1.Let (m+n*) = k, in which k is an element of N.

then

2.k* = (k+1) by Postulate II.

3.(k+1) = (m+n*)+1 by substitution.

4.(m+n*)+1 = 1+(m+n*) by Commutative Law.

5.1+(m+n*) = (1+m)+n* by Associative Law.

6.(1+m)+n* = (m+1)+n* by Commutative Law.

7.(m+1)+n* = m*+n* by i).

Hence by steps 1 to 7, the given postulate, and rules of addition, (m+n*)* = m* + n*.


How does this look as a proof? I know there probably some unnecessary steps, but I didn't want to make too many assumptions and I wanted to show and anotate every step. Would this be considered an adequate proof? Should I be doing something more or less to formalize this proof?

Once again, any comments are appreciated.
 
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  • #2
Looks good to me. That's the kind of "picky" little steps one has to go through when working from something as basic as Peano's axioms!
 
  • #3
I don't mind if I have to do the picky steps. I just like to know if it is the right way to do a formal proof. And if there is anything else I should consider when constructing such a proof.

Thanks Ivy,
 

1. What is the proof of the distributive property in Modern Abstract Algebra?

The proof of the distributive property in Modern Abstract Algebra, specifically for the operation of multiplication, is known as the "distributive law." It states that for any three elements a, b, and c in a ring R, the following equation holds: (a + b)c = ac + bc. This law can be extended to any number of elements and is a fundamental property of rings in abstract algebra.

2. How do you prove that (m+n*)* = m* + n* in Modern Abstract Algebra?

To prove that (m+n*)* = m* + n* in Modern Abstract Algebra, we can use the distributive law mentioned in the previous answer. By substituting (m + n*) for a and * for c, we get (m + n**)* = (m*)* + (n*)*. Then, using the associative property of rings, we can simplify (m*)* + (n*)* to m* + n*, giving us (m + n*)* = m* + n*.

3. Why is the distributive property important in Modern Abstract Algebra?

The distributive property is important in Modern Abstract Algebra because it allows us to perform calculations with multiple elements in a ring efficiently. It is a fundamental property that is used in many other proofs and can help simplify complex equations.

4. What is the difference between distributivity in Modern Abstract Algebra and elementary algebra?

The main difference between distributivity in Modern Abstract Algebra and elementary algebra is the level of abstraction. In elementary algebra, we are working with specific numbers and operations, whereas in abstract algebra, we are working with general structures and operations. Additionally, distributivity in abstract algebra applies to any number of elements, while in elementary algebra, it only applies to two elements.

5. Can the distributive property be generalized to other operations in Modern Abstract Algebra?

Yes, the distributive property can be generalized to other operations in Modern Abstract Algebra. For example, it can be extended to the operation of addition, where it states that a*(b + c) = a*b + a*c. It can also be extended to other algebraic structures, such as fields and vector spaces, with slight variations in the definition.

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