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**[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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