Justify each step using commutativity and associativity

[happyprimate]

Summary:: Justify each step using commutativity and associativity in proving the following identities.
(a-b)+(c-d) = (a+c)+(-b-d)

Exercise 3 Chapter 1 Basic Mathematics Serge Lang

Verifying my answer.

My answer:

(a-b)+(c-d) = (a+c)+(-b-d)

Let p = (a-b)+(c-d) We need to show that p = (a+c)+(-b-d)

(a-b)+(c-d)

a+(-b+(c-d)) Associativity

a+((-b+c)-d) Associativity

a+((c-b)-d) Commutativity

((a+c)-b)-d) Associativity

(a+c)+(-b-d) Associativity

Need feedback on the proof. I am currently working through basic mathematics book by Serge Lang. Thank you.

 

[fresh_42]

Summary:: Justify each step using commutativity and associativity in proving the following identities.
(a-b)+(c-d) = (a+c)+(-b-d)

Exercise 3 Chapter 1 Basic Mathematics Serge Lang

Verifying my answer.

My answer:

(a-b)+(c-d) = (a+c)+(-b-d)

Let p = (a-b)+(c-d) We need to show that p = (a+c)+(-b-d)

(a-b)+(c-d)

a+(-b+(c-d)) Associativity

a+((-b+c)-d) Associativity

a+((c-b)-d) Commutativity

((a+c)-b)-d) Associativity

(a+c)+(-b-d) Associativity

Need feedback on the proof. I am currently working through basic mathematics book by Serge Lang. Thank you.

Looks good.

 

[happyprimate]

Thank you.

 

[fresh_42]

Thank you.

Only a remark to the presentation. It would be better to write it as equations:
\begin{align*}
(a-b)+(c-d) &\stackrel{Ass.}{=}a+(-b+(c-d)) \\
&\stackrel{Ass.}{=} a+((-b+c)-d) \\
&\stackrel{Com.}{=}a+((c-b)-d) \\
&\stackrel{Ass.}{=} ((a+c)-b)-d) \\
&\stackrel{Ass.}{=} (a+c)+(-b-d)
\end{align*}

I liked that you wrote $a-b=a+(-b)$ because that is what it is: an addition. There is no subtraction on this level. Subtraction is only an abbreviation, not an operation.

 

[happyprimate]

Thanks a lot fresh_42.
Question! I was trying to work through a few more identities where the solution wasn't provided. Should I post it in the same thread due to the same theme ie proving identities.

 

[fresh_42]

Thanks a lot fresh_42.
Question! I was trying to work through a few more identities where the solution wasn't provided. Should I post it in the same thread due to the same theme?

It's usually better to start a new one. I admit that this rule is not always followed, and we let it slip if the questions are closely related, but a new one is better in case a debate evolves. If the only answer is "this is correct" then it appears that it could well be treated in one thread. Start a new.

If you want a related problem then solve the following:
Show that the group axioms are equivalent to: "For all $a,b \in G$ there is a unique solution to $a\cdot x=b$ and $x\cdot a=b$." (non commutative case)

 

[Office_Shredder]

My only objection is the 4th step is two sets of parentheses adjustments, so maybe should be written as two associativity steps. If you think doing it in one shot is fine, then you can condense the first and second step into one line also

 

[Mark44]

It would be better to write it as equations:

Strongly agree...