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

[happyprimate]

Homework Statement

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

Relevant Equations

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

I would like to verify my solution for the following: (a-b) + (c-d) = -(b+d) - (-a-c)

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

(a-b) + (c-d) = (-b+a) + (c-d) By commutativity.

= (-b+a) + (-d+c) By commutativity.

= -b+[a+(-d+c)] By associativity.

= -b+[(a+(-d))+c] By associativity.

= -b+[(a-d)+c] By associativity.

= -b+[(-d+a)+c] By commutativity.

= -b+[-d+(a+c)] By associativity.

= (-b+(-d)) + (a+c) By associativity.

= (-b-d) + (a+c) By associativity.

= Factoring the negative, we get -(b+d) + (a+c).

a+c can be expressed as -(-a-c)

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

This is from Basic Mathematics by Serge Lang. Exercise 7 from chapter 1. Thank you.

 

[PeroK]

Homework Statement:: Show that (a-b) + (c-d) = -(b+d) - (-a-c)
Relevant Equations:: (a-b) + (c-d) = -(b+d) - (-a-c)

I would like to verify my solution for the following: (a-b) + (c-d) = -(b+d) - (-a-c)

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

(a-b) + (c-d) = (-b+a) + (c-d) By commutativity.

= (-b+a) + (-d+c) By commutativity.

= -b+[a+(-d+c)] By associativity.

= -b+[(a+(-d))+c] By associativity.

= -b+[(a-d)+c] By associativity.

= -b+[(-d+a)+c] By commutativity.

= -b+[-d+(a+c)] By associativity.

= (-b+(-d)) + (a+c) By associativity.

All good to here.

= (-b-d) + (a+c) By associativity.

I don't see that follows from associativity.

= Factoring the negative, we get -(b+d) + (a+c).

a+c can be expressed as -(-a-c)

Are those axioms or theorems or what?

 

[fresh_42]

$a+(-d)=:a-d$ is by definition of the RHS.

$-(-a-c)=a+c$ has to be shown (or referenced in case it has been shown in the book). E.g.
$-(-a-c)$ solves the equation $x+(-a+(-c))=0$ and such an equation has a unique solution in a group, which also had to be shown first, or referenced.

 

[happyprimate]

$a+(-d)=:a-d$ is by definition of the RHS.

$-(-a-c)=a+c$ has to be shown (or referenced in case it has been shown in the book). E.g.
$-(-a-c)$ solves the equation $x+(-a+(-c))=0$ and such an equation has a unique solution in a group, which also had to be shown first, or referenced.

Can you show me the correct (formal) way of referencing it in the solution? Thank you.

 

[PeroK]

Can you show me the correct (formal) way of referencing it in the solution? Thank you.

Two results that you may have proved or seen proved already are:$$-(a +b) = (-a) + (-b)$$$$-(-a) = a$$

 

[fresh_42]

Can you show me the correct (formal) way of referencing it in the solution? Thank you.

I don't know the book, but as it is Bourbaki (S. Lang), I bet, everything has a number, lemma 3.1. or so. It can be informal for our purposes here since we only want to check the correctness and we believe you if you say something is shown in the book. We know the group axioms, but we do not know the "relevant equations" for the answer in your circumstances. That is why it is mentioned in the thread frame! It is the most important and the most neglected part of the frame! So either you break it down to group axioms and definitions, or you mention the results you already have available.

Formally, one would write something like ...
\begin{align*}
(a+c)+(-a+(-c))&=(c+a)+(-a+(-c))=c+(a+(-a+(-c))\\&=c+((a+(-a))+(-c))=c+(0+(-c))=c+(-c)=0\\
-(-a-c)+(-a+(-c))&=-(-a-c)+(-a-c)=0\text{ by definition }
\end{align*}
... shows that $a+c$ and $-(a-c)$ both solve the equation $x+(-a+(-c))=0$. By proposition 1.2 (iii), page 8, in [16], we know that solutions are unique, so $a+c= -(-a+(-c)).$

and [16] is listed in your sources at the end of your essay as [16] Lang, S., Algebra, Rev 3rd. ed. GTM 211, Springer, Paris 2000.

Important note: This is not necessary here. I only wrote it since you asked and I wanted you to give an example of what such a reference typically looks like. For us it is sufficient to tell us what you already know from the book in the section "relevant equations".

 

[happyprimate]

I don't know the book, but as it is Bourbaki (S. Lang), I bet, everything has a number, lemma 3.1. or so. It can be informal for our purposes here since we only want to check the correctness and we believe you if you say something is shown in the book. We know the group axioms, but we do not know the "relevant equations" for the answer in your circumstances. That is why it is mentioned in the thread frame! It is the most important and the most neglected part of the frame! So either you break it down to group axioms and definitions, or you mention the results you already have available.

Formally, one would write something like ...
\begin{align*}
(a+c)+(-a+(-c))&=(c+a)+(-a+(-c))=c+(a+(-a+(-c))\\&=c+((a+(-a))+(-c))=c+(0+(-c))=c+(-c)=0\\
-(-a-c)+(-a+(-c))&=-(-a-c)+(-a-c)=0\text{ by definition }
\end{align*}
... shows that $a+c$ and $-(a-c)$ both solve the equation $x+(-a+(-c))=0$. By proposition 1.2 (iii), page 8, in [16], we know that solutions are unique, so $a+c= -(-a+(-c)).$

and [16] is listed in your sources at the end of your essay as [16] Lang, S., Algebra, Rev 3rd. ed. GTM 211, Springer, Paris 2000.

Important note: This is not necessary here. I only wrote it since you asked and I wanted you to give an example of what such a reference typically looks like. For us it is sufficient to tell us what you already know from the book in the section "relevant equations".

That helps tremendously. Thanks a lot.