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

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In summary, the solution for the equation (a-b) + (c-d) = -(b+d) - (-a-c) was verified by showing that p = (a-b) + (c-d) can be expressed as -(b+d) - (-a-c). This was proven using the principles of commutativity and associativity, and referencing the results of -(a+b) = (-a) + (-b) and -(-a) = a from [16], Algebra, Rev 3rd. ed. GTM 211, Springer, Paris 2000.
  • #1
happyprimate
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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.
 
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  • #2
happyprimate said:
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.
happyprimate said:
= (-b-d) + (a+c) By associativity.
I don't see that follows from associativity.
happyprimate said:
= Factoring the negative, we get -(b+d) + (a+c).

a+c can be expressed as -(-a-c)
Are those axioms or theorems or what?
 
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  • #3
##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.
 
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  • #4
fresh_42 said:
##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.
 
  • #5
happyprimate said:
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:$$-(a +b) = (-a) + (-b)$$$$-(-a) = a$$
 
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  • #6
happyprimate said:
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".
 
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  • #7
fresh_42 said:
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.
 

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

1. What is the purpose of showing that (a-b) + (c-d) = -(b+d) - (-a-c)?

The purpose of showing this equation is to demonstrate the commutative property of addition, which states that the order in which numbers are added does not affect the result. This is an important concept in mathematics and is often used in solving equations.

2. How do you prove that (a-b) + (c-d) = -(b+d) - (-a-c)?

To prove this equation, we can use the distributive property of multiplication over addition, which states that a(b+c) = ab + ac. By applying this property to the left side of the equation, we get (a-b) + (c-d) = a + c - b - d. Then, by rearranging the terms, we get -(b+d) - (-a-c) = a + c - b - d, which is equivalent to the left side of the equation.

3. Can you provide an example to illustrate this equation?

Yes, for example, let's say we have the numbers a = 5, b = 3, c = 2, and d = 1. Substituting these values into the equation, we get (5-3) + (2-1) = -(3+1) - (-5-2), which simplifies to 2 + 1 = -4 - (-7), or 3 = 3. This shows that the equation holds true for these values.

4. How is this equation related to other properties of addition?

This equation is related to the associative property of addition, which states that the grouping of numbers does not affect the result. By rearranging the terms in the equation, we can see that it follows the same pattern as the associative property, where the grouping of (a-b) and (c-d) does not affect the result.

5. Can this equation be used in other mathematical applications?

Yes, this equation can be used in various mathematical applications, such as simplifying algebraic expressions, solving equations, and proving other mathematical properties. It can also be used in real-world scenarios, such as calculating profit and loss, where the order of transactions does not affect the final result.

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