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

  • Thread starter Thread starter badateverything
  • Start date Start date
AI Thread Summary
The discussion focuses on proving that the additive inverse of the sum of three variables, -(a + b + c), equals the sum of their individual inverses, -a + (-b) + (-c). Participants clarify that the goal is to demonstrate that -a + (-b) + (-c) is indeed the additive inverse of (a + b + c) using only the properties of associativity and commutativity. The proof involves showing that adding -a + (-b) + (-c) to a + b + c results in zero, confirming the validity of distributing the negative sign. It is emphasized that while rearranging equations can be useful, it may not always be appropriate in advanced mathematical contexts. Understanding the notation and the rules applied at each step is crucial for clarity in the proof.
badateverything
Messages
1
Reaction score
0

Homework Statement


show that -(a + b + c) = -a + (-b) + (-c) using associativity/commutativity

Homework Equations


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

The Attempt at a Solution


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

the solution according to lang
(a+b+c) + (-a) + (-b) + (-c) = 0
(a+b+c) + (-a) + (-b) + (-c) = a + b + c - a - b - c
= a - b - a + c - b - c
= a - a + b + c - b - c
= a - a + b - b + c - c
= (a-a) + (b-b) + (c - c)
= 0 + 0 + 0 = 0am i supposed to be rearranging until both sides look the same or am i supposed to solve until i get to 0?? he didn't really explain proofs. sorry if this is a dumb question.
 
Physics news on Phys.org
It doesn't matter - you are free to choose the application of the rules, but you should write a short sentence at each stage saying which rule you are using to get that step.
 
These problems can look a bit strange if you're unclear on what you're supposed to be showing and what the notation means. You also need to pay attention to what you're "allowed" to know in some cases. It can be a bit of a challenge to force yourself to discard some of your more closely held "beliefs" of how the symbol-pushing works.

##-(a+b+c)## is meant to denote the additive inverse of the sum of ##a##, ##b##, and ##c##; the unique element satisfying ##-(a+b+c)+(a+b+c)=0##. And ##-a+(-b)+(-c)## is the sum of the additive inverses of ##a##, ##b## and ##c##, respectively.

The point of the exercise is, essentially, to show that ##-a+(-b)+(-c)## is the additive inverse of ##(a+b+c)##, thus verifying that it is a valid move to "distribute" the minus sign through. Note that you're meant to use only the associativity and commutativity of addition to accomplish this. You're not supposed to use distributivity of multiplication (multiplication ma not even be a defined operation depending on context), nor are you allowed to assume that ##-a=(-1)\cdot a##.

So Lang's proof shows that adding ##-a+(-b)+(-c)## to ##a+b+c## results in ##0##, demonstrating that ##-a+(-b)+(-c)## is the additive inverse of ##a+b+c##.

Also, while there is often nothing logically incorrect about rearranging both sides of an equation to derive a truth from an unknown, it is generally considered "bad form" in many cases once you reach a certain level of mathematical maturity.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
Back
Top