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A-(-b)=a+b How to prove

by Tyler.Smith
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Tyler.Smith
#1
Nov15-12, 06:05 PM
P: 6
Well as the subject states... How does one prove that a-(-a)=a+b ?

If "a" is any number
"b" is a positive number, being "-b" its corresponding negative








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haruspex
#2
Nov15-12, 06:44 PM
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With these seemingly trivial questions about the fundamentals of algebra, it is essential to use only the axioms and already established deductions. So you'll need to post those as "relevant equations".
HallsofIvy
#3
Nov16-12, 07:14 AM
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As haruspex says, how you would prove this depends upon what axioms you have to use. If this is a general group, you are asking to prove that "the additive inverse of the additive inverse of a is a".

b is the "additive inverse of a" if and only if a+ b= 0 and b+ a= 0. c is the additive inverse of b if and only if b+ c= c+ b= 0. Play around with those equation to arrive at "a= c". You might need the easily proved result that 0 is its own additive inverse.

But you use the term "number: so perhaps this is to be for a specific set of numbers. In that case, how you would prove this depends upon which set and which axioms you are using. For example are you to use Peano's axioms for the integers?

Tyler.Smith
#4
Nov18-12, 11:36 AM
P: 6
A-(-b)=a+b How to prove

Hello, I followed your advice, and use some axioms and deductions that had already established. I do not know if it is correct



Demostracion
I have a group G, which has an internal composition law ∘, satisfies the following axioms.
1. a+(b+c)=(a+b)+c, ∀a,b,c∈G
2. ∃e∈G : e+a=a-e=a
3. ∀a∈G ∃a⁻∈G : a+a⁻=e

.I'm going to try to prove that if a is an element of G and its corresponding opposite is -a
.Then -(-a)=a (ie, the opposite of the opposite, leaves unaltered the element)
.Indeed, by definition
a+(-a)=0
.Therefore it is evident that the corresponding opposite of -a is a (Since the sum of them results in zero)

.Therefore, we can write -(-a)=a

.Then
a-(-b)=a+[-(-b)]=a+b
Q.E.D?

---------------------------------------------------
if you want, I can write complete axioms and deductions.
-----------------------------------------------------
Tyler.Smith
#5
Nov19-12, 07:48 PM
P: 6
Quote Quote by HallsofIvy View Post
As haruspex says, how you would prove this depends upon what axioms you have to use. If this is a general group, you are asking to prove that "the additive inverse of the additive inverse of a is a".

b is the "additive inverse of a" if and only if a+ b= 0 and b+ a= 0. c is the additive inverse of b if and only if b+ c= c+ b= 0. Play around with those equation to arrive at "a= c". You might need the easily proved result that 0 is its own additive inverse.

But you use the term "number: so perhaps this is to be for a specific set of numbers. In that case, how you would prove this depends upon which set and which axioms you are using. For example are you to use Peano's axioms for the integers?
Hello, I followed your advice, and use some axioms and deductions that had already established. I do not know if it is correct



Demostracion
I have a group G, which has an internal composition law ∘, satisfies the following axioms.
1. a+(b+c)=(a+b)+c, ∀a,b,c∈G
2. ∃e∈G : e+a=a-e=a
3. ∀a∈G ∃a⁻∈G : a+a⁻=e

.I'm going to try to prove that if a is an element of G and its corresponding opposite is -a
.Then -(-a)=a (ie, the opposite of the opposite, leaves unaltered the element)
.Indeed, by definition
a+(-a)=0
.Therefore it is evident that the corresponding opposite of -a is a (Since the sum of them results in zero)

.Therefore, we can write -(-a)=a

.Then
a-(-b)=a+[-(-b)]=a+b
Q.E.D?

---------------------------------------------------
if you want, I can write complete axioms and deductions.
-----------------------------------------------------
Dick
#6
Nov19-12, 09:36 PM
Sci Advisor
HW Helper
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P: 25,228
Quote Quote by Tyler.Smith View Post
Hello, I followed your advice, and use some axioms and deductions that had already established. I do not know if it is correct



Demostracion
I have a group G, which has an internal composition law ∘, satisfies the following axioms.
1. a+(b+c)=(a+b)+c, ∀a,b,c∈G
2. ∃e∈G : e+a=a-e=a
3. ∀a∈G ∃a⁻∈G : a+a⁻=e

.I'm going to try to prove that if a is an element of G and its corresponding opposite is -a
.Then -(-a)=a (ie, the opposite of the opposite, leaves unaltered the element)
.Indeed, by definition
a+(-a)=0
.Therefore it is evident that the corresponding opposite of -a is a (Since the sum of them results in zero)

.Therefore, we can write -(-a)=a

.Then
a-(-b)=a+[-(-b)]=a+b
Q.E.D?

---------------------------------------------------
if you want, I can write complete axioms and deductions.
-----------------------------------------------------
Yes. Your axiom 3 is is using multiplicative notation instead additive. I'd write it as ∀a∈G ∃(-a)∈G : a+(-a)=0. And for 2 substitute 0 for e. So sure, since a+(-a)=0, an inverse -(-a) of (-a) is a. That's almost Q.E.D. If you want to go whole hog on this your axioms don't explicitly state that the identity or inverses are unique. I'm not sure if you are expected to prove that or not.
Tyler.Smith
#7
Nov21-12, 10:48 AM
P: 6
Quote Quote by Dick View Post
Yes. Your axiom 3 is is using multiplicative notation instead additive. I'd write it as ∀a∈G ∃(-a)∈G : a+(-a)=0. And for 2 substitute 0 for e. So sure, since a+(-a)=0, an inverse -(-a) of (-a) is a. That's almost Q.E.D. If you want to go whole hog on this your axioms don't explicitly state that the identity or inverses are unique. I'm not sure if you are expected to prove that or not.
Hello, I made the corrections you mentioned, and based on the theory of abelian group, I try to prove that the identity element and inverse element are unique. I think that maybe i got it.
-------------------------------------------------------------------------------------
I have a group G, which has an internal composition law ∘, satisfies the following Axioms.
A1 a+(b+c)=(a+b)+c, ∀a,b,c∈G
A2 ∃0∈G : 0+a=a+0=a
A3 ∀a∈G, ∃(-a)∈G : a+(-a)=(-a)+a=0

..Theorem 1 - Identity element, in G, is unique.
Proof: If 0 and f are two identity elements of G. Then:
0=0+f (A2)
0=f (A2)

..Theorem 2 - Inverse element, in G, are unique
Proof: If (-a) and (-a)′ are two inverses of an element a of G. Then:
(-a)=(-a)+0 (A2)
(-a)=(-a)+[a+(-a)′] (A3)
(-a)=[(-a)+a]+(-a)′ (A1)
(-a)=0+(-a)′ (A3)
(-a)=(-a)′ (A2)
--------------------------------------------
Now i'm going to try to prove that if a is an element of G and its corresponding opposite is (-a)
Then: -(-a)=a (ie, the opposite of the opposite, leaves unaltered the element)
.Indeed, by definition
a+(-a)=0
.Therefore it is evident that the corresponding opposite of (-a) is a (Since the sum of them results in zero)

Therefore, we can write -(-a)=a
.Then
a-(-b)=a+[-(-b)]=a+b
Q.E.D?
Dick
#8
Nov21-12, 11:11 AM
Sci Advisor
HW Helper
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P: 25,228
Quote Quote by Tyler.Smith View Post
Hello, I made the corrections you mentioned, and based on the theory of abelian group, I try to prove that the identity element and inverse element are unique. I think that maybe i got it.
-------------------------------------------------------------------------------------
I have a group G, which has an internal composition law ∘, satisfies the following Axioms.
A1 a+(b+c)=(a+b)+c, ∀a,b,c∈G
A2 ∃0∈G : 0+a=a+0=a
A3 ∀a∈G, ∃(-a)∈G : a+(-a)=(-a)+a=0

..Theorem 1 - Identity element, in G, is unique.
Proof: If 0 and f are two identity elements of G. Then:
0=0+f (A2)
0=f (A2)

..Theorem 2 - Inverse element, in G, are unique
Proof: If (-a) and (-a)′ are two inverses of an element a of G. Then:
(-a)=(-a)+0 (A2)
(-a)=(-a)+[a+(-a)′] (A3)
(-a)=[(-a)+a]+(-a)′ (A1)
(-a)=0+(-a)′ (A3)
(-a)=(-a)′ (A2)
--------------------------------------------
Now i'm going to try to prove that if a is an element of G and its corresponding opposite is (-a)
Then: -(-a)=a (ie, the opposite of the opposite, leaves unaltered the element)
.Indeed, by definition
a+(-a)=0
.Therefore it is evident that the corresponding opposite of (-a) is a (Since the sum of them results in zero)

Therefore, we can write -(-a)=a
.Then
a-(-b)=a+[-(-b)]=a+b
Q.E.D?
Q.E.D. Looks fine to me. BTW you didn't use that the group is abelian. I think you meant 'additive group'.
Tyler.Smith
#9
Nov21-12, 11:48 AM
P: 6
Quote Quote by Dick View Post
Q.E.D. Looks fine to me. BTW you didn't use that the group is abelian. I think you meant 'additive group'.
Finally -"Q.E.D"-. Thanks for the feedback & help!


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