# Prove the given properties - Ring Theory

• I
• chwala
In summary: For ##5##,We want to prove,## (-a)(b-c)=(-a)b-(-a)c \\∀ a,b,c ∈ \mathbb{R}##Using distributive property on lhs,##(-a)(b-c)=(-a)b-(-a)c+(-a)(-c)####=(-a)b-(-a)c+(--(ac))=(-a)b-(-a)c+(ac)####=(-a)b-(-a)c+ac=(-a)b+(-a)(-c)####=(-a)b+(-a)(-c)=(-a)b-(-a)c##therefore ##(-a)(b-c)=(-a)b-(-a)c##In
chwala
Gold Member
TL;DR Summary
I am looking at the basics here...Consider the attachment below. My intention is to try and prove ##2.11## from ##2## to ##4## and i would appreciate any constructive feedback.

Ok for ##1##, we also have,
##a⋅0=a⋅(0+0)=a⋅0 + a⋅0 ## We know that ##a⋅0=0 ## by additive cancellation.

For ##2.11##, Number ##2##;
We first show and prove that
##-b=-1⋅b##
##-b+b=0## for the lhs
##-1⋅b +1⋅b=b(-1+1)=b(0)=0## for the rhs
therefore,
##(-a)b=(-1⋅a)b=-1(ab)=-(ab)## using distributive property...
also,
##a⋅(-b)=a⋅-1⋅b=-1⋅(a⋅b)=-1(ab)=-(ab)##
Will look at (3) and (4) later...

For number 2) you need to show that, for example:
$$(-a) b + a b = a b + (-a) b = 0$$Hence, by definition:$$(-a)b = -(ab)$$ is the additive inverse of ##ab##.

chwala
So in general, we are just checking as to whether the ring additive axioms are satisfied, right?...I can see that you have used commutative property and the additive inverse property to show this...in my case I could say that I made use of the distributive property...
In showing these proofs, there isn't any general rule right? The way is to check how the problem fits/satisfies the ring property axioms...

I also read on the Field which basically has more axioms satisfied ( ie For Real, rational and complex numbers)as compared to the Ring. (Integers)...

The problem with what you did was that you didn't fully recognise that ##-(ab)## is, by notational definition, the additive inverse of ##ab##. And that we have to show that ##(-a)b##, which is the additive inverse of ##a## multiplied by ##b##, is also the additive inverse of ##ab##.

PeroK said:
The problem with what you did was that you didn't fully recognise that ##-(ab)## is, by notational definition, the additive inverse of ##ab##. And that we have to show that ##(-a)b##, which is the additive inverse of ##a## multiplied by ##b##, is also the additive inverse of ##ab##.
Thanks, yes that's true...I can see that now.

@PeroK hi, by the way could i also be right in using,
if ##a=b## and ##b=c## then it follows that, ##a=c##
##a-b=0## and ##b-c=0##, ##⇒a-b+b-c=0##

therefore in our problem we shall have,
##(-a)b=-(ab)##
##(-a)b- - (ab)=0##

also,
##-(ab)=a(-b)##
##-(ab)-a(-b)=0##,

##(-a)b- -(ab)+-(ab)-a(-b)=0##
##(-a)b+(ab)-(ab)-a(-b)=0##
##(-a)b-a(-b)=0##
##(-a)b=a(-b)## thus completing our proof.

chwala said:
therefore in our problem we shall have,
##(-a)b=-(ab)##
##(-a)b- - (ab)=0##
You can't start with what you are asked to prove. Try starting with $$(-a)b + ab = \dots$$

PeroK said:
You can't start with what you are asked to prove. Try starting with $$(-a)b + ab = \dots$$
ok, i may have, ##(-a)b+ab=b(-a+a)=b⋅0=0##
also, ##-(ab)+ab=0##
also,##(a)-b+ab=a(-b+b)=a⋅0=0## therefore ##(-a)b=-(ab)=(a)-b##

Last edited:
chwala said:
ok, i may have, ##(-a)b+ab=b(-a+a)=b⋅0=0##
A ring is not necessarily commutative, so you can't swap the order like that:$$(-a)b+ab=(-a+a)b=(0)b=0$$
chwala said:
also, ##-(ab)+ab=0## ##⇒(-a)b=-(ab)##
That's not quite right either. $$(-a)b+ab = 0 \ \Rightarrow \ (-a)b = -(ab)$$
chwala said:
also,##(a)-b+ab=a(-b+b)=a⋅0=0## therefore our proof is complete.
Okay, but ##a(-b)## is better than ##(a)-b##

chwala
PeroK said:
A ring is not necessarily commutative, so you can't swap the order like that:$$(-a)b+ab=(-a+a)b=(0)b=0$$

That's not quite right either. $$(-a)b+ab = 0 \ \Rightarrow \ (-a)b = -(ab)$$

Okay, but ##a(-b)## is better than ##(a)-b##
cheers gd day...

chwala said:
cheers gd day...
It is abstract algebra; not slapdash algebra!

chwala
PeroK said:
A ring is not necessarily commutative, so you can't swap the order like that:$$(-a)b+ab=(-a+a)b=(0)b=0$$

That's not quite right either. $$(-a)b+ab = 0 \ \Rightarrow \ (-a)b = -(ab)$$

Okay, but ##a(-b)## is better than ##(a)-b##
noted on this; $$(-a)b+ab=(-a+a)b=(0)b=0$$ thanks...i will read more on that...i had thought that for a set to be considered a ring, then it has to satisfy commutativity...which you are saying it is not always the case...

chwala said:
noted on this; $$(-a)b+ab=(-a+a)b=(0)b=0$$ thanks...i will read more on that...i had thought that for a set to be considered a ring, then it has to satisfy commutativity...which you are saying it is not always the case...
Ring multiplication is not necessarily commutative. Ring addition is commutative, of course.

chwala
For ##3##,
We want to prove,
##(-a)(-b) =ab \\∀ a,b ∈ \mathbb{R}##

Using distributive property and using ##-(ab)## on lhs;
##(-a)(-b) -(ab)=-(a)(-b)-(ab)=[--(ab)]-(ab)=(ab)-(ab)=0##

Also using ##-(ab)## on rhs;
##ab-(ab)=ab - ab=0##,

therefore,
##(-a)(-b) =ab ∀ a, b ∈ \mathbb{R}##

Last edited:
For ##4##,
We want to prove,
##(-1)(a) =-a \\∀ a ∈ \mathbb{R}##

For lhs, using distributive property;
##(-1)a + 1a=(-1+1)a=(0)a=0##
also for rhs, ##-a=-(1a)##
##⇒(-1)a- -a=(-1)a+a=(-1)a+1a=0##
thus ##(-1)a=-a##

## 1. What is Ring Theory?

Ring Theory is a branch of abstract algebra that studies the properties and structures of rings, which are mathematical objects consisting of a set of elements and two binary operations, usually addition and multiplication.

## 2. What are the basic properties of a ring?

The basic properties of a ring include closure under addition and multiplication, associativity of addition and multiplication, distributivity of multiplication over addition, and the existence of an additive identity and additive inverse for each element.

## 3. How do you prove that a set is a ring?

To prove that a set is a ring, you must show that it satisfies all of the basic properties of a ring. This can be done by verifying that the set is closed under the operations of addition and multiplication, the operations are associative and distributive, and the identity and inverse elements exist for each element in the set.

## 4. What is the difference between a commutative ring and a non-commutative ring?

A commutative ring is one in which the operation of multiplication is commutative, meaning that a*b = b*a for all elements a and b in the ring. In a non-commutative ring, the operation of multiplication is not commutative, and a*b may not be equal to b*a.

## 5. How does Ring Theory relate to other areas of mathematics?

Ring Theory has many applications in other areas of mathematics, such as number theory, algebraic geometry, and representation theory. It also serves as a foundation for other algebraic structures, such as fields, modules, and algebras.

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