a*0 = 0 in a ring proof

[Greglar]

I fully understand the proof but have a further question on it.

In this version of the proof:

a*0 = a*(0+0) = a*0 + a*0

so a*0 = a*0 + a*0

then a*0 - a*0 = a*0 + a*0 - a*0

therefore 0 = a*0

My question is: why must we cancel an a*0 on both sides?

Isn't the equation a*0 = a*0 + a*0 sufficient?

I mean, let x = a*0

then x = x + x

In a group under addition, isn't the additive identity the only element that could satisfy the equation x = x + x?

My instructor said that it's because x = x + x only proves that x acts as the additive identity for itself, but you would have to prove that x + a = a for all a in the group for x to be the additive identity.

I didn't like that explanation because I feel like I'm using different reasoning. I'm curious as to what, besides 0, x could be in the equation x = x + x in an additive group. Also, a concrete counter example would be preferred to other forms of explanations.

 

[Ishuda]

I fully understand the proof but have a further question on it.

In this version of the proof:

a*0 = a*(0+0) = a*0 + a*0

so a*0 = a*0 + a*0

then a*0 - a*0 = a*0 + a*0 - a*0

therefore 0 = a*0

My question is: why must we cancel an a*0 on both sides?

Isn't the equation a*0 = a*0 + a*0 sufficient?

I mean, let x = a*0

then x = x + x

In a group under addition, isn't the additive identity the only element that could satisfy the equation x = x + x?

My instructor said that it's because x = x + x only proves that x acts as the additive identity for itself, but you would have to prove that x + a = a for all a in the group for x to be the additive identity.

I didn't like that explanation because I feel like I'm using different reasoning. I'm curious as to what, besides 0, x could be in the equation x = x + x in an additive group. Also, a concrete counter example would be preferred to other forms of explanations.

I'm confused. The symbol '0' is, by definition, the additive identity. So, what is there to prove?

We have that the most basic set with one operation is a Group [a set with an operator satisfying some conditions]. Examples generally start with something like the integers with + being the operation. One of the properties is that there may exit an identity element '0' such that '0'+x=x+'0'=x for all x belonging to the group. The concept of 'zero divisors' are actually presented here but not discussed.

After that comes a Ring which adds another operator satisfying the properties of a Group and some another operator
[*] satisfying certain properties [about the same as the 'addition operator' but defined differently, i.e. multiplication]. Examples generally start with something like the integers and the second operation may have an identity element '1' such that '1'*x=x*'1'=x for all x in the Ring. The concept of 'zero divisors' are discussed in (more) detail here.

Next is a Field which is an Ring with certain additional quantities.

 

[pka]

I fully understand the proof but have a further question on it.

In this version of the proof:

a*0 = a*(0+0) = a*0 + a*0

so a*0 = a*0 + a*0

then a*0 - a*0 = a*0 + a*0 - a*0

therefore 0 = a*0

My question is: why must we cancel an a*0 on both sides?

Isn't the equation a*0 = a*0 + a*0 sufficient?

I mean, let x = a*0

then x = x + x

In a group under addition, isn't the additive identity the only element that could satisfy the equation x = x + x?

My instructor said that it's because x = x + x only proves that x acts as the additive identity for itself, but you would have to prove that x + a = a for all a in the group for x to be the additive identity.

I didn't like that explanation because I feel like I'm using different reasoning. I'm curious as to what, besides 0, x could be in the equation x = x + x in an additive group. Also, a concrete counter example would be preferred to other forms of explanations.

A Ring is a set of elements with two operations defined: an addition,\(\displaystyle +\) , and a multiplication, \(\displaystyle \cdot~\).
Here is a list of the eight axioms for a Ring.
1: For each a and b in the ring, a+b is in the ring.
2: For each a and b in the ring, a+b= b+a
3: For each a, b and c in the ring, (a+b)+c=a+(b+c).
4: There is an element 0 in the ring such that a+0=a, for all a in the ring.
5: For each a in the ring, there is an element, -a, in the ring and a+(-a)=0.
6: For each a and b in the ring, ab is in the ring.
7: For each a, b and c in the ring, (ab)c =a(bc).
8: For each a, b and c in the ring, a(b+c)= ab+ ac

Theorem1: For each a in the ring, a0 = 0 = 0a, note \(\displaystyle ~\cdot~\) that need not be commutative.
\(\displaystyle \begin{array}{*{20}{lcll}} 0& = &{a \cdot 0 + ( - [a \cdot 0])}&{\text{by }5} \\ {}& = &{a \cdot [0 + 0] + ( - [a \cdot 0])}&{\text{by }4} \\ {}& = &{[a \cdot 0 + a \cdot 0] + ( - [a \cdot 0])}&{\text{by }8} \\
{}& = &{a \cdot 0 + [a \cdot 0 + ( - [a \cdot 0])]}&{\text{by }3} \\ {}& = &{a \cdot 0 + 0}&{\text{by }5} \\ {}& = &{a \cdot 0}&{\text{by }4}\end{array}\)

Now note that only proves that \(\displaystyle a\cdot 0=0\) it is yet to prove that \(\displaystyle 0\cdot a=0\) because because rings are not necessarily commutative.