Difference between revisions of "Ring"

This is a classic "suppose there are two" proof, and we will do the same.

Suppose that $0\in R$ is such that $\forall x\in R[0+x=x+0=x]$

Suppose that $0'\in R$ with $0'\ne 0$ and also such that: $\forall x\in R[0'+x=x+0'=x]$

We will show that $0=0'$, contradicting them being different! Thus showing there is no other "zero"

Proof:

$$0+0'=0$$ by the property of $0$

$$0+0'=0'+0$$ by the commutivity of addition

$$0'+0=0'$$ by the property of $0'$

Thus $$0=0'$$

This contradicts that $0\ne 0'$ so the claim they are distinct cannot be, we have only one "zero element", which herein we shall denote as "$0$"

Suppose that $a+c=b+c$

By the additive inverse property, $$\exists x\in R:c+x=0$$

First notice that $$(a+c)+x=(b+c)+x$$ (using $$a+c=b+c$$ )

• Let us take $$(a+c)+x$$

  By associativity of addition, $$(a+c)+x=a+(c+x)=a+0=a$$

• Let us take $$(b+c)+x$$

  By associativity of addition, $$(b+c)+x=b+(c+x)=b+0=b$$

We see that $$a=a+c+x=b+c+x=b$$

Which is indeed just $$a=b$$

As claimed.

Note:

Note that $$c+a=b+c\implies a=b$$ , this can be proved identically to the above (but adding x to the left) or by:

$$c+a=a+c$$ and </math>b+c=c+b</math> and then apply the above.

TODO:

Let $x\in R$ be given.

Proof of: $x0=0$

Note that $x=x+0$ then

$xx=x(x+0)=xx+x0$ by distributivity

Note that $xx=xx+0$ then

$xx+0=xx+x0$

By the cancellation laws: $\implies 0=x0$

So we have shown $\forall x\in R[x0=0]$

Proof of: $0x=0$

Note that $x=x+0$ then

$xx=(x+0)x=xx+0x$ by distributivity

Note that $xx=xx+0$ then

$xx+0=xx+0x$

By the cancellation laws: $\implies 0=0x$

So we have shown $\forall x\in R[0x=0]$

So $$\forall x\in R[0x=0\wedge x0=0]$$ or simply $$\forall x\in R[0x=x0=0]$$

This completes the proof.