1. ## Using Field Axioms

I don't know if this is where this type of question should be, so sorry if this isn't it.

I need to show the following using the Field Axioms:

1. That the identity numbers 0 and 1 are unique.
2. That 0*a=0.
3. That the inverses -a and 1/a are unique.
4. That -1*a = -a
5. That (-1)(-1)=1
6. That if $\displaystyle a_3\neq 0$, then
$\displaystyle \frac{a_1+a_2}{a_3}=\frac{a_1}{a_3}+\frac{a_2}{a_3 }$

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Could the first be demonstrated like this?
1. Show that the identity numbers 0 and 1 are unique.

Suppose there is a number $\displaystyle 0'$, an additive identity, different from zero such that $\displaystyle 0'+a=a$. Then $\displaystyle 0'+a-a=a-a \Longleftrightarrow 0'+0=0$ by the definition of additive inverse and $\displaystyle 0'+0=0 \Longleftrightarrow 0'=0$ by the definition of additive identity which is clearly a contradiction. Therefore, the additive identity$\displaystyle 0$ is unique.

Suppose there is a number $\displaystyle 1'$, a multiplicative identity, different from one such that $\displaystyle 1'\cdot a = a$. Then $\displaystyle 1'\cdot a = a \Longleftrightarrow 1'\cdot a \cdot \frac{1}{a}$ by the definition of multiplicative inverse and $\displaystyle 1'\cdot 1 = 1 \Longleftrightarrow 1'=1$ by the definition of multiplicative identity which is a contradiction. Therefore, the multiplicative identity $\displaystyle 1$ is unique.

2. Basically a Field $\displaystyle F$ has $\displaystyle <F,+>$ an abelian group and $\displaystyle <F,\cdot >$ and abelian group. So it is necessary and sufficient to prove the following.

Theorem: In a Group $\displaystyle G$ the identity elemenet is unique.

Proof: Let $\displaystyle e$ be this identity element. And $\displaystyle e'$ another identity element. Then $\displaystyle e=ee'=e'$

3. Perfect! Thanks

4. Note, this does not show that $\displaystyle 1\not = 0$.

For example, consider $\displaystyle R=\{a\}$ defined as $\displaystyle a+a=a$ and $\displaystyle aa=a$. Then we see that $\displaystyle a=0$ and $\displaystyle a=1$. But this example is reffered to as trivial ring. And we do not consider it to be a field by convention, just to avoid this boring counterexample.

5. I simply want to get a feel for how to demonstrate something step by step. Here's what I've done so far but I'm having trouble justifying the step with the ***** in problem 2. I have no idea how to start with problem 6. Please look over them if possible and correct any mistakes.

2) Show that $\displaystyle 0\cdot a=0$.

$\displaystyle (0+\frac{b}{a})\cdot a = 0\cdot a + b$ by distributive law of multiplication over addition and multiplicative inverse
$\displaystyle 0\cdot a + b = b$ ****
$\displaystyle 0\cdot a + b - b = b - b \Longleftrightarrow 0\cdot a + 0 = 0$ by definition of additive inverse
$\displaystyle 0\cdot a = 0$ by definition of additive identity

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3) Show that in a group $\displaystyle G$, the inverse of each element $\displaystyle a$ is unique.

Let $\displaystyle G$ be a group with binary operation * and identity $\displaystyle e$.

Suppose there is some element $\displaystyle a$ with two inverses denoted by $\displaystyle a^{-1}$ and $\displaystyle a'$. Then

$\displaystyle (a^{-1} * a) * a' = a^{-1} * (a * a')$ by the associative law
$\displaystyle e*a'=a^{-1}*e$ by definition of inverse
$\displaystyle a'=a^{-1}$ by definition of identity

Therefore, the inverse of each element in a group is unique.

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4) Show that $\displaystyle -1\cdot a = -a$.

$\displaystyle 0=(1-1)\cdot a \Longleftrightarrow 0 = (1\cdot a) + (-1\cdot a)$ by distributive law
$\displaystyle 0=a +(-1\cdot a)$ by multiplicative identity
$\displaystyle 0-a=a-a+(-1\cdot a) \Longleftrightarrow -a=0+(-1\cdot a)$ by additive inverse
$\displaystyle -a=-1\cdot a$ by additive identity

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5) Show that $\displaystyle (-1)(-1)=1$.

$\displaystyle 0=(1-1)(-1) \Longleftrightarrow 0=-1+(-1)(-1)$ by distributive law and multiplicative identity
$\displaystyle 0+1=-1+1+(-1)(-1) \Longleftrightarrow 1=(-1)(-1)$ by additive inverse

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Does anyone know the latex for an actual negative that's shorter and higher than the minus sign and for the Real number sign?

6. Originally Posted by rualin
2) Show that $\displaystyle 0\cdot a=0$.
$\displaystyle 0a=(1-1)a=a-a=0$

3) Show that in a group $\displaystyle G$, the inverse of each element $\displaystyle a$ is unique.
Say $\displaystyle a$ has $\displaystyle b,c$ as inverses.
So $\displaystyle ab=e=ac$. Thus, $\displaystyle ab=ac$, use the left-cancellation law to get $\displaystyle b=c$.

4) Show that $\displaystyle -1\cdot a = -a$.
$\displaystyle -a$ means the additive inverse of $\displaystyle a$. So show $\displaystyle (-1)a+a=0$. By distributive law.

5) Show that $\displaystyle (-1)(-1)=1$.
Try this yourself.

7. Originally Posted by ThePerfectHacker

4) Show that $\displaystyle -1\cdot a=-a$.

$\displaystyle -a$ means the additive inverse of $\displaystyle a$. So show $\displaystyle (-1)a+a=0$. By distributive law.
How did you arrive at $\displaystyle (-1)a+a=0$?

Originally Posted by ThePerfectHacker
5) Show that $\displaystyle (-1)(-1)=1$.

Try this yourself.
$\displaystyle (-1)(-1)+(-1)=0 \Longleftrightarrow (-1)(-1)+(-1)+1=1 \Longleftrightarrow (-1)(-1)=1$

8. Originally Posted by rualin
How did you arrive at $\displaystyle (-1)a+a=0$?
Because,
$\displaystyle 0a=(-1+1)a=(-1)a+a$

$\displaystyle (-1)(-1)+(-1)=0 \Longleftrightarrow (-1)(-1)+(-1)+1=1 \Longleftrightarrow (-1)(-1)=1$
Correct! You just need to justify,
$\displaystyle (-1)(-1)+(-1)=0$
Which is the distributive law,
$\displaystyle (-1)[(-1)+1]=(-1)(0)=0$

9. Here is a PDF file that I use with students on your question.
I hope it helps.

10. Nice! Thanks, Plato.

I still am not able to follow the Double inverse theorem... maybe you could help make it clearer.
Let $\displaystyle *$ be an associative binary operation on a set $\displaystyle A$ with identity $\displaystyle e$, and let $\displaystyle x\in A$. If $\displaystyle x$ is invertible for $\displaystyle *$, let [tex]x^{-1}[/math denote the (unique) inverse for $\displaystyle x$. Then $\displaystyle x^{-1}$ is invertible and $\displaystyle (x^{-1})^{-1} = x$.

Proof: If $\displaystyle y$ is the inverse for $\displaystyle x$, then
$\displaystyle y*x=e=x*y$.

But this is exactly the condition for $\displaystyle x$ to be the inverse for $\displaystyle y$.
What confuses me about that proof is the last statement. I don't understand how that is sufficient.