# Using Field Axioms

• Jul 6th 2007, 06:07 AM
rualin
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.
• Jul 6th 2007, 06:59 AM
ThePerfectHacker
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 $\displaystyle e=ee'=e'$
• Jul 6th 2007, 07:04 AM
rualin
Perfect! Thanks
• Jul 6th 2007, 07:52 AM
ThePerfectHacker
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.
• Jul 6th 2007, 08:09 AM
rualin
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?
• Jul 6th 2007, 08:44 AM
ThePerfectHacker
Quote:

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

$\displaystyle 0a=(1-1)a=a-a=0$

Quote:

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 $\displaystyle b=c$.

Quote:

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.

Quote:

5) Show that $\displaystyle (-1)(-1)=1$.
Try this yourself.
• Jul 6th 2007, 09:04 AM
rualin
Quote:

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 $\displaystyle (-1)a+a=0$?

Quote:

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

Try this yourself.

$\displaystyle (-1)(-1)+(-1)=0 \Longleftrightarrow (-1)(-1)+(-1)+1=1 \Longleftrightarrow (-1)(-1)=1$
• Jul 6th 2007, 11:18 AM
ThePerfectHacker
Quote:

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

Because,
$\displaystyle 0a=(-1+1)a=(-1)a+a$

Quote:

$\displaystyle (-1)(-1)+(-1)=0 \Longleftrightarrow (-1)(-1)+(-1)+1=1 \Longleftrightarrow (-1)(-1)=1$
Correct! You just need to justify,
$\displaystyle (-1)(-1)+(-1)=0$
Which is the distributive law,
$\displaystyle (-1)[(-1)+1]=(-1)(0)=0$
• Jul 6th 2007, 03:20 PM
Plato
Here is a PDF file that I use with students on your question.
I hope it helps.
• Jul 6th 2007, 04:14 PM
rualin
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.