Using Field Axioms

**rualin** · July 6th 2007, 07:07 AM

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 $a_3\neq 0$ , then
$\frac{a_1+a_2}{a_3}=\frac{a_1}{a_3}+\frac{a_2}{a_3 }$

-----------

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

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

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

**ThePerfectHacker** · July 6th 2007, 07:59 AM

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

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

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

**rualin** · July 6th 2007, 08:04 AM

Perfect! Thanks

**ThePerfectHacker** · July 6th 2007, 08:52 AM

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

For example, consider $R=\{a\}$ defined as $a+a=a$ and $aa=a$ . Then we see that $a=0$ and $a=1$ . 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.

**rualin** · July 6th 2007, 09:09 AM

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 $0\cdot a=0$ .

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

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

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

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

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

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

---

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

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

---

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

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

---

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

**ThePerfectHacker** · July 6th 2007, 09:44 AM

Originally Posted by rualin

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

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

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

Say $a$ has $b,c$ as inverses.
So $ab=e=ac$ . Thus, $ab=ac$ , use the left-cancellation law to get $b=c$ .

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

$-a$ means the additive inverse of $a$ . So show $(-1)a+a=0$ . By distributive law.

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

Try this yourself.

**rualin** · July 6th 2007, 10:04 AM

Originally Posted by ThePerfectHacker

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

$-a$ means the additive inverse of $a$ . So show $(-1)a+a=0$ . By distributive law.

How did you arrive at $(-1)a+a=0$ ?

Originally Posted by ThePerfectHacker

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

Try this yourself.

$(-1)(-1)+(-1)=0 \Longleftrightarrow (-1)(-1)+(-1)+1=1 \Longleftrightarrow (-1)(-1)=1$

**ThePerfectHacker** · July 6th 2007, 12:18 PM

Originally Posted by rualin

How did you arrive at $(-1)a+a=0$ ?

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

$(-1)(-1)+(-1)=0 \Longleftrightarrow (-1)(-1)+(-1)+1=1 \Longleftrightarrow (-1)(-1)=1$

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

**Plato** · July 6th 2007, 04:20 PM

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

**rualin** · July 6th 2007, 05:14 PM

Nice! Thanks, Plato.

I still am not able to follow the Double inverse theorem... maybe you could help make it clearer.

Let $*$ be an associative binary operation on a set $A$ with identity $e$ , and let $x\in A$ . If $x$ is invertible for $*$ , let [tex]x^{-1}[/math denote the (unique) inverse for $x$ . Then $x^{-1}$ is invertible and $(x^{-1})^{-1} = x$ .

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

But this is exactly the condition for $x$ to be the inverse for $y$ .

What confuses me about that proof is the last statement. I don't understand how that is sufficient.

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