Basically a Field has an abelian group and and abelian group. So it is necessary and sufficient to prove the following.
Theorem: In a Group the identity elemenet is unique.
Proof: Let be this identity element. And another identity element. Then
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 , then
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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 , an additive identity, different from zero such that . Then by the definition of additive inverse and by the definition of additive identity which is clearly a contradiction. Therefore, the additive identity is unique.
Suppose there is a number , a multiplicative identity, different from one such that . Then by the definition of multiplicative inverse and by the definition of multiplicative identity which is a contradiction. Therefore, the multiplicative identity is unique.
Note, this does not show that .
For example, consider defined as and . Then we see that and . 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.
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 .
by distributive law of multiplication over addition and multiplicative inverse
****
by definition of additive inverse
by definition of additive identity
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3) Show that in a group , the inverse of each element is unique.
Let be a group with binary operation * and identity .
Suppose there is some element with two inverses denoted by and . Then
by the associative law
by definition of inverse
by definition of identity
Therefore, the inverse of each element in a group is unique.
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4) Show that .
by distributive law
by multiplicative identity
by additive inverse
by additive identity
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5) Show that .
by distributive law and multiplicative identity
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?
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 with identity , and let . If is invertible for , let [tex]x^{-1}[/math denote the (unique) inverse for . Then is invertible and .What confuses me about that proof is the last statement. I don't understand how that is sufficient.
Proof: If is the inverse for , then
.
But this is exactly the condition for to be the inverse for .