# Field and Sets Proof

Need some clarification on this one.

Question:

Is the set [0, infinity) with the usual addition and multiplication axioms, a field?

Attempted solution:

Yes it is a field.

let 0 <= r, s < infinity

r + s = s + r

s + r = r + s

and by commutative multiplication:

rs = sr

sr = rs

This is pretty much restating the axioms so I'm unclear as to whether this is the proof or maybe I'm completely wrong. Any help is appreciated thanks.

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Have checked that it is a group with respect to addition? – Mariano Suárez-Alvarez Jan 30 '11 at 22:56

You forgot a lot of axioms!

A field is a set $F$ together with two binary operations, $\oplus$ and $\odot$, such that:

1. $\oplus$ is associative: for every $r,s,t\in F$, $r\oplus(s\oplus t)=(r\oplus s)\oplus t$.
2. $\oplus$ is commutative: for every $r,s\in F$, $r\oplus s=s\oplus r$.
3. There exists an element $z\in F$ such that for all $x\in F$, $z\oplus x=x$ (additive identity).
4. For every $x\in F$ there exists $y\in F$ such that $x\oplus y = z$ (every element has an additive inverse).
5. $\odot$ is associative: for every $r,s,t\in F$, $r\odot(s\odot t) = (r\odot s)\odot t$.
6. $\odot$ is commutative: for every $r,s\in F$, $r\odot s=s\odot r$.
7. $\odot$ distributes over $\oplus$: for every $r,s,t\in F$, $r\odot(s\oplus t) = (r\odot s)\oplus(r\odot t)$.
8. There exists an element $e\in F$, $e\neq z$ ($z$ is the element from item 3) such that $\forall r\in F_4, e\cdot r=r$ (multiplicative identity).
9. $\forall r\in F$, if $r\neq z$ ($z$ the element from item 3) there exists $s\in F$ such that $r\odot s = e$ ($e$ the element from item 8) (multiplicative inverses for all nonzero elements).

You only checked (2) and (6) (twice each, for some reason). So, what are you still missing?

First, you would need to show that addition and multiplication is associative (yes, both are, so this is not a problem).

Then you need to show that there is an identity element for the sum (yes, there is one in your set: namely, $0$, but you need to show it).

Then you need to show that every element in your set has an additive inverse: for every $r$, there exists $s$ such that $r+s = 0$. (This might prove difficult here).

Then you need to show that multiplication distributes over the sum: that for all $r,s,t$ in you set, $r(s+t) = rs + rt$ (this holds in this case).

Then you need to show that there is an element different from the additive identity, which is a multiplicative identity: that is, an element $e$ such that $er = r$ for all $r$ (there is one, but you need to show it).

And you also need to show that every element other than the additive identity has a multiplicative inverse (for every $r$, $r\neq 0$, there exists $s$ such that $rs=e$, where $e$ is the multiplicative identity from the previous part). (This is not a problem here either).

So, you are missing quite a bit there!

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There are one or two numbers that do not have additive inverses.

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+1 for the modesty of your claim... – Georges Elencwajg Jan 30 '11 at 23:22