1. ## Field

Tell whether or not $\mathbb{Q}$ with the two operations ⊕ and $\cdot$ (in fact I want to type the $\cdot$ but with a circle as in ⊕ but I don't know how to) is a field.
Looking to my class notes, a field $\mathbb{K}$ is a set with 2 operations " $+$" and " $\cdot$" such that
1) $x+y=y+x$
2) $(x+y)+z=x+(y+z)$
3) $\exists{!}$ element " $0$" such that $x+"0"=x$
4) $\forall x\in \mathbb{K}$, $\exists{!} "-x"$ such that $x+(-x)=0$
5) $x\cdot y=y\cdot x$
6) $(x\cdot y)\cdot z=x\cdot (y\cdot z)$
7) $\exists{!}$ element $1$ such that $x\cdot 1=x$
8) $\forall x\neq 0$, $\exists{!} x^{-1}$ such that $x\cdot x^{-1}=1$
9) $x\cdot (y+z)=x\cdot y+x\cdot z$.

So to solve my problem, I have to check if any $x$ and $y$ elements of $\mathbb{Q}$ respect the 9 rules. But I'm not able to start. How can I check if any element $x$ and $y$ of $\mathbb{Q}$ are such that $x+y=y+x$? I thought about writing down $\frac{1}{a}+\frac{1}{b}=\frac{b+a}{a\cdot}$ and $\frac{1}{b}+\frac{1}{a}=...$ But it doesn't solve the problem. (Note that $a$ and $b \in \mathbb{Z}$).
I also see that a field is a set with 3 elements as minimum. (unless $x$, $y$ an $z$ can be the same element.)

2. Surely you understand that $\mathbb{Q}$ is the set of rational numbers.
What can one say about the rationals with respect to the listed axioms?

3. What can one say about the rationals with respect to the listed axioms?
I think one can say that the rationals respect the listed axioms. This is what I have to show.
EDIT :How do I show that $\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}$?
$\frac{a}{b}+\frac{c}{d}=\frac{d\cdot a+b\cdot c}{b\dot d}$ and $\frac{c}{d}+\frac{a}{b}=\frac{b\cdot c+d\cdot a}{d\cdot b}$. I know that they are equal, but I have to show it...

4. Originally Posted by arbolis
I think one can say that the rationals respect the listed axioms. This is what I have to show.
Well do it. Are 0 & 1 rational?
Can you show that they are unique?
There is a good bit of work ahead of you.

5. Originally Posted by Plato
Well do it. Are 0 & 1 rational?
Can you show that they are unique?
There is a good bit of work ahead of you.
Ok, I'll try to do it.
Yes they are rational ( $0=\frac{0}{a}$ with $a \in \mathbb{Z} -\text{{0}}$ and $1=\frac{a}{b}$ where $a=b\neq 0$). I'll try to prove they are unique (I think it's not that hard by reductio ad absurdum).
But even then, I will just have almost proved part 3) and 7).
I'm not sure I understand well the problem. For part 1... Have I to show that $\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}$? Or just assume it? If I have to prove it, how can I do so?

6. Originally Posted by arbolis
Ok, I'll try to do it.
Yes they are rational ( $0=\frac{0}{a}$ with $a \in \mathbb{Z} -\text{{0}}$ and $1=\frac{a}{b}$ where $a=b\neq 0$). I'll try to prove they are unique (I think it's not that hard by reductio ad absurdum).
direct proofs will suffice.

assume there are other elements with the same property and show that they have to be equal to the additive and multiplicative identity elements that you already defined.

But even then, I will just have almost proved part 3) and 7).
I'm not sure I understand well the problem. For part 1... Have I to show that $\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}$? Or just assume it? If I have to prove it, how can I do so?
you must show it. add the fractions on both sides as you would normally. of course, addition of fractions in $\mathbb{Q}$ is defined as:

$\frac ab + \frac cd = \frac {ad + bc}{bd}$

and we know that two fractions, say $\frac mn$ and $\frac pq$ are equivalent (by definition) if $mq = pn$

7. Thanks Jhevon!
Ok for part 3 and 7, I'll try later. I prefer to do it following the order.

You said that
of course, addition of fractions in is defined as:

so what I did in post 3 :
and
is in fact wrong because in the numerator I have $d\cdot a$ instead of $a\cdot d$? I know that both are equal but I can't assume it since one property I have to show is precisely that $x\cdot y=y\cdot x$! That's really limitating me. Maybe I should learn precisely the definition of the $\mathbb{Q}$ set, so I can think faster about how things work.

8. Originally Posted by arbolis
Thanks Jhevon!
Ok for part 3 and 7, I'll try later. I prefer to do it following the order.

You said that so what I did in post 3 : is in fact wrong because in the numerator I have $d\cdot a$ instead of $a\cdot d$? I know that both are equal but I can't assume it since one property I have to show is precisely that $x\cdot y=y\cdot x$! That's really limitating me. Maybe I should learn precisely the definition of the $\mathbb{Q}$ set, so I can think faster about how things work.
yes, you have to prove commutativity in addition and multiplication first. it will make proving the harder claims easier when you have those basic properties under your belt

you might have to use it if you do my method as well. i'm too lazy to expand and see. it is the standard way though.