# Thread: Is it a field? set of real numbers in the form...

1. ## Is it a field? set of real numbers in the form...

Hi,

problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.

(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?

attempt:

(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?

(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..

Thanks

2. Originally Posted by Mollier
Hi,

problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.

(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?

attempt:

(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?

(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..

Thanks
just go through each property for a field and make sure that they all hold. you have to verify these explicitly.

and no, you cannot say any real number can be expressed in that form. how would you express $\pi$ that way, for instance?

3. Originally Posted by Mollier
Hi,

problem:
Let $Q(\sqrt{2})$ be the set of all real numbers of the form
$\alpha+\beta\sqrt{2}$, where $\alpha\;and\;\beta$ are rational.

(a) Is $Q(\sqrt{2})$ a field?
(b) What if $\alpha\;and\;\beta$ are required to be integers?

attempt:

(a)
I look at the axioms for a field and check whether any of them fail. I am unable to find one that does.
Is it ok to say that I can express any real number in the form $\alpha+\beta\sqrt{2}$ as long as $\alpha\;and\;\beta$ are rational?

(b)
I am unable to find an axiom that "fails", but say I want to write the real number $\frac{\sqrt{2}}{2}$. I don't think I can do this if $\alpha\;and\;\beta$ are integers..

Thanks
Firstly, if every real number was of the form $\alpha+\beta \sqrt{2}$ with $\alpha, \beta \in \mathbb{Q}$ then we can easily define a bijection $\mathbb{R} \rightarrow \mathbb{Q} \times \mathbb{Q}$, $\alpha + \beta \sqrt{2} \mapsto (\alpha, \beta)$. As there exists a bijection between $\mathbb{Q}$ and $\mathbb{Q} \times \mathbb{Q}$ we have that $|\mathbb{R}| = |\mathbb{Q}|$, a contradiction.

Secondly, in both of your problems you know that "most" of the axioms for fields holds (both are clearly rings, and multiplication in both cases is commutative). The one to concentrate on will be multiplicative inverses.

What is the inverse of $\alpha+\beta \sqrt{2}$ with $\alpha$ and $\beta$ rationals?

What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\alpha + \beta \sqrt{2}$. That is, every rational number of the form $\frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\}$ can be written as $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$. As we are, allegedly, in a field we have that $\frac{n}{\gamma}$ is also of this form, with $n \in \mathbb{N}$. Thus, every rational number is of the form $\alpha + \beta \sqrt{2}$. This is silly!)

4. Originally Posted by Jhevon
and no, you cannot say any real number can be expressed in that form. how would you express $\pi$ that way, for instance?
Thanks

Originally Posted by Swlabr
The one to concentrate on will be multiplicative inverses.

What is the inverse of $\alpha+\beta \sqrt{2}$ with $\alpha$ and $\beta$ rationals?

What if both were integers, does that imply that the inverse has integer coefficients? (Alternatively, think about what the problem is saying...take $\beta = 0$. Then if your ring was a field every element in this subset has an inverse in the ring. This subset is just the integers. So, every integer has an inverse of the form $\alpha + \beta \sqrt{2}$. That is, every rational number of the form $\frac{1}{\gamma}, \gamma \in \mathbb{Z} \setminus \{0\}$ can be written as $\alpha + \beta \sqrt{2}, \alpha, \beta \in \mathbb{Z}$. As we are, allegedly, in a field we have that $\frac{n}{\gamma}$ is also of this form, with $n \in \mathbb{N}$. Thus, every rational number is of the form $\alpha + \beta \sqrt{2}$. This is silly!)
It would be $\frac{1}{\alpha+\beta\sqrt{2}}$ .
Now, the inverse would have to be in $Q(\sqrt{2})$, so it would have to be in the form $\alpha+\beta\sqrt{2}$.
Since $\frac{1}{\alpha+\beta\sqrt{2}}$ is not in that form, $Q(\sqrt{2})$ is not a field.
Is this totally off? I think I need to get (a) right, before I move on to (b).

Thanks guys!

5. Originally Posted by Mollier
Thanks

It would be $\frac{1}{\alpha+\beta\sqrt{2}}$ .
Now, the inverse would have to be in $Q(\sqrt{2})$, so it would have to be in the form $\alpha+\beta\sqrt{2}$.
Since $\frac{1}{\alpha+\beta\sqrt{2}}$ is not in that form, $Q(\sqrt{2})$ is not a field.
Is this totally off? I think I need to get (a) right, before I move on to (b).

Thanks guys!
You want to try and write $\frac{1}{\alpha + \beta \sqrt{2}}$ as $\alpha^{\prime} + \beta^{\prime}\sqrt{2}$. I mean, $\sqrt{2} = \frac{2}{\sqrt{2}}$...numbers are strange things...just because something isn't obviously in a certain form does not mean you cannot write it in that form.

I shall give you a tantalising hint: It is a field.

6. $
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}
$

$\alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)$

$\beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)$

Since both $\alpha'\;and\;\beta'$ are rationals, $Q(\sqrt{2})$ is a field?

Thanks mate!

7. Originally Posted by Mollier
$
\frac{1}{\alpha+\beta\sqrt{2}}=\frac{\alpha-\beta\sqrt{2}}{\alpha^2-2\beta^2}=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right) - \left(\frac{\beta}{\alpha^2-2\beta^2}\right)\sqrt{2}
$

$\alpha'=\left(\frac{\alpha}{\alpha^2-2\beta^2}\right)$

$\beta'=\left(\frac{\beta}{\alpha^2-2\beta^2}\right)$

Since both $\alpha'\;and\;\beta'$ are rationals, $Q(\sqrt{2})$ is a field?

Thanks mate!
Yeah, that's it! Although to be thorough you would need to make sure there is no division by 0 going on.

8. So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..

As for (b), can I use a similar argument by saying that since $\alpha'$ and $\beta'$ are rational numbers, $Q(\sqrt{2})$ is not a field when $\alpha,\beta\in Z$?

Thank you very much.

9. Originally Posted by Mollier
So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..

As for (b), can I use a similar argument by saying that since $\alpha'$ and $\beta'$ are rational numbers, $Q(\sqrt{2})$ is not a field when $\alpha,\beta\in Z$?

Thank you very much.
For part (b) you need to find a counter-example - $\alpha^{\prime}$ and $\beta^{\prime}$ may look like non-integers, but this may not always be the case.

So, to complete the problem simply take an integer value of $\alpha$ and an integer value of $\beta$ such that $\alpha^{\prime} \notin \mathbb{Z}$ or $\beta^{\prime} \notin \mathbb{Z}$. For simplicity, let one of them be zero.

10. Originally Posted by Mollier
So, $\alpha^2-2\beta^2\neq 0$,
then $\beta \neq \frac{1}{\sqrt{2}}\alpha$, but $\beta$ is then not a rational number anyways..
No, this is not a proof. You want to assume that $\alpha^2 -2\beta^2=0$ and look for a contradiction.

11. $\alpha^2-2\beta^2\neq 0$ because of the irrationality of $\sqrt{2}$, it can be easily proved in number theory.

12. Originally Posted by Swlabr
No, this is not a proof. You want to assume that $\alpha^2 -2\beta^2=0$ and look for a contradiction.
$\alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta$
Contradiction since $\alpha\in Q$

13. Originally Posted by Mollier
$\alpha^2-2\beta^2=0 \Rightarrow \alpha=\sqrt{2}\beta$
Contradiction since $\alpha\in Q$
Precisely!

Sorry to be pedantic about it, but in maths the proof is the king...

14. Please don't apologize. I have a BS in mechanical engineering from a horrible school, so I have now decided to actually learn some math. You being "pedantic" is exactly what I need! Thanks!