order prpoperties

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January 18th 2011, 07:59 PM
mremwo

order prpoperties

Let k:= {s+t(sqrt)2 : s, t exist in Q (rational numbers). Show that K satisfies the following:
1. If x,y exist in K then x+y exists in K and xy exists in K
2. If x is not 0 and x is in K then 1/x is in K

(Thus the set K is a subfield of R. With the order inherited from R the set K is an ordered field that lies between Q and R)
Also im not sure what this little tidbit means either

Thank you!
January 18th 2011, 08:23 PM
FernandoRevilla

Quote:

Originally Posted by mremwo

(Thus the set K is a subfield of R. With the order inherited from R the set K is an ordered field that lies between Q and R)
Also im not sure what this little tidbit means either

$K$ is an ordered field with the order inherited from $\mathbb{R}$ because we verify:

$(i)\;x\leq y\Rightarrow x+z\leq y+z\quad ( \forall{z}\in K)$

$(ii)\;x\leq y\Rightarrow xz\leq yz\quad (\forall{z}\in K\wedge z\geq 0)$

Fernando Revilla
January 19th 2011, 02:48 AM
emakarov

So, $K=\{s+t\sqrt{2}\mid s,t\in\mathbb{Q}\}$ . (One usually says "s is in Q," not "s exists in Q.") Suppose $s+t\sqrt{2}\in K$ . Then $\displaystyle\frac{1}{s+t\sqrt{2}}=\frac{s-t\sqrt{2}}{(s+t\sqrt{2})(s-t\sqrt{2})}=<br /> \frac{s-t\sqrt{2}}{s^2-2t^2}=\frac{s}{s^2-2t^2}-\frac{t}{s^2-2t^2}\sqrt{2}\in K$ . Checking point 1 is easier.

Quote:

(Thus the set K is a subfield of R. With the order inherited from R the set K is an ordered field that lies between Q and R)
Also im not sure what this little tidbit means either

$K$ is a subfield of $\mathbb{R}$ : every element of K is a real number and $K$ satisfies the axioms of a field.

With the order inherited from $\mathbb{R}$ the set $K$ is an ordered field: with the regular order, $K$ satisfies the axioms of an ordered field. Fernando showed two axioms of an ordered field that come on top of axioms of a field.

...lies between $\mathbb{Q}$ and $\mathbb{R}$ : $\mathbb{Q}\subset K\subset\mathbb{R}$ .
January 19th 2011, 01:07 PM
mremwo

Im still not understanding! :(
January 19th 2011, 01:24 PM
emakarov

What exactly are you not understanding?

You need to prove

Quote:

1. If x,y exist in K then x+y exists in K

What should the first phrase of the proof be?
January 20th 2011, 06:16 AM
mremwo

i dont get how you know if it is in K
just if it is in the form of "s+tsqrt2"?

Also, would i make y (in the original problem it is xsub1) ssub1 + tsub1sqrt2 ? and then say that s+tsqrt2,ssub1 + tsub1sqrt2 is in k?
January 20th 2011, 07:56 AM
emakarov

Quote:

Originally Posted by mremwo

i dont get how you know if it is in K
just if it is in the form of "s+tsqrt2"?

What is "it" in your question? The notation $\{s+t\sqrt{2}\mid s,t\in\mathbb{Q}\}$ is used to denote a set that has all possible elements of the form $s+t\sqrt{2}$ for $s,t\in\mathbb{Q}$ and only such element. So, something of the form $u+v\sqrt{2}$ is in this set, called K here, and every element of K has this form.

Quote:

Also, would i make y (in the original problem it is xsub1) ssub1 + tsub1sqrt2 ?

First, I am not sure what you mean by the "original problem." For me, the original problem is the first post of this thread, and it does not have $x_1$ . Yes, if $y\in K$ , then y has the form $s_1+t_1\sqrt{2}$ for some rational $s_1,t_1$ .

Quote:

and then say that s+tsqrt2,ssub1 + tsub1sqrt2 is in k?

Is is not clear to me what s and t are, how they were introduced. In any case, if $s,s_1,t,t_1$ are rational, then yes, $s+t\sqrt{2}\in K$ and $s_1+t_1\sqrt{2}\in K$ , by definition of K. I am not sure how this advanced the proof, though.

To answer my question from the previous post:

Quote:

What should the first phrase of the proof be?

It should be: "Suppose some arbitrary x, y are in K." In general, a proof of a statement of the form "If A, then B" starts with "Suppose A." Then, since $x,y\in K$ , there exists $s,t,s_1,t_1\in\mathbb{Q}$ such that $x=s+t\sqrt{2}$ and $y=s_1+t_1\sqrt{2}$ . Note that I introduced $s_1,t_1$ by saying "there exist". Also, the point was not to say that $s_1+t_1\sqrt{2}\in K$ , which is a triviality; the point was to say that $y=s_1+t_1\sqrt{2}$ . After that, you can see what x + y looks like.

Hint: it is customary to write x1 or x_1 instead of xsub1. On the other hand, it is better to write sqrt(2) than sqrt2.