Question

Small math question:

Answer 1

Let x,y \(\in \mathbb{R} / x<y\) prove that the following affirmations are true:
(1) In the density of \(\mathbb{Q} \) in \(\mathbb{R} \) there exists \(a \in \mathbb{Q} \) such that \(x<a<y\)
(2) In the density \(\mathbb{R} / \mathbb{Q} \) in \(\mathbb{R} \) there exists \(\alpha \in \mathbb{R} / \mathbb{Q} \) such that \(x< \alpha < y\)

Answer 2

Im thinking if we should let

\[a=\frac{ x+y }{ 2 }\]

Answer 3

Hmm

Let \[a=\frac{ x+y }{ 2 } {\space}, where {\space} a \in Q\]

Now, I'm sure if you want to prove this, then we can assume other simpler theorems to apply.

So we can use the fact that \[x+y\] is rational and that the product of rationals \[\left( x+y, {\space} \frac{ 1 }{ 2 } \right)\] is also rational.

Also,
since it states that,
\[x<y\]
then adding x to both sides means \[2x<x+y\]
And similarly, by adding y to both sides we get:
\[x+y<2y\]

So therefore we can bound these restrictions essentially into one domain:
\[2x<x+y<2y\]

This implies that
\[x<\frac{ x+y }{ 2 }<y\] , where \[a=\frac{ x+y }{ 2 }\]

Answer 4

now for part 2, its probs a bit tricky

Answer 5

perhaps set

\[a=\frac{ x+y }{ \sqrt2 }\]

Now one tricky bit here is that we can make this value rational if \[x=-y\]
which means with some algebra, we get \[a=0\]

So consider two cases:
Either,
\[x \ge 0\] or
\[y \le0\]

so we can take
\[a=\frac{ x+y }{ \sqrt2 }\]

whereas if \[x<0<y\], then we can have it set to
\[a=\frac{ y }{ \sqrt2 }\]

and by now we can see 'a' is necessarily irrational and therefore proof is complete since we can assume an earlier theorem where

\[If {\space} x \in Q {\space} and {\space} y \in R/Q, {\space} then {\space} x+y \in R/Q\]

Answer 6

Probably a good idea to use that assumption above as well as
\[If {\space} x \in Q {\space} and {\space} y \in R/Q, {\space} then {\space} xy \in R/Q {\space} provided {\space} x \neq0\]

Answer 7

sorry if this is a bit general, my real analysis knowledge is a bit distant now

Answer 8

probs best if i stick with studying my chem eng courses atm haha

Answer 9

Well, this looks good but thing is bounding out the domain would cause troubles when wanting to actually analyze the density of Q in R...
I was thinking about calling "n" : \(n \in \mathbb{N} \) such that \(\frac{ 1 }{ n }< y-x \), this will generate a rational, bounded to Q in R such that calling it "a" would be:
\[a:=\frac{ E(nx)+1 }{ n }\]
And this should work smoothly since "E" would be the whole function defined by:
\[E(x):=\max \left\{ p \in \mathbb{Z} :p \le x \right\}\]
Now this should be enough to constate that inside these conditions \(x<a<y \) should be verified because of E(x) and it's behaviour.
I have to research that on the library in the afternoon, but thanks a lot for giving this question some time haha.

Answer 10

yes very curious to see!

Answer 11

let me try and think maybe haha good though process though

Answer 12

I based my conclusion on:
\[\frac{ E(nx) }{ n } \le x< \frac{ E(nx)+1 }{ n }\]
\[x<a \le x+\frac{ 1 }{ n } < x+(y-x)=y\]

Answer 13

interesting

Answer 14

im kinda stuck now haha

Answer 15

Haha, I am also a lil' stuck but I think it looks right, I am also quite rusty when it comes to real analysis.

Answer 16

perhaps share me the verdict when you find out? I wouldn't mind delving back into real analysis for a bit haha

Answer 17

good question though!

i'm off to bed. catchya