Fun challenge.

Knowing only that between any two real numbers is a rational number. Show that between any two real numbers is an irrational number.

Question

Fun challenge.

Knowing only that between any two real numbers is a rational number. Show that between any two real numbers is an irrational number.

perl · Answer

$\large   a+\frac{\sqrt{2}(b-a)}{2}$  is between $ a$ and $ b$ and is irrational

rsadhvika · Answer

Consider any two real numbers $r_1$ and $r_2$ 

we know that there exists a rational number $a$ in $(\pi+r_1, \pi + r_2)$.
Then $a-\pi \in (r_1,r_2)$.

zzr0ck3r · Answer

this needs pi is irrational, which is not so trivial. But sqrt(2) is...

rsadhvika · Answer

Ahh right, any irrational number will do !

zzr0ck3r · Answer

Also needs rational + irrational is irrational

zzr0ck3r · Answer

That is how I answered it today at our tudor table, but I really liked the question and thought it was very cute

zzr0ck3r · Answer

outside of the famous proofs, I think that is one of my favs (for the little guys).

I also like that identities are unique proof. $e_1=e_1e_2=e_2$ very cute as well.

perl · Answer

identity for a group?

zzr0ck3r · Answer

yeah or monoid or ....

zzr0ck3r · Answer

You technically won, but I was not gonna medal a 99er

zzr0ck3r · Answer

:)

zzr0ck3r · Answer

Would love to see other ideas also.

perl · Answer

rsadvika's proof is neat ;)

across · Answer

Here's a lengthier proof:

Let $r_1$ and $r_2$ be any two different real numbers, let $q$ be a rational number between $r_1$ and $r_2$, and let $k$ be an irrational number. Since both the sum and the product of a rational number and an irrational number is an irrational number, it follows that $\left\{q+k/n\right\}_{n=1}^\infty$ is a sequence of irrational numbers converging to $q$. This proves that there are infinitely-many irrational numbers between $r_1$ and $r_2$.

across · Answer

I should say countably infinite...

zzr0ck3r · Answer

way to much analysis :) wth does converge mean

across · Answer

It just means that$$\lim_{n\to\infty}q+\frac kn=q.$$

across · Answer

@perl's proof breaks down for $a=0$ and $b=\sqrt2$.

zzr0ck3r · Answer

I was joking about the sequence thing. I was more pointing out the tools needed for that proof.

zzr0ck3r · Answer

and rsadhvika gives us density also, because we just do it countably infinitely many times:)

zzr0ck3r · Answer

but I like it. Keep em coming !!!

across · Answer

You could also say that if there are no irrational numbers in $\left[r_1,r_2\right]$, then it is countably infinite, which is a contradiction.

zzr0ck3r · Answer

but you need cantor for the contradiction. love me some cantor

across · Answer

I know, I know, xD.

zzr0ck3r · Answer

you are way to quick though :)

zzr0ck3r · Answer

Cantor might be to much for most of the people here, but I bet the enumeration of the rationals would not be. The zig zag thingy

kainui · Answer

If you choose the arbitrary real numbers $a=(2+\sqrt{2})c$ and $b=\sqrt{2}d$ then perl's condition:
$$a + \frac{\sqrt{2}(b-a)}{2} =c+d $$
So in general there are infinitely many holes in that solution. It looks like you might accidentally end up with a limit if you're not careful with this kind of reasoning though if you suggest wiggling in the bounds to some other number, what's to guarantee that next set of numbers to also not be a miss and accidentally fall into an infinite regress?