Proof that in any open interval there is infinitley many irrationals

**hmmmm** · September 27th 2011, 05:32 AM

Is this proof ok?

If we take the open interval $(a,b)$ and shift it $(ax,bx)$ . Where $x=\sqrt(p)$ where p is a prime.

Then we have that as the rationals are dense in the reals ( we can have a rational arbitrarly close to any real) we have:

$ax<q<bx\ \ s.t. q\in\mathbb{Q}\rightarrow a<\frac{q}{x}<b$ .

Therefore as $\frac{q}{x}$ and we have $\frac{q}{x}\in(a,b)$
we have an irrational in any open interval.

Now as there are infinitely many primes we have infinitley many irrationals of the form $\sqrt(p)$ , so we have infinitley many irrationals in the interval.

thanks for any help.

**Plato** · September 27th 2011, 06:36 AM

Originally Posted by hmmmm

If we take the open interval $(a,b)$ and shift it $(ax,bx)$ . Where $x=\sqrt(p)$ where p is a prime.
Then we have that as the rationals are dense in the reals ( we can have a rational arbitrarly close to any real) we have:
$ax<q<bx\ \ s.t. q\in\mathbb{Q}\rightarrow a<\frac{q}{x}<b$ .
Therefore as $\frac{q}{x}$ and we have $\frac{q}{x}\in(a,b)$
we have an irrational in any open interval.
Now as there are infinitely many primes we have infinitley many irrationals of the form $\sqrt(p)$ , so we have infinitley many irrationals in the interval.

Here is one comment. Let $\mathcal{I}=\mathbb{R}\setminus\mathbb{Q}$
You have proven that any open interval contains an irrational.
Once you find $i_1\in (a,b)\cap\mathcal{I}$ then find $i_2\in (a,i_1)\cap\mathcal{I}$ .
If $n\ge 3$ find $i_n\in (a,i_{n-1})\cap\mathcal{I}$ .

**hmmmm** · September 27th 2011, 12:22 PM

Yeah thanks for the help there, thought that it was a bit poor at the end there thanks.

**Drexel28** · September 27th 2011, 03:37 PM

Originally Posted by hmmmm

Is this proof ok?

If we take the open interval $(a,b)$ and shift it $(ax,bx)$ . Where $x=\sqrt(p)$ where p is a prime.

Then we have that as the rationals are dense in the reals ( we can have a rational arbitrarly close to any real) we have:

$ax<q<bx\ \ s.t. q\in\mathbb{Q}\rightarrow a<\frac{q}{x}<b$ .

Therefore as $\frac{q}{x}$ and we have $\frac{q}{x}\in(a,b)$
we have an irrational in any open interval.

Now as there are infinitely many primes we have infinitley many irrationals of the form $\sqrt(p)$ , so we have infinitley many irrationals in the interval.

thanks for any help.

Another possibility would be to note that if there were finitely many irrationals in $(a,b)$ then $(a,b)$ would be countable...but I'm sure you this isn't true.

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