Proof that in any open interval there is infinitley many irrationals

Is this proof ok?

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

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

Therefore as $\displaystyle \frac{q}{x}$ and we have $\displaystyle \frac{q}{x}\in(a,b)$

we have an irrational in any open interval.

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

thanks for any help.

Re: Proof that in any open interval there is infinitley many irrationals

Quote:

Originally Posted by

**hmmmm** If we take the open interval $\displaystyle (a,b)$ and shift it $\displaystyle (ax,bx)$. Where $\displaystyle x=\sqrt(p)$ 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:

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

Therefore as $\displaystyle \frac{q}{x}$ and we have $\displaystyle \frac{q}{x}\in(a,b)$

we have an irrational in any open interval.

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

Here is one comment. Let $\displaystyle \mathcal{I}=\mathbb{R}\setminus\mathbb{Q}$

You have proven that any open interval contains an irrational.

Once you find $\displaystyle i_1\in (a,b)\cap\mathcal{I}$ then find $\displaystyle i_2\in (a,i_1)\cap\mathcal{I}$.

If $\displaystyle n\ge 3$ find $\displaystyle i_n\in (a,i_{n-1})\cap\mathcal{I}$.

Re: Proof that in any open interval there is infinitley many irrationals

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

Re: Proof that in any open interval there is infinitley many irrationals

Quote:

Originally Posted by

**hmmmm** Is this proof ok?

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

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

Therefore as $\displaystyle \frac{q}{x}$ and we have $\displaystyle \frac{q}{x}\in(a,b)$

we have an irrational in any open interval.

Now as there are infinitely many primes we have infinitley many irrationals of the form $\displaystyle \sqrt(p)$, 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 $\displaystyle (a,b)$ then $\displaystyle (a,b)$ would be countable...but I'm sure you this isn't true.