There is a rational and an irrational number in any interval

**james_bond** · April 11th 2008, 08:26 AM

Prove that in any interval there is a rational and an irrational number.

**Aryth** · April 11th 2008, 11:35 AM

Dr. Math: Go here

This is a proof of just what you are asking.

**Plato** · April 11th 2008, 12:16 PM

Lemma: If $x-y>1$ then $\left( {\exists k \in Z} \right)\left[ {y < k < x} \right]$ .
Proof:
By the floor function, $\left\lfloor y \right\rfloor \leqslant y < \left\lfloor y \right\rfloor + 1\,\mbox{or}\,\left\lfloor y \right\rfloor + 1 \leqslant y + 1 < x \Rightarrow \quad y < \left\lfloor y \right\rfloor + 1 < x$ .

Between any two numbers there is a rational number.
Proof:
If $a<b$ then $\left( {\exists K \in Z^ + } \right)\left[ {\frac{1}<br /> {{b - a}} < K} \right]$ .
But that means that $1 < Kb - Ka$ so by the lemma $\left( {\exists J \in Z} \right)\left[ {Ka < J < Kb} \right] \Rightarrow \quad a < \frac{K}{J} < b$ .

Between any two numbers there is a irrational number.
Proof:
If $a<b$ then $a\sqrt 2 < b\sqrt 2$ so by the above $\left( {\exists r \in Q} \right)\left[ {a\sqrt 2 < r < b\sqrt 2 } \right]$ .
But $a < \frac{r}{{\sqrt 2 }} < b$ and $\frac{r}{{\sqrt 2 }}$ is irrational .

**JaneBennet** · April 28th 2008, 05:02 PM

Originally Posted by Plato

But $a < \frac{r}{{\sqrt 2 }} < b$ and $\frac{r}{{\sqrt 2 }}$ is irrational.

$\frac{r}{\sqrt{2}}$ is rational if $r=0$ .

**Isomorphism** · April 29th 2008, 01:06 AM

Originally Posted by JaneBennet

$\frac{r}{\sqrt{2}}$ is rational if $r=0$ .

True.
But if a < 0 < b, then we can apply density of rationals to $a\sqrt{2}<0$ to obtain a 'r' such that $a\sqrt{2} < r < 0$ ......

**JaneBennet** · April 30th 2008, 06:17 AM

Originally Posted by Plato

Between any two numbers there is a irrational number.

Proof:

If $a<b$ then $a\sqrt 2 < b\sqrt 2$ so by the above $\left( {\exists r \in Q} \right)\left[ {a\sqrt 2 < r < b\sqrt 2 } \right]$ .

But $a < \frac{r}{{\sqrt 2 }} < b$ and $\frac{r}{{\sqrt 2 }}$ is irrational .

Originally Posted by JaneBennet

$\frac{r}{\sqrt{2}}$ is rational if $r=0$ .

Originally Posted by Isomorphism

True.
But if a < 0 < b, then we can apply density of rationals to $a\sqrt{2}<0$ to obtain a 'r' such that $a\sqrt{2} < r < 0$ ......

Bah! The last part should be done this way.

If $a<b$ then $a+\sqrt 2 < b+\sqrt 2$ so by the above $\left( {\exists r \in \mathbb{Q}} \right)\left[ {a+\sqrt 2 < r < b+\sqrt 2 } \right]$ .

But $a < r-\sqrt 2 < b$ and $r-\sqrt 2$ is irrational.

Thread: There is a rational and an irrational number in any interval

LinkBack

Thread Tools

Display

There is a rational and an irrational number in any interval

Similar Math Help Forum Discussions

Need help changing an irrational number into a rational number, involving polar coord

Prove that e is an irrational number!�ee is is an irrational number!

Continuous Functions, Rational number/ Irrational number.

Is -2 3/5 a rational or irrational number?

Number of Rational vs irrational numbers

Search Tags