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

1. There is a rational and an irrational number in any interval

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

2. Dr. Math: Go here

This is a proof of just what you are asking.

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

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

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

4. Originally Posted by Plato
But $\displaystyle a < \frac{r}{{\sqrt 2 }} < b$ and $\displaystyle \frac{r}{{\sqrt 2 }}$ is irrational.
$\displaystyle \frac{r}{\sqrt{2}}$ is rational if $\displaystyle r=0$.

5. Originally Posted by JaneBennet
$\displaystyle \frac{r}{\sqrt{2}}$ is rational if $\displaystyle r=0$.
True.
But if a < 0 < b, then we can apply density of rationals to $\displaystyle a\sqrt{2}<0$ to obtain a 'r' such that $\displaystyle a\sqrt{2} < r < 0$......

6. Originally Posted by Plato
Between any two numbers there is a irrational number.

Proof:

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

But $\displaystyle a < \frac{r}{{\sqrt 2 }} < b$ and $\displaystyle \frac{r}{{\sqrt 2 }}$ is irrational .
Originally Posted by JaneBennet
$\displaystyle \frac{r}{\sqrt{2}}$ is rational if $\displaystyle r=0$.
Originally Posted by Isomorphism
True.
But if a < 0 < b, then we can apply density of rationals to $\displaystyle a\sqrt{2}<0$ to obtain a 'r' such that $\displaystyle a\sqrt{2} < r < 0$......
Bah! The last part should be done this way.

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

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