Prove that in any interval there is a rational and an irrational number.
Dr. Math: Go here
This is a proof of just what you are asking.
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 .
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.