Between two Rational Numbers exists Irrational Number/Lemma 1
Jump to navigation
Jump to search
Lemma for Between two Rational Numbers exists Irrational Number
Let $\alpha \in \Q \setminus \set 0$ and $\beta \in \R \setminus \Q$.
Then:
- $\alpha \cdot \beta \in \R \setminus \Q$
Proof
Aiming for a contradiction, suppose $\alpha \cdot \beta \in \Q$.
By the definition of rational numbers:
- $\exists n, m, p, q \in \Z: \alpha = \dfrac n m$
- $\alpha \cdot \beta = \dfrac p q$
Thus:
- $\beta = \dfrac p q \cdot \dfrac 1 \alpha = \dfrac p q \cdot \dfrac m n$
By Rational Multiplication is Closed, we have $\beta \in \Q$, which contradicts the statement that $\beta \in \R \setminus \Q$.
Therefore $\alpha \cdot \beta \in \R \setminus \Q$.
$\blacksquare$