1. Irrational number

At Dr. Math, I read about irrational number being not a closed system. The writer proved it by showing that $\sqrt{2}$ multiplied by itself produces a non-irrational number, and that $(\pi-1)+(6-\pi)=5$ is also a non-irrational number.

Now, if I multiply $\sqrt{2}$ by 2, I get an irrational number $2\sqrt{2}$, and when I divide $\sqrt{2}$ by 2, I also get an irrational number $\frac{\sqrt{2}}{2}$.

The question remains that if I have $b\sqrt{a}$ or $\frac{\sqrt{a}}{b}$, where $a,b \in \mathbb{Z}$, will I definitely get an irrational number? Is there an existing theorem in regard to this?

2. Originally Posted by novice
The question remains that if I have $b\sqrt{a}$ or $\frac{\sqrt{a}}{b}$, where $a,b \in \mathbb{Z}$, will I definitely get an irrational number? Is there an existing theorem in regard to this?
Yes. An irrational number multiplied by a rational number will always be irrational. The proof is simple: suppose that there are three numbers a, b, c such that a and c are rational, b is irrational, and $ab = c$.
Then $a = \frac{m}{n}$ for some integers m, n and $c = \frac{p}{q}$ for some integers p, q. Therefore, $ab = c$ becomes $\frac{m}{n} \cdot b = \frac{p}{q}$; $b = \frac{np}{mq}$, thus b is rational, which is a contradiction.

3. Originally Posted by novice
At Dr. Math, I read about irrational number being not a closed system. The writer proved it by showing that $\sqrt{2}$ multiplied by itself produces a non-irrational number, and that $(\pi-1)+(6-\pi)=5$ is also a non-irrational number.

Now, if I multiply $\sqrt{2}$ by 2, I get an irrational number $2\sqrt{2}$, and when I divide $\sqrt{2}$ by 2, I also get an irrational number $\frac{\sqrt{2}}{2}$.

The question remains that if I have $b\sqrt{a}$ or $\frac{\sqrt{a}}{b}$, where $a,b \in \mathbb{Z}$, will I definitely get an irrational number? Is there an existing theorem in regard to this?

Yes, the theorem that says that rational numbers is a field: if, for example, $b\sqrt{a}=q\in\mathbb{Q}\Longrightarrow \sqrt{a}=\frac{q}{b}\in\mathbb{Q}$ , and this is nonsense unless a is a perfect square or UNLESS $b=0$ ...

Tonio