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Thread: Irrational number

  1. #1
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    Irrational number

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

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

    The question remains that if I have $\displaystyle b\sqrt{a}$ or $\displaystyle \frac{\sqrt{a}}{b}$, where $\displaystyle a,b \in \mathbb{Z}$, will I definitely get an irrational number? Is there an existing theorem in regard to this?
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  2. #2
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    Quote Originally Posted by novice View Post
    The question remains that if I have $\displaystyle b\sqrt{a}$ or $\displaystyle \frac{\sqrt{a}}{b}$, where $\displaystyle a,b \in \mathbb{Z}$, 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 $\displaystyle ab = c$.
    Then $\displaystyle a = \frac{m}{n}$ for some integers m, n and $\displaystyle c = \frac{p}{q}$ for some integers p, q. Therefore, $\displaystyle ab = c$ becomes $\displaystyle \frac{m}{n} \cdot b = \frac{p}{q}$; $\displaystyle b = \frac{np}{mq}$, thus b is rational, which is a contradiction.
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  3. #3
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    Quote Originally Posted by novice View Post
    At Dr. Math, I read about irrational number being not a closed system. The writer proved it by showing that $\displaystyle \sqrt{2}$ multiplied by itself produces a non-irrational number, and that $\displaystyle (\pi-1)+(6-\pi)=5$ is also a non-irrational number.

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

    The question remains that if I have $\displaystyle b\sqrt{a}$ or $\displaystyle \frac{\sqrt{a}}{b}$, where $\displaystyle a,b \in \mathbb{Z}$, 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, $\displaystyle 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 $\displaystyle b=0$ ...

    Tonio
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