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Thread: product of rational and irrational is always irrational

  1. #1
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    product of rational and irrational is always irrational



    Am I assuming too much here?
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  2. #2
    Member Black's Avatar
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    I don't understand why you're setting $\displaystyle t=\frac{r}{s}$ in your proof if $\displaystyle t$ is irrational.

    To prove these, you just use the fact that $\displaystyle \mathbb{Q}$ is a field, i.e, closed under addition, subtraction, multiplication, and division.

    If $\displaystyle a+t \in \mathbb{Q}$, then by closure, $\displaystyle (a+t)-a=t \in \mathbb{Q}$. Contradiction.

    If $\displaystyle at \in \mathbb{Q}$, then by closure, $\displaystyle \frac{at}{a}=t \in \mathbb{Q}$. Contradiction.
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  3. #3
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    Quote Originally Posted by Black View Post
    I don't understand why you're setting $\displaystyle t=\frac{r}{s}$ in your proof if $\displaystyle t$ is irrational.
    He did NOT assume $\displaystyle t= \frac{r}{s}$. He showed, first, that if you assume a/t is a rational number, then you arrive at t= r/s, a ration number and so a contradiction.

    He then showed that if you assume at is rational, [b]then[b] you arrive at t= r/s, again a contradiction. His proof is perfectly valid.

    To prove these, you just use the fact that $\displaystyle \mathbb{Q}$ is a field, i.e, closed under addition, subtraction, multiplication, and division.

    If $\displaystyle a+t \in \mathbb{Q}$, then by closure, $\displaystyle (a+t)-a=t \in \mathbb{Q}$. Contradiction.

    If $\displaystyle at \in \mathbb{Q}$, then by closure, $\displaystyle \frac{at}{a}=t \in \mathbb{Q}$. Contradiction.
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