product of rational and irrational is always irrational

**davismj** · January 30th 2010, 06:07 PM

Am I assuming too much here?

**Black** · January 30th 2010, 09:00 PM

I don't understand why you're setting $t=\frac{r}{s}$ in your proof if $t$ is irrational.

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

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

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

**HallsofIvy** · January 31st 2010, 05:40 AM

Originally Posted by Black

I don't understand why you're setting $t=\frac{r}{s}$ in your proof if $t$ is irrational.

He did NOT assume $t= \frac{r}{s}$ . 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 $\mathbb{Q}$ is a field, i.e, closed under addition, subtraction, multiplication, and division.

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

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

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