# product of rational and irrational is always irrational

• Jan 30th 2010, 06:07 PM
davismj
product of rational and irrational is always irrational
http://i46.tinypic.com/2pooa34.png

Am I assuming too much here?
• Jan 30th 2010, 09:00 PM
Black
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.
• Jan 31st 2010, 05:40 AM
HallsofIvy
Quote:

Originally Posted by Black
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.

Quote:

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.