I need help with this proof. We have to prove the following statement by contradiction:

"Show that the product of a non-zero rational number and an irrational number is irrational."

This is what I have so far, based off of examples we did in class:

We proceed by way of contradiction. Assume that the product of a non-zero rational number and an irrational number is rational. Let p and r be rational numbers, and q be an irrational number. By definition, there exist integers a and b, b not equal to 0, and integers c and d, d not equal to 0, for which p = (a/b) and r = (c/d). So, pq = r, which implies (a/b)q = (c/d). Without loss of generality, we assume (a/b) and (c/d) are in lowest terms.

I don't know where to go from here. If anyone could help, I would greatly appreciate it.

2. Originally Posted by Lprdgecko
I need help with this proof. We have to prove the following statement by contradiction:

"Show that the product of a non-zero rational number and an irrational number is irrational."

This is what I have so far, based off of examples we did in class:

We proceed by way of contradiction. Assume that the product of a non-zero rational number and an irrational number is rational. Let p and r be rational numbers, and q be an irrational number. By definition, there exist integers a and b, b not equal to 0, and integers c and d, d not equal to 0, for which p = (a/b) and r = (c/d). So, pq = r, which implies (a/b)q = (c/d). Without loss of generality, we assume (a/b) and (c/d) are in lowest terms.

I don't know where to go from here. If anyone could help, I would greatly appreciate it.
Let p be a nonzero rational and q be irrational and suppose pq = r is rational. Then r/p = q is irrational. But $\mathbb{Q}$ is closed under multiplication. Contradiction.

3. Hello, Lprdgecko!

I think you're there . . .

$\text{Show that the product of a non-zero rational number}$
$\text{and an irrational number is irrational.}$

Suppose that: . $\text{(rational)} \times \text{(irrational)} \;=\;\text{(rational)}$

$\text{That is: }\;p \cdot q \:=\:r\quad\text{ where: }\;\begin {Bmatrix} p &=& \text{rational} \\ q &=& \text{irrational} \\ r &=& \text{rational}\end{Bmatrix}$

$\text{Then }p = \dfrac{a}{b},\;\text{ where }a\text{ and }b\text{ are integers and }b \ne 0.$

$\text{And }r = \dfrac{c}{d},\;\text{ where }c\text{ and }d\text{ are integers and }d \ne 0.$

$\text{Hence, we have: }\;\dfrac{a}{b}\cdot q \:=\:\dfrac{c}{d} \quad\Rightarrow\quad q \;=\;\dfrac{bc}{ad}$

$\text{Since }bc\text{ and }ad\text{ are integers, }\dfrac{bc}{ad}\text{ is rational.}$

Therefore, $\,q$ is rational . . . We have reached a contradiction.

4. Thank you both. I keep forgetting that I can go back to what I have stated in the proof and use that to complete the proof. Thanks again