Given a rational number and an irrational number, both greater than 0, prove that the product between them is irrational.

Question

Does this proof I made make sense?

Proof//

$\mathbf a$ is the rational number, $\mathbf b$ is the irrational number. Assume that $\mathbf {a * b}$ is rational due to proof by contradiction. Therefore, $\mathbf {a * b = P}$ for some rational number $\mathbf P$. In another word, $\mathbf {rational * irrational = rational}$.

If you divide the rational number, from both side, we get $\mathbf {b = \frac P a}$ which also translates to $\mathbf {irrational = \frac {rational} {rational}}$ which then becomes $\mathbf {irrational = rational}$. However this reaches a contradiction because $\mathbf {irrational \neq rational}$. Therefore, proving that the product between an irrational number and rational number is equal to an irrational number. $\Box$

$\mathbf {EDIT:}$ I would like to apologize for the tough wording of the proof, I just completed this on a test, and I tried to write it word for word as I did from the test to see if I was on the right track.

Answer 1

Yes: though make sure you deal with the case where a=0 explicitly in your proof i.e. Assuming both a and b are greater than zero with a rational and b irrational,...

Answer 2

The premise is almost correct. The wording is pretty tough to read. But more important than the wording, is a missed an edge case.

You tried to use the fact (implicitly) that the quotient of two rationals is rational. But that's not exactly true. The denominator cannot be $0$.

You need to cite that the original rational $a$ is non-zero, then cite explicitly that the quotient of two non-zero rationals is rational. That'll at least make the proof correct.