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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.

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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.

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  • $\begingroup$ I didn't state that because it was already told in the question, would I still need to state that in the proof? $\endgroup$ – AyT Feb 17 '16 at 17:27
  • $\begingroup$ Absolutely you must state it. If you never state it in the proof, how would I know where you used the fact? You are citing a theorem that says "a quotient of non-zero rationals is rational." You are required to prove that the conditions of the theorem apply. That means mentioning that they are non-zero. $\endgroup$ – Zach Stone Feb 17 '16 at 17:43
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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,...

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  • $\begingroup$ However, the fact that a and b are greater than 0 are already stated in the question, would I still need to restate it in the proof? $\endgroup$ – AyT Feb 17 '16 at 17:33
  • $\begingroup$ It is unlikely you will lose significant points for not stating the assumption that was given in the statement of the problem, but in the future always anticipate questions concerning "special cases". In this case, zero doesn't have a multiplicative inverse, yet your argument relies on this fact, so it's important to be clear about that. $\endgroup$ – John Feb 17 '16 at 17:36

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