Proving either $x^2$ or $x^3$ is irrational if $x$ is irrational

Question

I had a test today in discrete mathematics and I am dubious whether or not my proof is correct.

Suppose $x$ is an irrational number. Prove that either $x^2$ or $x^3$ is irrational.

My Answer:

Suppose $x^2$ is irrational. If $x^2$ is irrational, then the above statement holds true. Now suppose $x^2$ is rational. If $x^2$ is rational and since $x$ is irrational, then according to the theorem that a rational number multiplied by an irrational number is irrational, $x^3=x(x^2)$ is irrational since $x$ is irrational and $x^2$ is rational. In both cases, the above statement is true, thus it is proved.

I was running out of time and did this really quick and I am not sure if this is really a correct way of doing a proof.

Answer 1

Observe the negation of $p\vee q$ is $\neg p \wedge \neg q$. Suppose both $x^3,x^2$ are rational. Then $x^3/x^2=x$ is?

Answer 2

Yes, it's corrrect: more generally if $\,y = 0,\ $ (yours is $\, y = x^2)$

$\qquad\qquad x\not\in \Bbb Q\,\Rightarrow\, y\not\in Q\,$ or $\,xy\not\in \Bbb Q,\ $ since, contrapositively

$\qquad\qquad y\in \Bbb Q,\ xy\in \Bbb Q\,\Rightarrow\, x\in\Bbb Q\ $ by $\ x = xy/y$

Remark $\ $ As you did, we could proceed directly using $\,x\not\in \Bbb Q,\, y\in\Bbb Q\,\Rightarrow\, xy\not\in\Bbb Q,\,$ i.e. using that a rational times an irrational is irrational if nonzero. This composition law has a natural interpretation as a complementary view of a subgroup.

Theorem $\ $ Let $\rm\,G\,$ be a nonempty subset of an abelian group $\rm\,H,\,$ with complement set $\rm\,\bar G = H\backslash G.\,$ Then $\rm\,G\,$ is a subgroup of $\rm\,H\iff G + \bar G\, =\, \bar G. $

Proof $\ $ $\rm\,G\,$ is a subgroup of $\rm\,H\iff G\,$ is closed under subtraction, so, complementing

$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\ fails\\ &\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\ &\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED} \end{eqnarray}$

Instances of this are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here.