Show that there are no rationals $r$ such that $r^3 = 6$

Question

Show that there are no rationals $r$ such that $r^3 = 6$

We were asked this for a real analysis assignment. I just assumed that it would be the same as if it were $r^2$, but now I'm second guessing myself.

Answer 1

Suppose that $r=\dfrac{a}{b}$ is a rational with $a,b$ coprime such that $r^3=6$

Then $a^3=6b^3$

The prime factorization of the LHS contains primes with powers which are multiple of $3$, and so does $b^3$ in the RHS.

But RHS contains factor $6=2\times 3$, therefore prime factors $2$ and $3$ in the RHS are NOT multiple of $3$ because their exponent is $3k+1;\;3h+1,\; k,h\in\mathbb{N}$

This is a contradiction so there is no rational $r$ such that $r^3=6$

Hope this helps

Answer 2

Suppose $r \in \Bbb Q$ satisfies

$r^3 = 6; \tag 1$

then we can take

$r = \dfrac{p}{q}, \tag 2$

$p, q, \in \Bbb Z$, with

$\gcd(p, q) = 1; \tag 3$

thus

$\dfrac{p^3}{q^3} = \left ( \dfrac{p}{q} \right )^3 = r^3 = 6, \tag 4$

whence

$p^3 = 6q^3 = 2 \cdot 3 \cdot q^3; \tag 5$

we have via (5) that

$2 \mid p^3; \tag 6$

it follows from (6) that $p$ is even; that is,

$p = 2s\tag 7$

for some $s \in \Bbb Z$; if this were not the case, so that $p$ were odd, we could write

$p = 2s + 1, \tag 8$

whence

$p^3 = (2s + 1)^3 = 8s^3 + 12s^2 + 6s + 1 = 2(4s^3 + 6s^2 + 3s) + 1, \tag 9$

which contradicts (6); thus (7) binds, and

$8s^3 = (2s)^3 = p^3 = 2 \cdot 3 \cdot q^3, \tag{10}$

or

$4s^3 = 3 \cdot q^3, \tag{11}$

which in turn implies

$2 \mid 3 \cdot q^3. \tag{12}$

(12) forces $q$ to be even as well; that is, $q = 2t$ where $t \in \Bbb Z$; if this were not the case, if $q = 2t + 1$, then in a manner similar to the above we would have

$q^3 = 8t^3 + 12t^2 + 6t + 1, \tag{13}$

whence

$3 \cdot q^3 = 24t^3 + 36t^2 + 18t + 3 = 2(12t^3 + 18t^2 + 9t + 1) + 1, \tag{14}$

which contradicts (12); thus $q$ is even and we may take

$q = 2t. \tag{15}$

(7) and (15) together now contradict (3); thus we cannot find $p, q$ such that (1) and (2) hold; so,

$r \notin \Bbb Q. \tag{16}$

Note: I have, in the above argument, explicitly tried to avoid invoking the fact that $2$ (or $3$ for that matter) is prime, which fact would immediately allow the conclusions (7) from (6) and (15) from (12) without the need to carry through the reductio ad absurdum demonstration embodied in (8)-(9) and (13)-(14); but I wanted to avoid using that for primes $k$

$[k \mid ab] \Longrightarrow [k \mid a \vee k \mid b], \tag{17}$

since I anticipate that many of the readers of this answer may not have encountered the notions of irreducible vs. prime and of principle ideal domain which links them in $\Bbb Z$. So I tried to keep things as elementary as I could see my way to. End of Note.

Answer 3

Just to give a very different answer from the usual ones for this kind of problem I find.

Let $\dfrac{p^3}{q^3}=6\lt2^3\Rightarrow \dfrac pq\lt 2$. Put then $\dfrac pq=2-h$ where $h$ is positive. It follows $$6=8-12h+6h^2-h^3\iff h^3-6h^2+12h-2=0$$ By Eisenstein's criterion for irreducibility with the prime $2$, the real root $h$ of this equation is not rational ($h\approx 0.1829$). Consequently $2-h$ is not rational. Contradiction, $\dfrac pq$ can not be a rational.

Answer 4

Having $r=p/q$ written in lowest terms, then $p^3/q^3$ is in lowest terms too. But $p^3/q^3=r^3=6/1$ and since the fractions can be written in lowest terms in exactly one way (with positive denominator), then $p^3=6$. Clearly no integer satisfies this equality.

Note: Showing the uniqueness of lowest terms invokes use of unique prime factorization, but it is commonly known that this holds for rational numbers.

Answer 5

The question is do you know the prime factorization theorem or not.

It's not significantly different than showing there is no rational $r$ so that $r^2 = 2$ but....

if $r = \frac mn; m,n \in \mathbb Z$ and $\frac mn$ is in "lowest terms" so that $r^3 = 6$ then $6n^3 = m^3$ which if we accept the PFT then the prime factors of $m^3$ are all to multiples of $3$ powers whereas the powers of $2$ and $3$ of $6n^3$ will be one more than multiples of $3$.

... which is how we prove that if $r^n = k \in \mathbb Z$ then $r$ is an integer are $k$ is a perfect $n$ power.

But if we can't use PFT, the replicating the old square root of $2$ prove. $m^3$ is even. If $m$ is odd then $m^3$ is odd. So $m$ is even. So $8|6n^3$ so $4|3n^3$ so $3n^2 $ is even. If $n$ is odd, so is $3n^3$ so $n$ is even. But that contradicts that $\frac mn$ was in "lowest terms".

And .... the factor of $3$ didn't really have much to do with it.

Of course, we could have used that $6$ is a multiple of $3$ rather than it was even: $6n^3 = m^3$ so $3|m^3$ if $m = 3k + i$ then $m^3 = 27k^3 + 27k^2 + 9k^3 + i^3$ and if $i = 1, 2$ then $3\not \mid i^3$ so $3|m$ so $27|6n^3$ and $3|2n^3$ and if $n = 3j +i$ then $2n^3 = 54j^3 + 54j^2 + 18k^3 + 2i$ and if $i= 1,2$ then $3 \not \mid 2i^3$ so $3|n$ and ... not in lowest terms.

It... gets pretty repetitive. Best learn the PFT once and for all.

Answer 6

Another way to attack such problems is using polynomials with integer coefficients and Rational root theorem (RTT). In this case the polynomial is $$P(x)=x^3-6$$ Now, let's assume $r \in \mathbb{Q}$. From RTT, $r$ as a solution of $P(x)$, can only be an integer (or irrational, but we skip this part since we assumed $r \in \mathbb{Q}$), since the leading coefficient of $P(x)$ is $1$, so $r \in \mathbb{Z}$. Now, we have an integer $r$ s.t. $$r^3=6=2\cdot 3 \tag{1}$$ or $3 \mid r\cdot r\cdot r$ which (from Euclid's lemma and since $3$ is prime) means $3 \mid r$. In another words, $\exists q\in \mathbb{Z}, q\ne0: r=3\cdot q$. Applying this to (1) we have $$3\cdot q \cdot 3\cdot q \cdot 3\cdot q=6 \Rightarrow 3^2\cdot q^3=2$$ which means $3 \mid 2$ - contradiction. Thus $r \notin \mathbb{Q}$.