How to show that there is no positive rational number a such that $a^3$ = 2? I'm stuck on this one.
So far, I've tried this:
\begin{align}
a^3 &= 2 \\
a &= \left(\frac mn\right) \\ 
a^3 &= \left(\frac mn\right)^3 = 2 \\
m^3 &= 2n^3 \\
m &= 2p \\
m^3 &= (2p)^3
\end{align}
I'm really confused about what to do after this—the book answer says that $m^3$ becomes $m^3 = 2(4p)^3$. And $2n^3 = 2(4p)^3$. I'm super confused here and can't really understand what's going on here.
Can someone tell what exactly is going on here and how should I prove this?
 A: Your proof is on the right lines but there are no words in it. You need to explain what you are doing to your readers. Here is how I would prove this:

*

*Assume, for the sake of contradiction, that there is a rational number $a$ such that $a^3=2$.

*Since $a$ is rational, we can write it in the form $m/n$, where $m$ and $n$ are coprime integers.

*This means that $$\frac{m^3}{n^3}=2 \, .$$

*From this we see that $m^3=2n^3$, and so $m^3$ is even. That implies that $m$ is even, i.e. it can be written in the form $2p$, where $p$ is an integer. It follows that

*$$\frac{8p^3}{n^3}=2$$
which implies $2n^3=8p^3$ and so $n^3=4p^3=2(2p^3)$. Since $n^3$ is even, $n$ must also be even. But earlier we said that $m$ and $n$ are coprime, which is a contradiction. Hence, our original assumption—that there is a positive rational number $a$ such that $a^3=2$—must be false, meaning that no rational number satisfies the desired property.

A: Note that $m^3 = 2n^3$ implies that $m$ is even (is it clear why?). So you can take $m=2p$ for some $p$. Therefore:
$$
m^3 = 2n^3 \implies (2p)^3 = 2n^3 \implies 4p^3 = n^3.
$$
Because of the same reason, $n=2q$ for some $q$. Finally, use that $m$ and $n$ do not have common factors (you did not mention it in your question, but that is a key point).
A: Alternative route:
Let $\alpha^3=2$, and assume $\alpha\in\Bbb Q$, then we have $$\frac{p^3}{q^3}=2\\p^3=2q^3\\p^3=q^3+q^3$$
The last line contradicts Fermat's Last Theorem.


Let $x,y,z$ be non-zero integers. Then if $x^3+y^3=z^3$ we have $xyz=0$.

Proof: We can assume that $x,y,z$ are relatively prime, since if they were not we could divide by $\gcd(x,y,z)^3$.
Reducing the equation modulo $9$, and noticing that the only cubes are $0,-1,1$ we conclude that exactly one of $x,y,z$ is divisible by $9$. Without loss of generality we may assume that $9\mid z$.
We can factor the equation is question as $$z^3=(x+y)(x^2-xy+y^2)$$
Since $9\mid z^3$ we have that $3$ divides one of the factors $x+y, x^2-xy+y^2$. In fact, $3$ divides both of them as $$x^2-xy+y^2=(x+y)^2-3xy$$ That is, $3\mid x+y\iff 3\mid x^2-xy+y^2$. Since $3^3\mid z^3$ one of these factors are divisible by $9$ which will yield a contradiction:

*

*If $9\mid x^2-xy+y^2$ then $9\mid (x+y)^2-3xy$, so $9\mid 3xy$ which is impossible, as $x,y$ are assumed not to be divisible by $3$.


*If $9\mid x+y$, let $k$ be the highest power such that $3^k\mid x+y$, $k\geq 2$. Then we have $$3^{3k}\mid (x+y)^3=x^3+y^3+3xy(x+y)$$ As $3^{3k}\mid x^3+y^3=z^3$, we have $$3^{3k}\mid 3xy(x+y)$$ This is impossible, since then $3^{2k-1}\mid xy$, and $2k-1\geq 3$.
A: Assume that $a=p/q$, where $p$ and $q$ are not both even (or make the stronger assumption that $\gcd(p,q)=1$), and that $a^3=2$.  Then $(p/q)^3=2$, or $p^3=2q^3$.  Since the right-hand side is even, the left-hand side must be as well; and since the cube of an odd number is odd, $p$ must be even.  So $p=2p'$ for some integer $p'$.  Now $8(p')^3=2q^3$, or $q^3=4(p')^3$, and hence $q$ must be even as well.  This contradicts the original assumption, completing the proof.
A: Assume $ a=\frac pq $ with $gcd(p,q)=1$
$$p^3=2q^3\implies$$
$$(p-q)(p^2+pq+q^2)=q^3 \implies$$
$$q|( p^2+pq+q^2)\implies$$
$$q|p$$
A: This takes some setting up but, once that's done, the actual proof is pretty simple and easily visualized.
Let $\{p_1, p_2, p_3, \dots \}$ be the set of all positive prime numbers. We will assume that the $p_i$ are listed in increasing order.
Then every integer $n > 1$ can be expressed, uniquely, as an infinite product
$$n = \prod_{i=1}^\infty p_i^{\nu_i}$$
where the $\nu_i$ are non negative integers and only a finite number of them are not equal to $0$.
For example, $45 = 2^0 \cdot 3^2 \cdot 5^1 \cdot 7^0 \cdot 11^0 \cdot 13^0 \cdots$
Suppose

*

*$m$ and $n$ are positive integers

*$m = \prod_{i=1}^\infty p_i^{\mu_i}$

*$n = \prod_{i=1}^\infty p_i^{\nu_i}$

*$m^3 = 2 n^3$
The powers of $2$ in $m^3$ are $2^{3\mu_1}$.
The powers of $2$ in $2n^3$ are $2^{3\nu_1+1}$.
Hence $2^{3\mu_1} = 2^{3\nu_1+1}$. Hence $3\mu_1 = 3\nu_1+1$.
There do not exists integers $\mu_1$ and $\nu_1$
that will make $3\mu_1 = 3\nu_1+1$ true.
Hence $\sqrt[3] 2$ is an irrational number.
