A positive real number $x$ with the property $x^3=3$ is irrational. I have the following problems:
1) There exists a positive real number $x$ such that $x^3=3$.
2) A positive real number $x$ with the property $x^3=3$ is irrational.
My Idea for 1) would be (there might be a few mistakes here):
Let $S = \{ x  \geq  0 |  x^3 \leq  3\}$, with $1\ge0$ and $1^3=1$ we have $1 \in S$, so $S \neq \emptyset$. With that we have $\forall x \in S, x < 3$, so S is bounded above. With the upper bound property of real numbers, $S$ has a least upper bound $s$: 
$s=sup(S)$.
Since $1$ is in $S $, we know that $s>1$. 
Now $s$ either is the solution, or one of the follwing two cases are true: 
I) $s^3<3$
Let: $\varepsilon = \frac{3-s^3}{3s+1}$. By assumption $0<\varepsilon<1$, so that:
$(s+\varepsilon)^3=s^3+3s^2\varepsilon+3s\varepsilon^2+\varepsilon^3 \le s^3+3s^2\varepsilon+3s\varepsilon^2+\varepsilon^2=s^3+\frac{3-s^3}{3s+1}(3s+1)=3$.
Hence, $s + \varepsilon$  is also in $S$, in which case $s$ can not be an upper bound for $S$. This is a contradiction, so this case is not possible. 
II) $s^3>3$
Let: $\varepsilon = \frac{s^3-3}{3s}$. Again $\varepsilon>0$, so that:
$(s-\varepsilon)^3=s^3-3s^2\varepsilon+3s\varepsilon^2-\varepsilon^3\ge s^3-3s^2\varepsilon+3s\varepsilon^2=s^3-3s\frac{s^3-3}{3s}=3.$
Hence, $s -\varepsilon$  is another upper bound for $S$, so that $s$ is not the least upper bound for $S$. This is a contradiction, so that this case is not possible.
Having eliminated these two cases, we are left with $s^3 = 3$, which is what we wanted to prove.
2) However I don't know how to proof that a positive real number with the property $x^3=3$ is irrational. It would be really nice if someone could help!
Edit: Made a correction regarding $(s+\varepsilon)^3$ and $(s-\varepsilon)^3$ (Hope this is correct)

To prove it's irrational, proceed just like in the proof that $\sqrt{2}$ is irrational. Assume there are integers, in lowest terms, such that $\frac{a^3}{b^3} = 3$. So, $a^3 = 3b^3$. Show that $3$ must divide both $a$ and $b$. 
