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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)

MatheSt
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    For the first, do you not yet have access to the intermediate value theorem? For the second, do you know how to show that $\sqrt{2}$ is irrational (this is often one of the first proofs that students are taught in upper division classes)? The proof that $\sqrt[3]{3}$ is irrational is similar. – Xander Henderson Nov 18 '17 at 22:47
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    Try it by contradiction. Suppose there exists a rational and try to show that it cant happen. – Shri Nov 18 '17 at 22:49
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    Let $\frac ab$ be such that $a,b$ are integers and $(\frac ab)^3=3$ so $3b^3 = a^3$ so $3|a^3$.... is this looking familiar to you? – fleablood Nov 18 '17 at 22:51
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    Fun fact: If you had a value $\sqrt [n] {n}$ then the largest value of this form is obtained if you let $n = 3$ – Mr Pie Nov 18 '17 at 22:52
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    The very first time we prove the that $\sqrt 2$ is irrational we don't use the unique prime factorization as, in spirit, we are introducing the concept of the l.u.b. principal and the need for it. and the UPF undermines that in using the well-ordering principal. But can you use it now. If $3b^3 = a^3$ then the LHS is divisible by some power $3k+1$ of $3$ whereas the RHS is divisble by some power $3m$. $3m = 3k+1$ is not compatible. ... At any rate, proving irrational should not be the difficult part. – fleablood Nov 18 '17 at 23:03
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    Here is another set of examples as a source of inspiration https://math.stackexchange.com/questions/2482457/there-is-no-rational-number-r-with-the-property-r2-3/ or https://math.stackexchange.com/questions/2511028/show-that-there-are-no-rationals-r-such-that-r3-6/ – rtybase Nov 18 '17 at 23:14

3 Answers3

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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 $b$.

Duncan Ramage
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  • Ok: if $\frac ab$ is rational, we have $a \in \mathbb{Z}$ and $b \in \mathbb{N}$. With the definition of divisibility we have $a^3=3b^3 $and with that: $3|a^3$ and $3|a$. Since $a^3$ and $a$ are both divisible by 3, we can say that $a=3r$, so we have: $(3r)^3=3b^3 \to 27r^3=3b^3 \to 9r^3=b^3$. With that we see that b is also divisible by $3$, which is a contraidction. With that $x^3=3$ is irrational. – MatheSt Nov 19 '17 at 00:02
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    @MatheSt Perfect. – Duncan Ramage Nov 19 '17 at 04:00
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You don't justify with if $x^3<3$ then $x<3$; it is in fact very easy to check that if $x^3<3$, then $x<2$.

In case I, you want to deal with $s+\epsilon$, and not with $s-\epsilon$; otherwise, there is no contradiction.

I didn't check your inequalities carefully, but it should be something like that.

Regarding irrationality, you assume that $x$ is rational, write the equation, and you get a contradiction by looking at the prime decompositions.

Martin Argerami
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Alternative proof: assuming that the positive solution of $x^3=3$ is a rational number, we have that $x^3-3$ factors over $\mathbb{Q}$, hence it factors over any finite field $\mathbb{F}_p$. Let us consider $p=19$. The cubic residues $\!\!\pmod{19}$ are $0,\pm 1,\pm 7,\pm 8$. $3$ is not one of them, hence $x^3-3$ is irreducible over $\mathbb{F}_{19}$ and it is irreducible over $\mathbb{Q}$, too. It follows that $\sqrt[3]{3}\not\in\mathbb{Q}$.

The same approach for proving the irrationality of $\sqrt{2}$: $x^2-2$ is irreducible over $\mathbb{F}_7$, hence it is irreducible over $\mathbb{Q}$, too.

Jack D'Aurizio
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