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I am trying to follow the proof for the irrationality of $\sqrt[3]{6}$ to form a similar proof for $\sqrt[3]{16}$ (proof by contradiction). Going from $16=(a/b)^3$ (with $a,b \in \mathbb{Z}$ and $b\neq 0$ and GCD$(a,b)=1$) I can show that $2|a^3$ and then consequently $2|a \Rightarrow a=2c$.

But I'm unable to show $2|b$ following the step for $16b^3 = (2c)^3$ which leads to $2b^3 = c^3$ as this only shows that $c$ is even. I'm probably either making a mistake or overlooking a key fact. Any suggestions?

quid
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SymplecticForm
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    If $\sqrt[3]{16}=p/q$, then that implies $p^3=16q^3=(2q)^3+(2q)^3$, which contradicts Fermat's last theorem. – Arthur Aug 08 '16 at 22:11
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    If c is even then c = 2d and 2b^3 = 8d^3 so b^3 = 4d^3 so b is even. – fleablood Aug 08 '16 at 22:41
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    If the cube root was rational there would be no way to cancel the prime factors in the denominator with the prime factors in the numerator. Cubing does not help this situation. – Jacob Wakem Aug 08 '16 at 23:30
  • the proof here works: http://math.stackexchange.com/questions/1310014/what-is-the-most-rigorous-proof-of-the-irrationality-of-the-square-root-of-3 – Mark Joshi Aug 09 '16 at 00:03
  • @Arthur thanks, that is really neat. It's probably the first time I'll be using Fermat's Last Theorem! – SymplecticForm Aug 09 '16 at 15:21

5 Answers5

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The proof you want is a varaint of the Pythagorean proof of the irrationality of $\sqrt 2$

Proof by conradiction: Suppose $\frac pq$ is a fraction in lowest terms such that

$\frac pq = 16\\ p^3 = 16 q^3$

If this is true the $p$ must be even, which case we can say

$p = 2r\\ 8 r^3 = 16 q^3\\ r^3 = 2 q^3$

$r$ is also even

$r = 2s\\ 8s^3 = 2 q^3\\ 4 s^3 = q^3$

Which would imply that $q$ is even, but that would contradict our initial assumption that $\frac pq$ is in lowest terms.

Doug M
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  • "If this is true the $p$ must be even" Why? – AJJ Aug 08 '16 at 23:26
  • Because (odd)(odd)=(odd) which can be easily proven implies (odd)^3=odd @ArukaJ but we need $p^3=\text{even}$ – Ahmed S. Attaalla Aug 09 '16 at 00:10
  • @dougm thanks, so basically unlike the sqrt(2) case where p(even) led directly to q(even) and thus the contradiction, it had to be p(even) to r(even) to q(even). Re-application of the same method. Got it now. – SymplecticForm Aug 09 '16 at 15:13
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If $x=\sqrt[3]{16}$, we have $x^3-16=0$ . It is easy to check no rational roots exist by the rational root theorem.

Ahmed S. Attaalla
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  • Here you can find a proof of the theorem which is very useful for a lot of these irrationality problems https://www.artofproblemsolving.com/wiki/index.php?title=Rational_Root_Theorem. – Ahmed S. Attaalla Aug 08 '16 at 23:45
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"But I'm unable to show 2|b following the step for $16b^33=(2c)^3$ which leads to $2b^3=c^3$ as this only shows that c is even. "

So $c$ is even. So $c = 2d$ so $2b^3 = (2d)^3 = 8d^3$. So $b^3 = 4d^3$. So $4|b^3$ so $2|b^3$ so $2|b$.

fleablood
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Let $16^{1/3}=\frac ab$ where $gcd(a,b)=1$. Cube both sides and arrive at $16b^3=a^3, 2(2b)^3=a^3$ notice that on the right hand side in the prime factorisation the power of two is a multiple of 3 (e.g. $(2^pf)^3=2^{3p}f^3$ (where $f$ is relatively prime to 2)), whereas on the left hand side the power of 2 in the prime factorisation is of the form $3k+1$ which is a violation of the fundamental theorem of arithmetic and is hence a contradiction,

Aka_aka_aka_ak
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$x=16^{1/3}$ obviously isn't integral, since $2^3=8$, $3^3=27$, and the cube function is strictly increasing.

Let $k$ be the integer such that $k<x<k+1$

If $x$ were rational, then there would exist integers whose product with $x^2$ is also an integer. Let's denote by $n$ the smallest positive one.

Then, $n(x-k)$ is a positive integer smaller than $n$ which also verifies this property, which is a contradiction.

Therefore, $16^{1/3}$ is irrational.

Evariste
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