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I need to prove the cube root is irrational. I followed the proof for the square root of $2$ but I ran into a problem I wasn't sure of. Here are my steps:

  1. By contradiction, say $ \sqrt[3]{2}$ is rational
  2. then $ \sqrt[3]{2} = \frac ab$ in the lowest form, where $a,b \in \mathbb{Z}, b \neq 0$
  3. $2b^3 = a^3 $
  4. $b^3 = \frac{a^3}{2}$
  5. therefore, $a^3$ is even
  6. therefore, $2\mid a^3$,
  7. therefore, $2\mid a$
  8. $\exists k \in \mathbb{Z}, a = 2k$
  9. sub in: $2b^3 = (2k)^3$
  10. $b^3 = 4k^3$, therefore $2|b$
  11. Contradiction, $a$ and $b$ have common factor of two

My problem is with step 6 and 7. Can I say that if $2\mid a^3$ , then $2\mid a$. If so, I'm gonna have to prove it. How??

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    $\begingroup$ The steps are fine. Think the other way round: if $a$ were odd, the $a^{3}$ would still be odd- check it out, write $a = 2c+1$ and see what happens. $\endgroup$ – Geoff Robinson Feb 12 '15 at 17:13
  • $\begingroup$ it is true.you can use factorization of a or euclid lemma:if p|ab then p|a or p|b(p is prime. $\endgroup$ – ali Feb 12 '15 at 17:14
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    $\begingroup$ Suppose to the contrary that $2$ does not divide $a$. So $a$ is odd, say $a=2t+1$. Then $a^3=8t^3+12t^2+6t+1$, odd. Or else we can use more fancy stuff, if a prime divides a product, it divides one of the terms. $\endgroup$ – André Nicolas Feb 12 '15 at 17:15
  • $\begingroup$ @AlexSilva haha, thanks. gotta love basic math mistakes $\endgroup$ – Ashley Feb 12 '15 at 17:20
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    $\begingroup$ Did nobody answering this read beyond the title? The question asker was asking about a specific step in a specific proof, not for a déluge of alternative proofs... Not that offering an alternative proof is entirely irrelevant or useless, but come on, guys. $\endgroup$ – Jack M Feb 12 '15 at 18:11
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The Fundamental Theorem of Arithmetic tells us that every positive integer $a$ has a unique factorization into primes $p_1^{\alpha_1}p_2^{\alpha_2} \ldots p_n^{\alpha_n}$.

You have $ 2 \mid a^3$, so $2 \mid (p_1^{\alpha_1}p_2^{\alpha_2} \ldots p_n^{\alpha_n})^3 = p_1^{3\alpha_1}p_2^{3\alpha_2} \ldots p_n^{3\alpha_n}$.

Since primes are numbers that are only divisible by 1 and themselves, and 2 divides one of them, one of those primes (say, $p_1$) must be $2$.

So we have $2 \mid a^3 = 2^{3\alpha_1}p_2^{3\alpha_2} \ldots p_n^{3\alpha_n}$, and if you take the cube root of $a^3$ to get $a$, it's $2^{\alpha_1}p_2^{\alpha_2} \ldots p_n^{\alpha_n}$. This has a factor of 2 in it, and therefore it's divisible by 2.

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  • $\begingroup$ But doesn't that just show me that a^3 has a factor of 3? I know that. But does that mean a has a factor of 2? $\endgroup$ – Ashley Feb 12 '15 at 17:37
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    $\begingroup$ This is a bit circular: proving the uniqueness part of the Fundamental Theorem of Arithmetic requires knowing beforehand that if a prime divides a product then it divides at least one of the factors, which is what the OP is asking about. $\endgroup$ – Santiago Canez Feb 12 '15 at 17:40
  • $\begingroup$ @Ashley There's no reason that $a^3$ should have a factor of $3$. For instance, $2^3$ is $8$, and that does not have a factor of $3$. I'll edit the post so maybe it's clearer. $\endgroup$ – NoName Feb 12 '15 at 17:54
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This is not, probably, the most convincing or explanatory proof, and this certainly does not answer the question, but I love this proof.

Suppose that $ \sqrt[3]{2} = \frac p q $. Then $ 2 q^3 = p^3 $. This means $ q^3 + q^3 = p^3 $. The last equation has no nontrivial integer solutions due to Fermat's Last Theorem.

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    $\begingroup$ And yet FLT isn't strong enough to prove the irrationality of $\sqrt 2$! $\endgroup$ – wchargin Feb 12 '15 at 17:36
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    $\begingroup$ Are you sure this isn't circular? $\endgroup$ – Fengyang Wang Feb 12 '15 at 20:34
  • $\begingroup$ @FengyangWang nice note. I have no idea :( $\endgroup$ – lisyarus Feb 12 '15 at 21:13
  • $\begingroup$ @FengyangWang but even if if is cicular, there is no problem in proving it another way, and then using Fermat's Theorem. $\endgroup$ – lisyarus Feb 12 '15 at 21:43
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    $\begingroup$ @FengyangWang It is circular. $\endgroup$ – apnorton Feb 13 '15 at 4:22
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If $p$ is prime, and $p\mid a_1a_2\cdots a_n$ then $p\mid a_i$ for some $i$.

Now, let $p=2$, $n=3$ and $a_i=a$ for all $i$.

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  • $\begingroup$ but how and why can I say that p|ai for some i? $\endgroup$ – Ashley Feb 12 '15 at 17:31
  • $\begingroup$ There is a theorem that says if $\gcd(a,b)=1$ and $a\mid bc$ then $a\mid c$. This is a lemma to unique factorization. In particular, then, if $p\mid bc$, then $p\mid b$ or $p\mid c$. Use induction from there to privethe above. @Ashley $\endgroup$ – Thomas Andrews Feb 12 '15 at 17:34
  • $\begingroup$ Or you can see the result as a consequence of unique factorization - if none of the $a_i$ is divisible by $p$, then the prime factorization of $a_1\cdot a_n$ does not contain $p$, and so $p$ cannot be a divisor. But I prefer the method in the previous comment. $\endgroup$ – Thomas Andrews Feb 12 '15 at 17:35
  • $\begingroup$ So could I say $2|a^3 = 2|a1*a2*a3$, therefore $2|ai$ for some $i$, and $ai = a for all i$ due to the nature of powers, therefore $2|a$ $\endgroup$ – Ashley Feb 12 '15 at 17:45
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    $\begingroup$ @Ashley $2\mid a^3$ is the same as $2\mid a\cdot a\cdot a$. $\endgroup$ – Thomas Andrews Feb 12 '15 at 17:46
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Your proof is fine, once you understand that step 6 implies step 7:

This is simply the fact odd $\times$ odd $=$ odd. (If $a$ were odd, then $a^3$ would be odd.)

Anyway, you don't need to assume that $a$ and $b$ are coprime:

Consider $2b^3 = a^3$. Now count the number of factors of $2$ on each side: on the left, you get an number of the form $3n+1$, while on the right you get an a number of the form $3m$. These numbers cannot be equal because $3$ does not divide $1$.

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