# Prove the cuberoot of 2 is irrational

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 $\sqrt{2}$ is rational
2. then $\sqrt{2} = \frac ab$ 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??

• 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. Feb 12, 2015 at 17:13
• it is true.you can use factorization of a or euclid lemma:if p|ab then p|a or p|b(p is prime.
– ali
Feb 12, 2015 at 17:14
• 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. Feb 12, 2015 at 17:15
• @AlexSilva haha, thanks. gotta love basic math mistakes Feb 12, 2015 at 17:20
• 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. Feb 12, 2015 at 18:11

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 $\sqrt{2} = \frac p q$. 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.

• And yet FLT isn't strong enough to prove the irrationality of $\sqrt 2$! Feb 12, 2015 at 17:36
• Are you sure this isn't circular? Feb 12, 2015 at 20:34
• @FengyangWang nice note. I have no idea :( Feb 12, 2015 at 21:13
• @FengyangWang It is circular. Feb 13, 2015 at 4:22
• The general proof is, but I think there are proofs for n=3 that don't require it for 2. Might for 3 though Jul 16, 2019 at 5:22

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 $i$.

• but how and why can I say that p|ai for some i? Feb 12, 2015 at 17:31
• 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 Feb 12, 2015 at 17:34
• 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. Feb 12, 2015 at 17:35
• 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 $2|a$ Feb 12, 2015 at 17:45
• @Ashley $2\mid a^3$ is the same as $2\mid a\cdot a\cdot a$. Feb 12, 2015 at 17:46

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 $1$.

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.

• 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? Feb 12, 2015 at 17:37
• 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. Feb 12, 2015 at 17:40
• @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. Feb 12, 2015 at 17:54

For the sake of contradiction, assume $$\sqrt{2}$$ is rational.

We can therefore say $$\sqrt{2} = a/b$$ where $$a,b$$ are integers, and $$a$$ and $$b$$ are coprime (i.e. $$a/b$$ is fully reduced).

2=$$a^{3}/b^{3}$$

$$2b^{3} = a^{3}$$

Hence $$a$$ is an even integer.

Like all even integers, we can say $$a=2m$$ where $$m$$ is an integer.

2$$b^{3} = (2m)^{3}$$

$$2b^{3} = 8m^{3}$$

$$b^{3} = 4m^{3}$$

So $$b$$ is also even. This completes the contradiction where we assumed $$a$$ and $$b$$ were coprime.

Hence, $$\sqrt{2}$$ is irrational.

A different approach is using polynomials and the rational root theorem. Since $$\sqrt2$$ is a root of $$f(x)=x^3-2$$, it is enough to show that if $$f(x)$$ has no rational roots, then $$\sqrt2$$ is irrational.

By the rational root theorem, possible roots are $$x=\pm 1$$ or $$x=\pm2$$

Next check that $$f(-2)$$, $$f(-1)$$, $$f(1)$$, $$f(2)$$ ,$$\not= 0$$

$$f(-2)=-10\not= 0$$ $$f(-1)=-3\not= 0$$ $$f(1)=-1\not= 0$$ $$f(2)=6\not= 0$$

So since none of these possible rational roots are equal to zero, $$\sqrt2$$ is irrational.