Proof. Assume by way of contradiction that $\sqrt[3] 2$ is rational. Then $\sqrt[3] 2$ can be written as $a/b$ where $ a,b\in\mathbb{N},b\neq0$. Then $(a^3/b^3 = 2) = (a^3 = 2b^3)$. Thus $2|a^3$ and by the lemma, $2|a$. Then $a=2x $ where $x\in\mathbb{Z} $. Substituting $a=2x$ into $a^3 = 2b^3$ we get $8x^3 = 2b^3$, simplifying we get $4x^3 = b^3$. Which implies $4|b^3$ and then by the lemma, $4|b$. This is a contradiction because if a is a multiple of $2$ and $b$ is multiple of $4$, then they are not in lowest terms. Hence $\sqrt[3] 2$ is irrational.

I think I am good up til the $2|a^3$ part. I don't know if I can say that, "by the lemma $2|a$". Any tips or help is much appreciated!

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    $\begingroup$ If you like you can point that, were $a=2b+1$ odd, we'd have, $a^3=(2b+1)^3=8b^3+12b^2+6b+1$ which is also odd. $\endgroup$
    – lulu
    Feb 23, 2021 at 0:48
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    $\begingroup$ You can’t conclude that if $4\mid b^3$ then $4\mid b.$ For example, $b=2,$ $4\mid 2^3$ but we don’t have $4\mid 2.$ $\endgroup$ Feb 23, 2021 at 1:08
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    $\begingroup$ well, I think it is fine the part of $a$, by the comment above you can argue it can't be odd, I don't know which lema it is, but it is wrong to conclude $4|b$. Counter-example: $4|8$, $8= 2^3$, but 4 do not divide 2. $\endgroup$ Feb 23, 2021 at 1:09
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    $\begingroup$ Note: in my prior comment I should not have used the variable $b$, as you already have that as your supposed denominator. If $a$ is odd then we can write $a=2n+1$ for some integer $n$ and then the argument goes as before, with $n$ in place of $b$. $\endgroup$
    – lulu
    Feb 23, 2021 at 1:13
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    $\begingroup$ But if $4|b^3$, clearly $2|b^3$, then you can continue the proof. $\endgroup$ Feb 23, 2021 at 1:15

2 Answers 2


Since $2|a^3$, then $2|a$ because $2$ is prime (Euclid's Lemma).

Later on when you say $4|b^3$, you can argue that this implies $2|b^3$.

By the same argument as before, $2|b$ (as mentioned in the comments). This contradicts that $\text{gcd}(a,b)=1$ (you will need to state this as well in your proof because the fraction $\dfrac{a}{b}$ must be reduced to lowest terms).


The present problem may be resolved if we can show that for primes $p$ with $p \mid a^n$, $n \ge 1$, then $p \mid a$.

We note that if

$p \not \mid a \tag 0$


$\gcd(p, a) = 1; \tag 1$

for the only divisors of $p$ are $p$ and $1$; and if $p \not \mid a$, we are left with $1$ as the sole common divisor of $p$ and $a$; now in light of (1) we have, via Bezout's identity,

$xp + ya = 1, \tag 2$

for some

$x, y \in \Bbb Z; \tag 3$

multiplying (2) by $a^{n - 1}$ yields

$xpa^{n - 1} + ya^n = a^{n - 1}, \tag 4$

which since $p \mid a^n$ yields

$p \mid a^{n - 1}; \tag 5$

we now repeat the argument, replacing $a^{n - 1}$ with $a^{n - 2}$; then from (2),

$xpa^{n - 2} + ya^{n - 1} = a^{n - 2}; \tag 6$

now, again by virtue of (4) we find

$p \mid a^{n - 2}; \tag 7$

continuing in this manner we reach

$p \mid a \tag 8$

after $n - 1$ repititions. But this stands in contradiction to (0), which must thus be false; hence (8) binds.


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