# Prove that $2^{1/3}$ is irrational.

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!

• 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.
– lulu
Feb 23, 2021 at 0:48
• 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 $4\mid 2.$ Feb 23, 2021 at 1:08
• 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. Feb 23, 2021 at 1:09
• 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 $b$.
– lulu
Feb 23, 2021 at 1:13
• But if $4|b^3$, clearly $2|b^3$, then you can continue the proof. Feb 23, 2021 at 1:15

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

$$p \not \mid a \tag 0$$

then

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

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.