I need to prove the cube source is irrational. I followed the proof for the square root of $2$ yet I ran into a problem I wasn"t certain of. Below are mine steps:

By contradiction, to speak $ \sqrt<3>2$ is rationalthen $ \sqrt<3>2 = \frac ab$ in the shortest form, whereby $a,b \in \piersonforcongress.combbZ, b \neq 0$$2b^3 = a^3 $$b^3 = \fraca^32$therefore, $a^3$ is eventherefore, $2\mid a^3$,therefore, $2\mid a$$\exists k \in \piersonforcongress.combbZ, a = 2k$ sub in: $2b^3 = (2k)^3$$b^3 = 4k^3$, as such $2|b$ Contradiction, $a$ and $b$ have usual factor that two

My problem is with step 6 and 7. Have the right to I say the if $2\mid a^3$ , then $2\mid a$. If so, I"m gonna need to prove it. How??



This is not, probably, the most convincing or explanatory proof, and this absolutely does not answer the question, but I love this proof.

You are watching: Prove cube root of 3 is irrational

Suppose the $ \sqrt<3>2 = \frac ns q $. Then $ 2 q^3 = p^3 $. This method $ q^3 + q^3 = p^3 $. The critical equation has no nontrivial creature solutions because of Fermat"s last Theorem.


If $p$ is prime, and $p\mid a_1a_2\cdots a_n$ climate $p\mid a_i$ for part $i$.

Now, allow $p=2$, $n=3$ and $a_i=a$ for every $i$.


Your proof is fine, when you understand that action 6 indicates step 7:

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

Anyway, friend don"t need to assume the $a$ and also $b$ are coprime:

Consider $2b^3 = a^3$. Now count the number of factors of $2$ on each side: ~ above the left, you get an variety of the form $3n+1$, when on the ideal you obtain an a number of the type $3m$. This numbers cannot be equal because $3$ does not divide $1$.


The fundamental Theorem of Arithmetic tells united state that every hopeful integer $a$ has a distinctive factorization into primes $p_1^\alpha_1p_2^\alpha_2 \ldots p_n^\alpha_n$.

You have $ 2 \mid a^3$, so $2 \mid (p_1^\alpha_1p_2^\alpha_2 \ldots p_n^\alpha_n)^3 = p_1^3\alpha_1p_2^3\alpha_2 \ldots p_n^3\alpha_n$.

Since primes room numbers the are just divisible by 1 and themselves, and also 2 divides among them, among those primes (say, $p_1$) need to be $2$.

So we have actually $2 \mid a^3 = 2^3\alpha_1p_2^3\alpha_2 \ldots p_n^3\alpha_n$, and if you take the cube source of $a^3$ to obtain $a$, it"s $2^\alpha_1p_2^\alpha_2 \ldots p_n^\alpha_n$. This has a variable of 2 in it, and therefore it"s divisible by 2.

For the sake of contradiction, i think $ \sqrt<3>2$ is rational.

We can thus say $ \sqrt<3>2 = a/b$ wherein $a,b$ space integers, and $a$ and $b$ space coprime (i.e. $a/b$ is totally reduced).


$2b^3 = a^3$

Hence $a$ is an also integer.

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

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

$2b^3 = 8m^3$

$b^3 = 4m^3$

So $b$ is additionally even. This completes the contradiction where we presume $a$ and $b$ to be coprime.

Hence, $ \sqrt<3>2$ is irrational.

A different technique is making use of polynomials and the rational source theorem. Due to the fact that $\sqrt<3>2$ is a root of $f(x)=x^3-2$, the is enough to display that if $f(x)$ has no reasonable roots, climate $\sqrt<3>2$ is irrational.

By the rational source theorem, possible roots room $x=\pm 1$ 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 that these feasible rational roots space equal come zero, $\sqrt<3>2$ is irrational.

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