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).

2=$a^3/b^3$

$2b^3 = a^3$

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.

Thanks because that contributing an answer to piersonforcongress.comematics ridge Exchange!

Please be certain to answer the question. Carry out details and share her research!

But avoid

Asking for help, clarification, or responding to other answers.Making statements based on opinion; ago them increase with referrals or an individual experience.

Use piersonforcongress.comJax to format equations. Piersonforcongress.comJax reference.

See more: What Numbers Are Less Than -2, Negative Numbers & Greater Than And Less Than

To discover more, check out our tips on writing an excellent answers.

Not the answer you're looking for? Browse other questions tagged discrete-piersonforcongress.comematics proof-verification or ask your very own question.

Prove the complying with statement by proving its contrapositive: if $r$ is irrational, then $r ^ \frac 1 5$ is irrational
site architecture / logo © 2021 ridge Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.10.15.40479