I must prove the cube root is irrational. I followed the proof because that the square source of $2$ yet I ran into a trouble I wasn"t certain of. Right here are mine steps:

By contradiction, say $\sqrt<3>2$ is rationalthen $\sqrt<3>2 = \frac ab$ in the lowest form, where $a,b \in \yellowcomic.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$$\exists k \in \yellowcomic.combbZ, a = 2k below in: 2b^3 = (2k)^3$$b^3 = 4k^3$, as such $2|b$ Contradiction, $a$ and also $b$ have typical factor of two

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

This is not, probably, the most convincing or explanatory proof, and also this definitely does no answer the question, but I love this proof.

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Suppose that $\sqrt<3>2 = \frac p q$. Then $2 q^3 = p^3$. This method $q^3 + q^3 = p^3$. The last equation has actually no nontrivial creature solutions as result of Fermat"s last Theorem.

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

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

Your proof is fine, once you understand that step 6 suggests step 7:

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

Anyway, friend 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: top top the left, you obtain an number of the form $3n+1$, while on the best you acquire an a number of the type $3m$. These numbers can not be equal due to the fact that $3$ does not divide $1$.

The basic Theorem the Arithmetic tells us that every hopeful integer $a$ has a distinctive factorization right into primes $p_1^\alpha_1p_2^\alpha_2 \ldots p_n^\alpha_n$.

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

Since primes space numbers that are just divisible through 1 and also themselves, and also 2 divides one of them, one of those primes (say, $p_1$) need to be $2$.

So we have $2 \mid a^3 = 2^3\alpha_1p_2^3\alpha_2 \ldots p_n^3\alpha_n$, and if you take it the cube source of $a^3$ to acquire $a$, it"s $2^\alpha_1p_2^\alpha_2 \ldots p_n^\alpha_n$. This has actually a factor 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 therefore say $\sqrt<3>2 = a/b$ whereby $a,b$ space integers, and $a$ and $b$ room 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 have the right to 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 whereby we assumed $a$ and $b$ were coprime.

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

A different method is using polynomials and also the rational source theorem. Due to the fact that $\sqrt<3>2$ is a source of $f(x)=x^3-2$, the is sufficient to display that if $f(x)$ has actually no rational roots, climate $\sqrt<3>2$ is irrational.

By the rational root theorem, feasible roots space $x=\pm 1$ or $x=\pm2$

Next examine 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 the these feasible rational roots are equal come zero, $\sqrt<3>2$ is irrational.

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Prove the adhering to statement by proving its contrapositive: if $r$ is irrational, then $r ^ \frac 1 5$ is irrational