My trouble is with step 6 and 7. Deserve to I say the if $2\mid a^3$ , climate $2\mid a$. If so, I"m gonna need to prove it. How??
The fundamental Theorem the Arithmetic tells us that every hopeful integer $a$ has actually a distinct factorization into primes $p_1^\alpha_1p_2^\alpha_2 \ldots p_n^\alpha_n$.
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You have $ 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 are numbers that are just divisible through 1 and also themselves, and also 2 divides among them, among those primes (say, $p_1$) must 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 also if you take the cube source of $a^3$ to gain $a$, 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.
This is not, probably, the many convincing or explanatory proof, and also this definitely does no answer the question, yet I love this proof.
Suppose the $ \sqrt<3>2 = \frac ns q $. Then $ 2 q^3 = p^3 $. This method $ q^3 + q^3 = p^3 $. The last equation has actually no nontrivial integer solutions due to Fermat"s critical Theorem.
If $p$ is prime, and also $p\mid a_1a_2\cdots a_n$ then $p\mid a_i$ for some $i$.
Now, let $p=2$, $n=3$ and also $a_i=a$ for every $i$.
Your proof is fine, when you know that action 6 suggests step 7:
This is just the fact odd $\times$ odd $=$ odd. (If $a$ to be odd, then $a^3$ would be odd.)
Anyway, girlfriend don"t need to assume the $a$ and also $b$ room coprime:
Consider $2b^3 = a^3$. Currently count the variety of factors of $2$ on each side: ~ above the left, you obtain an variety of the form $3n+1$, if on the ideal you get an a number of the type $3m$. These numbers cannot be equal since $3$ does not divide $1$.
For the benefits of contradiction, assume $ \sqrt<3>2$ is rational.
We can thus say $ \sqrt<3>2 = a/b$ whereby $a,b$ room integers, and also $a$ and also $b$ room coprime (i.e. $a/b$ is totally reduced).
$2b^3 = a^3$
Hence $a$ is an also integer.
Like all also integers, we have the right to say $a=2m$ where $m$ is one 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 presume $a$ and also $b$ to be coprime.
Hence, $ \sqrt<3>2$ is irrational.
A different method is utilizing polynomials and also the rational root theorem. Due to the fact that $\sqrt<3>2$ is a source of $f(x)=x^3-2$, it is enough to present that if $f(x)$ has actually no rational roots, then $\sqrt<3>2$ is irrational.
By the rational root theorem, possible 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 to zero, $\sqrt<3>2$ is irrational.
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