My trouble is with action 6 and 7. Have the right to I say that if $2mid a^3$ , then $2mid a$. If so, I"m gonna have 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 p q $. Climate $ 2 q^3 = p^3 $. This method $ q^3 + q^3 = p^3 $. The critical equation has actually no nontrivial integer solutions as result of Fermat"s critical Theorem.

If $p$ is prime, and also $pmid a_1a_2cdots a_n$ climate $pmid a_i$ for some $i$.

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

Your proof is fine, when you recognize that action 6 means step 7:

This is just the truth *odd* $ imes$ *odd* $=$ *odd*. (If $a$ to be odd, then $a^3$ would certainly be odd.)

Anyway, girlfriend don"t *need* to assume the $a$ and $b$ space coprime:

Consider $2b^3 = a^3$. Currently count the variety of factors of $2$ on each side: on the left, you obtain an number of the type $3n+1$, when on the appropriate you acquire an a variety of the kind $3m$. This numbers can not be equal because $3$ does no divide $1$.

The fundamental Theorem the Arithmetic tells us that every confident integer $a$ has actually a distinct factorization into primes $p_1^alpha_1p_2^alpha_2 ldots p_n^alpha_n$.

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

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

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

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

We can because of this say $ sqrt<3>2 = a/b$ whereby $a,b$ room integers, and also $a$ and also $b$ are coprime (i.e. $a/b$ is completely reduced).

2=$a^3/b^3$

$2b^3 = a^3$

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 where we assumed $a$ and $b$ to be coprime.

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

A different strategy is using polynomials and also the rational root theorem. Since $sqrt<3>2$ is a root of $f(x)=x^3-2$, the is sufficient to show that if $f(x)$ has no reasonable roots, climate $sqrt<3>2$ is irrational.

By the rational source theorem, possible roots are $x=pm 1$ or $x=pm2$

Next check that $f(-2)$, $f(-1)$, $f(1)$, $f(2)$ ,$ ot= 0$

$$f(-2)=-10 ot= 0$$$$f(-1)=-3 ot= 0$$$$f(1)=-1 ot= 0$$$$f(2)=6 ot= 0$$

So because none that these possible rational roots space equal come zero, $sqrt<3>2$ is irrational.

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