I need to prove the cube root is irrational. I followed the proof for the square source of $2$ however I ran right into a difficulty I wasn"t sure of. Below are my steps:

By contradiction, speak $ sqrt<3>2$ is rationalthen $ sqrt<3>2 = frac ab$ in the shortest form, wherein $a,b in altoalsimce.orgbbZ, b eq 0$$2b^3 = a^3 $$b^3 = fraca^32$therefore, $a^3$ is eventherefore, $2mid a^3$,therefore, $2mid a$$exists k in altoalsimce.orgbbZ, a = 2k$ below in: $2b^3 = (2k)^3$$b^3 = 4k^3$, as such $2|b$ Contradiction, $a$ and also $b$ have usual factor the two

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.


Thanks for contributing response to altoalsimce.orgematics Stack Exchange!

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

But avoid

Asking for help, clarification, or responding to various other answers.Making statements based upon opinion; back them up with referrals or personal experience.

Use altoalsimce.orgJax to style equations. altoalsimce.orgJax reference.

See more: What Is The Gcf Of 42 And 70, Methods To Find Gcf Of 42 And 70

To learn more, see our tips on writing an excellent answers.


short article Your prize Discard

By clicking “Post your Answer”, friend agree come our regards to service, privacy policy and also cookie plan


Not the answer you're looking for? Browse various other questions tagged discrete-altoalsimce.orgematics proof-verification or ask your own question.


Prove the complying with statement by prove its contrapositive: if $r$ is irrational, climate $ r ^ frac 1 5 $ is irrational
site design / logo design © 2021 ridge Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.11.5.40661


her privacy

By click “Accept every cookies”, friend agree stack Exchange have the right to store cookies on your device and disclose info in accordance with our Cookie Policy.