I should prove the cube root is irrational. I adhered to the proof for the square root of $2$ but I ran into a trouble I wasn"t certain of. Here are my steps:

By contradiction, say $ sqrt<3>2$ is rationalthen $ sqrt<3>2 = frac ab$ in the lowest create, wright here $a,b in slrfc.orgbbZ, b eq 0$$2b^3 = a^3 $$b^3 = fraca^32$therefore, $a^3$ is eventherefore, $2mid a^3$,therefore, $2mid a$$exists k in slrfc.orgbbZ, a = 2k$ sub in: $2b^3 = (2k)^3$$b^3 = 4k^3$, therefore $2|b$ Contradiction, $a$ and $b$ have actually prevalent element of two

My difficulty is with action 6 and 7. Can I say that if $2mid a^3$ , then $2mid a$. If so, I"m gonna need to prove it. How??



This is not, probably, the many convincing or explanatory proof, and this definitely does not answer the question, but I love this proof.

You are watching: Prove cube root of 3 is irrational

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


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

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


Your proof is fine, once you understand also that step 6 means action 7:

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

Anymethod, you don"t need to assume that $a$ and $b$ are coprime:

Consider $2b^3 = a^3$. Now count the number of components of $2$ on each side: on the left, you get an number of the create $3n+1$, while on the best you obtain an a number of the develop $3m$. These numbers cannot be equal bereason $3$ does not divide $1$.


The Fundapsychological Theorem of Arithmetic tells us that eexceptionally positive integer $a$ has a distinct factorization right into primes $p_1^alpha_1p_2^alpha_2 ldots p_n^alpha_n$.

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

Due to the fact that primes are numbers that are just divisible by 1 and also themselves, and also 2 divides one of them, one of those primes (say, $p_1$) have to be $2$.

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

For the sake of contradiction, assume $ sqrt<3>2$ is rational.

We have the right to therefore say $ sqrt<3>2 = a/b$ wright here $a,b$ are integers, and $a$ and $b$ are coprime (i.e. $a/b$ is totally reduced).


$2b^3 = a^3$

Hence $a$ is an also integer.

Like all also integers, we deserve 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 likewise even. This completes the contradiction wright here we assumed $a$ and also $b$ were coprime.

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

A various technique is making use of polynomials and the rational root theorem. Because $sqrt<3>2$ is a root of $f(x)=x^3-2$, it is enough to display that if $f(x)$ has no rational roots, then $sqrt<3>2$ is irrational.

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

Next off inspect 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 since none of these feasible rational roots are equal to zero, $sqrt<3>2$ is irrational.

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