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.

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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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