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).

2=\$a^3/b^3\$

\$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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