# Proof of Easy Theorem?

I was reading the proof of this theorem and have a little trouble understanding one part of it:

Theorem: If $k > 2$ and $n$ are natural numbers, then $n^{\frac{1}{k}}$ is irrational unless $n$ is a perfect $k$th power.

Proof: Assume the contrary: $a^{k} = nb^{k}$, and some prime divisor of $n$ has an exponent that is not a multiple of $k$. Let $p$ be such a prime and note that the exponent of $p$ in $a^k$ is a multiple of $k$, but the exponent of $p$ in $b^{k}n$ is not a multiple of $k$. This violates the Fundamental Theorem of Arithmetic, so our assumption that $n$ is not a perfect kth power and $n^{\frac{1}{k}}$ is rational must be false.

The bold part is the part that is not very clear to me.

Thanks.

-

If all prime divisors of $n$ have an exponent wich is as multiple of $k$ we have that $n$ is a perfect $k$-power. We are assuming that the theorem is wrong, so that $n$ is not a $k$-power (i.e. there is a prime $p$ that have an exponent is not a multilple of $k$ ) but it is rational. The decomposition in prime numbers of $a^k$ and $nb^k$ is the same, so the total exponent of $p$ must be the same. In $a^k$ the exponent of $p$ is a multiple of $k$, instead in $nb^k$ it is the sum of the exponent of $p$ in $n$ (called $e_1$) and the exponent of $p$ in $b^k$ (called $e_2$ that can be also $0$). Clearly $e_1+e_2$ it is not divisible by $k$ because we are summing a multiple of $k$ and a number not divisible by $k$, so you have the contraddiction.

-
all the answers are good but this is most clearest thanks. –  mathnoob Oct 27 '12 at 14:18

If I understand correctly, in order to contradict the theorem, you need to find a number n such that its kth root is non-integer rational number. If that is the case, it can always be written in the form given. Also, for RHS, if some prime divisor of n was present with powers in multiple of k, it would be subsumed by b, hence it is safe to say that there exists some prime divisor of n whose power in n is not multiple of k. The proof then goes on to say that it is not possible if we consider LHS.

-

Suppose $\,n^{1/k}\,$ is rational, so taking a reduced fraction we get:

$$n^{1/k}=\sqrt[k] n=\frac{a}{b}\Longrightarrow b^kn=a^k$$

Let

$$\,n_1=\prod_{i=1}^rp_i^{\gamma_1}\;\;,\;a=\prod_{i=1}^sq_i^{\alpha_i}\;\;,\;b=\prod_{i=1}^tm_i^{\beta_i}$$

be the prime decompositions of the numbers $\,n,a,b\,$ ,so that

$$a^k=\prod_{i=1}^sq_i^{\alpha_ik}=b^kn=\prod_{i=1}^tm_i^{\beta_ik}\prod_{i=1}^rp_i^{\gamma_i}$$

Since every prime dividing the RHS also divides the LHS and the other way around, we get a straighforward contradiction...can you see why? Try to round and complete the argument now.

-

Suppose that $a^k=nb^k$. Let $a=p_1^{r_1}p_2^{r_2}\dots p_m^{r_m}$, where $p_1,\dots,p_m$ are distinct primes. Then $a^k=p_1^{kr_1}p_2^{kr_2}\dots p_m^{kr_m}$. This of course means that $nb^k=p_1^{kr_1}p_2^{kr_2}\dots p_m^{kr_m}$ as well.

Consider any prime factor $q$ of $nb^k$. Let $q^t$ be the highest power of $q$ that divides $n$, and let $q^s$ be the highest power of $q$ that divides $b$. Then the highest power of $q$ that divides $nb^k$ is $q^{t+ks}$: $nb^k$ gets $t$ factors of $q$ from $n$ and $s$ factors of $q$ from each of the $k$ factors of $b$ in $b^k$. By the fundamental theorem of arithmetic the prime factorization of $a^k=nb^k$ is unique, so $q^{t+sk}$ must be one of the factors $p_i^{kr_i}$, meaning that $q=p_i$ and $t+sk=kr_i$. But then $t=kr_i-ks=k(r_i-s)$ is a multiple of $k$.

In other words, if $n=q_1^{t_1}q_2^{t_2}\dots q_\ell^{t_\ell}$ is the prime factorization of $n$, then each $q_i$ is one of the primes $p_1,\dots,p_m$ and, more important, each $t_i$ is a multiple of $k$. Say $t_i=\alpha_ik_i$ for $i=1,\dots,\ell$; then $$n=q_1^{\alpha_1 t_1}q_2^{\alpha_2t_2}\dots q_\ell^{\alpha_\ell t_\ell}=\left(q_1^{\alpha_1}q_2^{\alpha_2}\dots q_\ell^{\alpha_\ell}\right)^k\;,$$

and $n$ is therefore a perfect $k$-th power.

The argument that you quoted is phrased as a proof by contradiction instead of a direct proof showing that if $n^{1/k}$ is rational, then $n$ is a perfect $k$-th power, but the reasoning is essentially the same.

-

Let the power of the prime $\rm\,p\,$ in $\rm\:A,B,N,\:$ be $\rm\:a,b,n\:$ resp. Comparing powers of $\rm\,p\,$ yields

$$\begin{eqnarray}\rm A^k &\,=\,&\rm N\times B^k\\ \Rightarrow\ \ \rm a\,k &\,=\,&\rm n\, +\, b\,k\\ \Rightarrow\quad\rm n &\,=\,&\rm (a\!-\!b)\,k \end{eqnarray}$$

So $\rm k\,$ divides $\rm\,n,\,$ i.e. all primes $\rm\,p\,$ dividing $\rm N$ occur to powers a multiple of $\rm k,\,$ so $\rm N$ is a $\rm\,k$-th power

$$\rm\, N\, =\, P^{\,JK}\cdots\, Q^{\,L K} =\, (P^{\,J}\cdots Q^{\,L})^K$$

Remark $\$ The argument depends crucially on the Fundamental Theorem of Arithmetic, i.e. the existence and uniqueness of prime factorizations of naturals.

-