The question came up in a conversation recently, and although it seemed intuitively true, I couldn't find a proof online. Many of you reading this are probably familar with the proof for the irrationality of √2. Well, this is a more general version for all

*n*integer roots of any prime.

**Theorem****Proof**

Euclid's first theorem states that for

*a*,*b*∈**Z**and any prime*p*:p|ab → p|a or p|b

An immediate corollary of it is, for

*a*,*1 ≤**n*∈**Z**and any prime*p*:p|a

^{n}→ p|aSince

p|a

^{n}= aa^{n-1}→ p|a or p|a^{n-1}but,

p|a

^{n-1}= aa^{n-2}→ p|a or p|a^{n-2}:

p|a

^{n-(n-2)}= a^{2}= aa → p|a or p|a→

**p|a**or**p|a**or**p|a**or...**p|a**(*n*times ) →**p|a*** * *

Now suppose for contradiction that

∃a,b,2≤n∈

**Z**, p∈**P**(p^{1/n}= a / b), where*a*nad*b*have no common factor.p

^{1/n}= a / b → p = a^{n}/ b^{n}→ pb

^{n}= a^{n}→ p|a^{n}by def. of*divides*, since b^{n}∈**Z**→ p|a by Corollary to Euclid's first theorem.

Hence

**a = pk**, k∈**Z**by def. of p|a. (1)But p

^{1/n}= a / b → a = p^{1/n}b = pk → pb^{n}= p^{n}k^{n}→ b

^{n}= p( p^{n-2}k^{n}) → p|b^{n}by def. of*divides*, since p^{n-2}k^{n}∈**Z**→ p|b by Corollary to Euclid's first theorem.

Hence

**b = pm**, m∈**Z**by def. of p|b. (2)Statements (1) and (2) jointly contradict the statement that "

*a*and*b*have no common factor".Hence ∀a,b,2≤n∈

**Z**, p∈**P**(p^{1/n}≠ a / b) q.e.d.
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