The n-th root of a prime number is irrational

Question

If $p$ is a prime number, how can I prove by contradiction that this equation $x^{n}=p$ doesn't admit solutions in $\mathbb {Q}$ where $n\ge2$

score 7 · Answer 1 · answered Oct 30 '13 at 11:13

7

Let $p=a^n/b^n$ as above, so $pb^n=a^n$. The number of the prime factors of $a^n$ is a multiple of $n$, whereas the number of prime factors of $pb^n$ isn't a multiple of $n$.

answered Oct 30 '13 at 11:13

Michael Hoppe

18,103
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Can anyone please elaborate on this? – user5826 Feb 03 '15 at 02:07
@AlJebr If $a=a_1\dots a_k$ is the prime decomposition of $a$ and $b=b_1\dots b_l$ that of $b$, then the prime decomposition of $a^n$ has $kn$ factors, and that of $b^n$ has $nl$ factors. Thus, $pb^n$ has $ln+1$ factors, while the other side of the equation has $kn$, which is a contradiction since $n$ divides the one but not the other (using uniqueness of prime decomposition). – Danu Jun 12 '17 at 11:06

Kortlek · Answer 2 · 2013-10-27T10:18:02.437

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Wlog, suppose $x=\frac{a}{b}$ such that $a, b\in \mathbb{Z}, b\neq 0$ and $(a,b)=1$. Then $pb^n=a^n$ so that $p|a$. Hence, we can write $a=a_1p$, $a_1\in \mathbb{Z}$. It follows that $pb^n=p^na_1^n$ so that $b^n=p^{n-1}a_1^n$. We see that $p|b$, which contradicts the fact that $(a,b)=1$.

edited Oct 27 '13 at 10:18

answered Oct 27 '13 at 10:12

Kortlek

1,103

I think you mean $a,b \in \mathbb{Z}$, not $\mathbb{Q}$. – Oliver Braun Oct 27 '13 at 10:13
@OliverBraun Thank you! – Kortlek Oct 27 '13 at 10:15

score 5 · Answer 3 · answered Oct 27 '13 at 10:18

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By Eisenstein's criterion, the polynomial $X^n-p\in\mathbb{Z}[X]$ is irreducible, so it's irreducible also in $\mathbb{Q}[X]$ by Gauss' lemma; therefore it can't have roots in $\mathbb{Q}$.

answered Oct 27 '13 at 10:18

egreg

238,574

score 0 · Answer 4 · edited Dec 25 '20 at 09:02

if possible, Let the n-th root of p prime number be rational. Therefore we can write p^1/n as
p^1/n = a/b
p = aⁿ/bⁿ
since a and b are coprime, aⁿ/bⁿ will be integer if bⁿ = 1
therefore, p = aⁿ .................................equation1
dividing both sides by a, we get:
p/a = a^n-1
Since a is integer, a^n-1 is also integer,
therefore, p/a is also integer. p is prime, so p/a is integer if a =1 or a=p
put 1 in equation1, we get a = 1 which is not possible. we then put a=p, and get n=1 but first root means the number itself. since there is no value of a/b for which p^1/nn is rational, it can be concluded that p^1/n is always irrational.

The n-th root of a prime number is irrational

4 Answers4