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If $n$ is a natural number then $n$ is a unique product of primes to integer powers

If $n$ is a perfect square then its prime factors will all be to even powers hence when taking the square root the results prime factors will be to integer powers hence the square root is an integer.

If $n$ is not a perfect square then at least one prime factor is to an odd power. Hence its square root will have a prime to a non integer power but all integers can be written as a product of primes to an integer power hence $\sqrt{n}$ cannot be an integer.

Is this sufficient to prove all square roots of non perfect squares is irrational?

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  • $\begingroup$ No, the product of irrational numbers could be rational. $\endgroup$ – Peter Jan 18 '19 at 9:30
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    $\begingroup$ @Peter Case in point: $\sqrt 5\times \sqrt 2 = \sqrt{10}$ which is irrational but $\sqrt 2\times \sqrt 8=\sqrt{16}=4$ which is not irrational $\endgroup$ – lioness99a Jan 18 '19 at 9:37
  • $\begingroup$ @lioness99a. Good examples. $\endgroup$ – DanielWainfleet Jan 18 '19 at 10:03
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I think there is one assumption that you are implicitly using: The prime factorization of $n$ could have more than one prime - let us say $p$ and $q$ - with an odd exponent. In this case you'd have to prove for instance that $p^{1/2} q^{1/2} \not \in \mathbb Z$.

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