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I'm learning mathematical proofs, and here I'm trying to prove that $\sqrt{p}$ is irrational, where $p$ is a prime number.

Usually, the statement is proven by contradiction, assuming $\sqrt{p}$ is rational and there exist $m,n \in N$ without common factors such that $\sqrt{p}=\frac{m}{n}$, then showing that $p$ must divide both numerator and denominator, which contradicts the assumption that $m,n$ have no common factors.

Here, my proof is slightly different:

  1. Assume $p$ is a prime, $\sqrt{p}$ is rational.
  2. Then there must be $m,n \in N$ without common factors such that $\sqrt{p}=\frac{m}{n}$.
  3. Square both sides. $p=\frac{m^2}{n^2}$
  4. All prime numbers are integers. Therefore, $p$ is an integer.
  5. If $p$ is an integer and $m,n$ have no common factors, then the denominator ($n^2$) must be equal to $1$.
  6. If $n^2=1$, then $p=m^2$.
  7. But that contradicts the assumption that $p$ is a prime number. Therefore, the assumption that $\sqrt{p}$ is rational must be false.
  8. Hence, $\sqrt{p}$ is irrational.

I'm not sure whether the steps 5-6 are logically correct. Are they? Overall, is the proof valid? If not, why?

g00dds
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    The proof is correct. But better is to begin with $pn^2=m^2$ to conclude $p\mid m$ and to continue analogue to the standard proof that $\sqrt{2}$ is irrational to show that also $p\mid n$ which leads to a contradiction. – Peter Aug 07 '23 at 15:18
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    Point (5) is not properly justified. – NDB Aug 07 '23 at 15:19
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    Just use the fact that any natural number can be expressed uniquely as a product of prime powers so by looking at the equation $pn^2=m^2$ and noticing that the exponent of $p$, is odd on the left and even on the right, you have an immediate contradiction. – P. Lawrence Aug 07 '23 at 15:49
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    @NDB, how to properly justify step 5? Would the following be a proper justification? If $p$ is an integer, then $\frac{m^2}{n^2}$ is also an integer. Therefore, $n^2$ must evenly divide $m^2$. Since $m,n$ have no common factors, the only natural number that evenly divides $m^2$ is 1. Therefore, $n^2$ must be 1. – g00dds Aug 07 '23 at 16:01
  • @P.Lawrence This is also possible , I wanted to pointout an easier approach. – Peter Aug 07 '23 at 16:39
  • I think that it is obvious that a fraction in lowest terms being a non-zero integer must have denominator $1$ (or $-1$ which here is ruled out). To prove that rigorously , assume $p=\frac{s}{t}$ is a positive integer and $s,t$ are positive coprime integers. If we would have $t>1$ , $t$ would have some prime factor , say $q$. But then , $s=pt$ would also have this prime factor $q$ , contradiction. – Peter Aug 07 '23 at 16:47
  • @g00dds Yes, your argument to justify step 5 is correct. – John Douma Aug 07 '23 at 17:50

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