Check my proof that sqrt{prime} is always irrational

The proof that is irrational is well-known and I never did the proof that extended this to all primes, so I will attempt this now.

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Assume towards a contradiction that . Thus , where and this fraction is in lowest terms.

By algebra, . If is not an integer, then clearly n is not an integer. This isn't possible by definition of n, so must be an integer.

Given this conclusion, p divided m^2 evenly. This could lead to two possibilities. (1) m=p and (2)m contains a factor of p. Since p is prime the only factor other than 1 is p itself. So m can be written as , where . If (1) is the case, then m/n is not in lowest terms by out initial assumption, thus cannot be true. If (2) is the case then .

Now we look back at our second statement, . Substituting in a new expression for n^2 yields . This implies that m and n have a common factor and thus possibility (2) cannot work. Since (1) and (2) are not possible, the initial assumption is false and is not in Q.

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I just read through it and I think it's ok. I'm sure there are ways to shorten it and I'm sure I didn't explain some things well enough. The big thing I think might get criticized is justifying when I say two things have a common factor. Anyway, let me know how I did and how I can improve.

Thanks