[gavinpc]

Your point is clear... but as to the example, unless I'm misunderstanding your notation, the premise is that

    2 = (p/q)^2

which is not the same as

    2 = p^2/q^2

as you have in (2).

[tomsmeding]

Those statements are equivalent. If p and q share no common factors, then p^2 and q^2 also don't, so p^2/q^2 is still irreducible.

[roywiggins]

They're the same, surely?

  (p/q)*(p/q)
  = p*(1/q)*p*(1/q)
  = p*p*(1/q)*(1/q)
  = (p*p)/(q*q)

Or

  (p/q)*(p/q) 
  = ((p/q)*p)/q
  = ((p*p)/q)/q
  = (p*p)/(q*q)