8 ms·
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 =
by gavinpc 8y ago
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 8y agoThose 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 8y agoThey'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)