- #1
LeBrad
- 214
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I saw a proof saying the root of a prime is always irrational, and it went something like this:
sqrt(r) = p/q where p/q is reduced
r = p^2/q^2
r*q^2 = p^2 therefore, r divides p
so define p = c*r
r*q^2 = (c*r)^2 = c^2*r^2
q^2 = c^2*r therefore, r divides q also
since r divides p and q, p/q is not reduced and we have a contradition so sqrt(r) is irrational.
Now my question is, assuming the above is correct, why does this work for prime r and not, say, r = 4.
sqrt(r) = p/q where p/q is reduced
r = p^2/q^2
r*q^2 = p^2 therefore, r divides p
so define p = c*r
r*q^2 = (c*r)^2 = c^2*r^2
q^2 = c^2*r therefore, r divides q also
since r divides p and q, p/q is not reduced and we have a contradition so sqrt(r) is irrational.
Now my question is, assuming the above is correct, why does this work for prime r and not, say, r = 4.