MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1
Generally, a ¡°solution¡± is something that would be acceptable if turned in in the form presented here, although the solutions given are often close to minimal in this respect. A ¡°solution (sketch)¡± is too sketchy to be considered a complete solution if turned in; varying amounts of detail would need to be ?lled in. Problem 1.1: If r ¡ô Q £Ü {0} and x ¡ô R £Ü Q, prove that r + x, rx ¡ô Q. Solution: We prove this by contradiction. Let r ¡ô Q£Ü{0}, and suppose that r +x ¡ô Q. Then, using the ?eld properties of both R and Q, we have x = (r + x) ? r ¡ô Q. Thus x ¡ô Q implies r + x ¡ô Q. Similarly, if rx ¡ô Q, then x = (rx)/r ¡ô Q. (Here, in add¡¦(»ý·«)
ition to the ?eld properties of R and Q, we use r = 0.) Thus x ¡ô Q implies rx ¡ô Q. Problem 1.2: Prove that there is no x ¡ô Q such that x2 = 12. Solution: We prove this by contradiction. Suppose there is x ¡ô Q such that x2 = 12. Write x = m in lowest terms. Then x2 = 12 implies that m2 = 12n2 .
