suppose x, u, and v are integers and let 0 be the additive identity in Z. prove
ID: 3210975 • Letter: S
Question
suppose x, u, and v are integers and let 0 be the additive identity in Z. prove for all x,u, and v in Z if x > 0 and u < v then xu < xv.Explanation / Answer
With z = x +iy and w = u + iv z + w = (x + iy) + (u + iv) = (x + u) + i(y + v) (Collecting like terms) If z + w is a non-zero real, (ie z + w = Re(z + w)), then Im(z + w) = y + v = 0 Also zw = (x + iy)(u + iv) = xu + ixv + iuy - i^2vy = (xu + vy) +i(xv + uy) There are two cases to consider here: 1) Case y = v = 0 (from Im(z + w) = 0) If this is so then z and w are just real numbers 2) Case y + v = 0 => v = -y In this case from zw = = (xu + vy) +i(xv + uy) we can write zw = (xu - y^2) +i(-xy + uy) = (xu - y^2) +iy(u - x) If Im(zw) = y(u - x) = 0, we know that if y = 0 ( and so v), then both z and w must be real. But we could have x = u, which would make Im(zw) = 0 but would NOT imply that z an w are real. For here we could have y (and also v) non-zero. Indeed if w = the complex conjugate of z - that is u = x and v = -y, so that w = x - iy z + w = (x + iy) + (x - iy) = 2x = 2Re(z) and y can be non-zero Further zw = (x + iy)(x - iy) = x^2 + y^2 = |z|^2 which is real, but again, y need not be zero So the claim: "z + w is real and zw is real implies z and w real" is false.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.