Free web servers - 4.2.2 ANSWERS TO EXERCISES 571 4. Yes; let
4.2.2 ANSWERS TO EXERCISES 571 4. Yes; let l/u x o = w, where v is large. 5. Not always; in decimal arithmetic take u = w = 9. 6. (a) Yes. (b) Only for b+p 5 4 (try u = 1-bPp). [W. M. Kahan observes that the identity does hold whenever b-l < fU 5 b-l/ . It follows that 1 @ (1 @ (1 @ u)) = 1 @ u for all u.] 7. If u and w are consecutive floating binary numbers, u $ v = 2u or 27,~. When it is 2v we often have u@@J@ < 2v@. For example, u = (.lO . . . 001)2, v = (.lO . . . OlO)s, u @v = 2v, and u@ + v@ = (.lO.. 011)2. 8. (a) -, ZZ; (b) -, zz; (c) -, M; (d) -; (e) -. 9. ]u -w] 2 ]u -v] + ]w - w] 5 ~1 min(beUeq, bevpq) + ~2 min(beuyq, beW-q) 5 elbeUmq + E2bewpq < (~1 + ~2) max(beUeq, bewpq ). The result cannot be strengthened in general, since for &le we might have e, very small compared to both e, and ezu, and this means that u -w might be fairly large under the hypotheses. 10. We have (.a~. . . ap--lap)* @ (.9.. .99)b = (.a~. up-l(ap -l))b if aP 2 1; here 9 stands for b -1. Furthermore, (.a~. . . up-iap)b @ (1.0.. . O)b = (.a,. u,-~O)~, so the multiplication is not monotone if b > 2 and up 2 2. But when b = 2, this argument can be extended to show that multiplication is monotone; obviously the certain computer had b > 2. 11. Without loss of generality, let z be an integer, ]z] < bP. If e 5 0, then t = 0. If 0 < e 5 p, then z -t has at most p f 1 digits, the least significant being zero. If e > p, then z -t = 0. [The result holds also under the weaker hypothesis Jt] < 2be.] 12. Assume that e, = p, e, 5 0, u > 0. Case 1, u > bp- . Case (la), w = u + 1, w 2 f, e, = 0. Then u = u or u+ 1, v = 1, u = u, w = 1 or 0. Case (lb), w = u, (v] 5 f. Then u = u, v = 0, u = u, v = 0. If ]v] = f and more general rounding is permitted we might also have u = u* 1, v = Fl. Case (lc), w = u-1, v 5 -4, e, = 0. Then u = u or u -1, v = -1, u = u, w = -1 or 0. Case 2, u = bpel. Case (2a), w = u + 1, v 2 f, e, = 0. Like (la). Case (2b), w = u, Iv] 5 f, u 2 u. Like (lb). Case (2c), w = u, ]v] 5 +, u < u. Then u = u -j/b where u = j/b + VI and ]vi I 5 f b-l for some positive integer j 5 4 b; we have v = 0, u = u, v = j/b. Case (2d), w < u. Then w = u -jfb where v = ---j/b + v1 and ]vi] < ab- for some positive integer j 5 b; we have (v , u ) = (--j/b, u), and (u , v ) = (u, --j/b) or (u-l/b, (1 --j)/b), the latter case only when vi = abE . In all cases ueu = u-u , v~~ =v-v ,~~~ =u-~ ,v~w =~-~ ,round(w-u-w)=w-~-v. 13. Since round(z) = 0 iff z = 0, we want to find a large set of integer pairs (m, n) with the property that m @ n is an integer iff m/n is. Assume that ]m], In] < bP. If m/n is an integer, then m @ n = m/n is also. Conversely if m/n is not an integer, but m @ n is, we have l/In] < Im @n -m/n] < ~lm/nlbl-P, hence ]m] > 2bP- . Our answer is therefore to require ]m] < 2bp- and 0 < In] < bP. (Slightly weaker hypotheses are also possible.) 14. I(~@~)c3w–vwl I I(u~ u)@w-(uc3 u)wI+IwI Iu@v-uvl I S(,,,),,$ be,-+, 6U@U I (1-t b)6(UB,,)B,. Now le(,B,)B, -e,m(,B,,)] 5 2, so we may take E = ;(l + b)b2-p.