MATH 211 Problems WorkSheet, Fall 2010

Worksheet I- Why is f not a function from R to R? 1 a) f(z) == b) f(z) =4V 27 +1 a) flo) & net difered b) an cbenend oe R Bem tee images II- Are the following functions from Z to Z one-to-one? a) f(n)=n-1 b) f(n)=n? +1 pente ‘ £On - Flr, 4) am + we 4 _ = AL b) ne since F(-1) = (Ayelet h - C1) III- page 147 exercise 36 Let f be a function from the set A to the set B, and let S and T be two subsets of A. Show that: a)f(SUT) = f(S)U f(T). d)F(SNT) € Ff(S)N F(T). a) t+ we F(SvT) > sane Sut such thet lm = xE SUT so xES oO “et cole tS) nm FO) e FC) G(x)su € £(S) U f(T) +Revefore | Fur) < ees) u #7) | Noeove, Lt y’ € F(QU f(T) ye FS) or ye F(T) nO sod x7eS Fi )ay so axveTt ,f day seo 3xE SvuT such ted k(m)ey +R, means tat ye £ (gut) | ee\ usa) £ © (set) | L) da ye £05 MT) » one sin such Thad Fox) =¥ ond ~ ET So) gen £OME £1) So FOR ay Ee £cCs)nect) IV- Given two mappings f and g from N to N defined by: on f:N—+N > g N—N neoan+1 n> max{0,n— 1} (a) Calculate f(n) and g(n) for n = 0, 1, 2, 3, 4, 73. (b) Is f an injection? a surjection? (c) Is g an injection? a surjection? (d) Show that f og # id, but 9° f =id,. a) F(e) = | 4(o)= 0 by fe injecting. ( proved ) ty aoor ° BY : wy ok we Since (3) <1 = wa § LM=s b3)= 2 qiven 1< IN NI £033) AY Jone 3 b. an KE Fr) <3 4(93) 92 we Con . ‘euch that Fe) = 4 eX mtl= 4 *="\~' tha o £4) = 4 Cc to not inpech ve (o)= c\) Dp tat yay wa yo d) fog lo) = P(gloi)= Fle) =I 40 fag + ed) got (x) < q (x+!) = max fo,x+t 0% = MaxtaAx ex Sinte K QO = id ny V- Show that if go f is a bijection then g is surjective and f is injective. To shnuothetd a & s reckue A 4 8 $e wat eed 3128 v\O such that gCb)=a So toke cE C ehve Bae A cuchthat gef(a)=c so nee gok 0 soy