1)let g:B->A f:A->B be functions, and gof is the identity function, prove/disprove the following: g=f^-1;f is onto;f is 1-1;g is onto;g is 1-1. iv'e showed that the first the third and the last are correct, and the others are wrong, am i right? 2)prove that f(x)=cos(x)+sqrt(3)sin(x) is a bounded function and find its supremum and infimum. well to prove that it's a bounded function was easy, it's bounded by: -1-sqrt(3)<=f(x)<=1+sqrt(3) now to find the supremum and infimum, the supremum is 2, and i have proven it this way: first we need to show that it's an upper bound, i.e cos(x)+sqrt(3)sinx<=2 we have 2-sqrt(3)sin(x)>=1 and we can suppose that cos(x) is positive, bacause the upper bound should be greater also from the least value of f(x) which can be obtained when cos(x)<0, so we can suppose that cos(x)>0 (is this assumption valid?) then we have sqrt(1-sin^2x)<=2-sqrt(3)sin(x) 1-sin^2x<=4+3sin^2x-4sqrt(3)sin(x) 4sin^2x-4sqrt(3)sin(x)+3>=0 2sin^2x-sqrt3)^2>=0 which is ofcourse correct. now we need to show that for every c>=f(x) 2<c, (in order to show that's least upper bound), so we suppose that c<2, then c>=cos(x)+sqrt(3)sin(x) again we assume cos(x)>0 so we have c^2-2csqrt(3)sin(x)+3sin^2(x)>=1-sin^2x 4sin^2x-2sqrt3sin(x)+c^2-1>=0 [/tex]\delta=4*3c-16c^2+16=-4c^2+16<=0[/tex] so we have (2-c)(2+c)<=0 c>=2 or c<=-2 but this clearly contradicts our assumptions, is my appraoch correct? p.s just something im interested about, how do you prove that 0.123456789101112.... (all the natural numbers) is irrational?