CSH 1.2 Sets

2969 days ago by CSH2012

#1) T={a,b} #1) subsets: {},{a},{b},{a,b} 
       
#2) {1,2,3} #2) subsets: {},{1},{2},{3},{1,2},{2,3},{1,3},{1,2,3} #2) {l,m,n,o} #2) subsets: {},{l},{m},{n},{o},{l,m},{l,n},{l,o} #3) subsets: {m,n},{m,o},{n,o},{l,m,o},{l,m,n},{l,n,o},{m,n,o},{l,m,n,o} #3) cadinality of n --> 2^n 
       
#3){2,7,16,17,19} --> {a,b,c,d,e} IS A FUNCTION #3) 2 --> a #3) 7 --> d #3) 16 --> e #3) 17 --> b #3) 19 --> c 
       
#4){2,7,16} --> {a,b} IS A FUNCTION! #4) 2 --> a #4) 7 --> b #4) 16 --> a 
       
#5) {2,7,16} --> {a,b,c,d} NOT A FUNCTION! #5) 2 --> a #5) 7 --> b #5) 16 --> c #5) 16 --> d 
       
#6) Z --> {-1, 0, +1} #6) <0 --> -1 #6) 0 --> 0 #6) >0 --> +1 
       
#6a) f(n)=n mod 3 print(2+7) print(2-7) print(222*77777) print(16/3.n(digits=10)) print(16//3) print(16%3) print(len(str(777^2222))) 
       
9
-5
17266494
5.333333333
5
1
6423
9
-5
17266494
5.333333333
5
1
6423
#7) cardinality of A is at least 1 and at most 17 
       
#8) A={1,2,3},B={a,b,c} #8) f(1,2,3)=(a,a,a), f(1,2,3)=(a,a,b), f(1,2,3)=(a,a,c) #8) f(1,2,3)=(b,b,b), f(1,2,3)=(b,b,a), f(1,2,3)=(b,b,c) #8) f(1,2,3)=(c,c,c), f(1,2,3)=(c,c,a), f(1,2,3)=(c,c,b) #8) f(1,2,3)=(a,b,a), f(1,2,3)=(a,b,b), f(1,2,3)=(a,b,c) #8) ... there are 27 possible functions! 
       
#9) f(n)=2*n+1 
       
#10) f(word) = 1st letter in word def worder(s): return s[0] worder("gregarious") 
       
'g'
'g'
#11) output the distance from the origin (0,0) to the point (x,y) #11) f(x,y)=sqrt(x^2+y^2) def cheat(x,y): return sqrt(x^2+y^2) show(cheat(.3,.4)) show(cheat(1,0)) show(cheat(.5,.866)) 
       
\newcommand{\Bold}[1]{\mathbf{#1}}0.500000000000000
\newcommand{\Bold}[1]{\mathbf{#1}}1
\newcommand{\Bold}[1]{\mathbf{#1}}0.999977999757995
\newcommand{\Bold}[1]{\mathbf{#1}}0.500000000000000
\newcommand{\Bold}[1]{\mathbf{#1}}1
\newcommand{\Bold}[1]{\mathbf{#1}}0.999977999757995
#12) if 0<x<=1, y=1/x with be in 1<=y<inf def flip(num): return 1/num show(flip(4/5)) 
       
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{5}{4}
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{5}{4}