let A= {6, 7, 8}. For each relation between A and B given as subset of A x B, de

Question

let A= {6, 7, 8}. For each relation between A and B given as subset of A x B, decide whether it is a function mapping A into B. If it is a function, decide whther it is one to one and whether it is onto B.

a) {(1,7), (2,6), (3,6)}

b) {(1,6), (2,7), (3,8)}

c) {(1,6), (1,7), (3,8)}

determine whether the following statements are true or false. If a statement is true, prove it. If it is not give a counterexample:

a) for all intergers m, if m > 2 then m^2 -4 is composite

b) for all integers n and m, if n-m is even then n^2-m^2 is even

c) if *is binary operation on any set S then a *a =a for all a in S

d) Let n and m be intergers such that n<m and m not equal to 0. Then (n/m)^3 <= (n/m)^2

use mathemcatical induction to prove that: 1+3+3^2.....+ 3^n = 3 ^n+1-1 for n E Z^+

2

Explanation / Answer

(a)function, not one-one, not onto

(b)function , one-one, onto

(c)not a function

(a)false : take m = 3

(b)true: let n-m = 2k for integer k, => n^2-m^2 = (n-m)(n+m) = 2k(n+m) => n^2-m^2 is an even integer . thus proved

(c)false : multiplication is an binary operation on R, but 2*2 !=2

(d)true: n<m => n/m < 1=> (n/m)*(n/m)^2 < (n/m)^2 => (n/m)^3 < (n/m)^2 ,

thus proved

statement is true for n = 1 since 1+3 = (3^2-1)/2

let the statement be true for n

let Sn = 1+3+3^2 ...+ 3^n

Sn+1 = 1+3+3^2+....+3^n + 3^n+1

= (3^(n+1) - 1)/2 +3^(n+1) = (3*3^(n+1)-1)/2 = (3^(n+2)-1)/2

=> the statement is true for n+1

thus it is proved by induction

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