Define F: ℤ → ℤ by the rule F(n) = 2 − 3n, for each integer n. (i) Prove that F is one-to-one. Proof: Suppose n1 and n2 are any integers, such that F(n1) = F(n2). Substituting from the definition of F gives that 2 − 3n1 = . Solving this equation for n1 and simplifying the result gives that n1 = . Therefore, F is one-to-one.

Step-by-step explanation:

(i)

To prove that  F    is   1-1 we have to show the following.

         If   F(n) = F(m)     then      n=m.   For any integers n,m.

So,  suppose that in fact

    2 - 3n  = 2 - 3m     ,  subtracting 2 on both sides  we get  

    -3n = -3m ,  dividing both sides of the inequality by 3 we get

   n = m        as we wanted.

The general idea behind a one to one function is that every image has a unique pre-image.

Let's think about a classic example of a function that is NOT one to one.

    is NOT one to one because    

So it NOT the preimage of a single element.

That's why the condition states

If   f is one to one then

if    f(n) = f(m)      then      n=m

b

step-by-step explanation:

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %