Question #203175

Prove that the set of integers is countable.


1
Expert's answer
2021-06-08T12:58:38-0400

An infinite set is countable if and only if it is possible to list the elements of the set in a sequence.

To list the integers in a sequence 0,-1,1,-2,2….

A function f from the set of natural numbers to the set of integers defined by

f(n)={(n1)/2,  n  is  oddn/2,  n  is  evenf(n)=\left\{\begin{matrix} (n-1)/2, \; n \;is\;odd & \\ n/2, \; n \;is \; even& \end{matrix}\right.

To show that this function is countable.

That is to prove that the function is a bijection.

(1) To prove that f is injective:

Let m and n be two even numbers, then

f(m)=m2f(n)=n2f(m)=f(n)m2=n2m=nf(m) = \frac{m}{2} \\ f(n) = \frac{n}{2} \\ f(m) = f(n) \\ \frac{m}{2} = \frac{n}{2} \\ m=n

Therefore f is injective.

(2) To prove that f is surjective:

Let t be negative then t appears even position in the sequence

f(2k)=2k2=tf(2k)= -\frac{2k}{2}=t

with t= -k

This implies that for every negative value of t in Z there is a natural number 2k.

Let t be positive then t appears odd position in the sequence

f(2k1)=2k112=2k22=tf(2k-1) = \frac{2k-1-1}{2}=\frac{2k-2}{2}=t

with t=k

This implies that for every positive value t in Z there is a natural number 2k-1.

Therefore f is surjective.

Then f is both injective and surjective implies that f is bijection.

There the set of integers is countable.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS