Question

Find the permutations of \({\rm{\{ 1,2,3,4,5\} }}\) that correspond to these integers with respect to the correspondence between Cantor expansions and permutations as described in the preamble to Exercise 14 .

a) 3

b) 89

c) 111

Short answer

a. The permutation is 23145.

b. The permutation of \({\rm{\{ 1,2,3,4,5\} }}\) that corresponds to cantor integer 89 is 35421 .

c. The permutation is 52431 .

Step-by-step solution

Definition of Concept

Functions: It is a expression ,rule or law which defines a relationship between one one variable and another variables.

Find the permutation

(a)

Considering the given information:

The set \({\rm{\{ 1,2,3,4,5\} }}\)& cantor integer 3 .

Using the following concept:

If cantor digits are \({{\rm{a}}_{\rm{1}}}{\rm{,}}{{\rm{a}}_{\rm{2}}}{\rm{,}}{{\rm{a}}_{\rm{3}}}{\rm{\& }}{{\rm{a}}_{\rm{4}}}\)then cantor expansion is

\({{\rm{a}}_{\rm{1}}}{\rm{(1!) + }}{{\rm{a}}_{\rm{2}}}{\rm{(2!) + }}{{\rm{a}}_{\rm{3}}}{\rm{(3!) + }}{{\rm{a}}_{\rm{4}}}{\rm{(4!) = }}\)Canter integer.

Because there are five symbols like \({\rm{1, 2,3, 4, 5}}\)

The cantor digits are 4 i.e. \({{\rm{a}}_{\rm{1}}}{\rm{,}}{{\rm{a}}_{\rm{2}}}{\rm{,}}{{\rm{a}}_{\rm{3}}}{\rm{,}}{{\rm{a}}_{\rm{4}}}\)

By cantor expansion.

\(\begin{array}{l}{\rm{ = }}{{\rm{a}}_{\rm{1}}}{\rm{(1) + }}{{\rm{a}}_{\rm{2}}}{\rm{(2) + }}{{\rm{a}}_{\rm{3}}}{\rm{(6) + }}{{\rm{a}}_{\rm{4}}}{\rm{(24) = 3}}\\{\rm{ = }}{{\rm{a}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{a}}_{\rm{2}}}{\rm{ = 1,}}{{\rm{a}}_{\rm{3}}}{\rm{ = 0\& }}{{\rm{a}}_{\rm{4}}}{\rm{ = 0}}\\{{\rm{a}}_{\rm{1}}}{\rm{ = 1}}\end{array}\)

The number of integers less than 2 and follows 2 are 1

So, the permutation is \({\rm{21}}....\) (1)

\({{\rm{a}}_{\rm{2}}}{\rm{ = 1}}\)

The number of integers between 3 and 1 that are less than or equal to 3.

As a result, the permutation is \({\rm{31}}...{\rm{(2)}}\).

Using equation (1) & (2), we have the permutation \({\rm{231}}....\) (3)

\({{\rm{a}}_{\rm{3}}}{\rm{ = 0}}\)

Number of integer less than \({\rm{4 \& }}\)follows 4 are 0

So, the permutation is \({\rm{2314}}.....{\rm{(4)}}\)

\({{\rm{a}}_{\rm{4}}}{\rm{ = 0}}\)

Number of integers less than 5 & follows 5 are 0

The required number of permutations is\({\rm{ = 23145}}\).

Therefore, the required permutation of \({\rm{\{ 1,2,3,4,5\} }}\) that corresponds to cantor integer 3 is 23145 .

Find the permutation

(b)

Considering the given information:

The set \({\rm{\{ 1,2,3,4,5\} }}\)& cantor integer 89 .

Using the following concept:

If cantor digits are \({{\rm{a}}_{\rm{1}}}{\rm{,}}{{\rm{a}}_{\rm{2}}}{\rm{,}}{{\rm{a}}_{\rm{3}}}{\rm{\& }}{{\rm{a}}_{\rm{4}}}\)then cantor expansion is

\({{\rm{a}}_{\rm{1}}}{\rm{(1!) + }}{{\rm{a}}_{\rm{2}}}{\rm{(2!) + }}{{\rm{a}}_{\rm{3}}}{\rm{(3!) + }}{{\rm{a}}_{\rm{4}}}{\rm{(4!) = }}\)Canter integer.

By using cantor expansion.

\(\begin{array}{l}{{\rm{a}}_{\rm{1}}}{\rm{(1!) + }}{{\rm{a}}_{\rm{2}}}{\rm{(2!) + }}{{\rm{a}}_{\rm{3}}}{\rm{(3!) + }}{{\rm{a}}_{\rm{4}}}{\rm{(4!) = 89}}\\{{\rm{a}}_{\rm{1}}}{\rm{(1) + }}{{\rm{a}}_{\rm{2}}}{\rm{(2) + }}{{\rm{a}}_{\rm{3}}}{\rm{(6) + }}{{\rm{a}}_{\rm{4}}}{\rm{(24) = 89}}\\{\rm{1(1) + 2(2) + 2(6) + 3(24) = 89}}\\{{\rm{a}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{a}}_{\rm{2}}}{\rm{ = 2,}}{{\rm{a}}_{\rm{3}}}{\rm{ = 2,}}{{\rm{a}}_{\rm{4}}}{\rm{ = 3}}\end{array}\)

Number of integers less than 2 is 1 , so string is \({\rm{21}}.....\) (1)

\({{\rm{a}}_{\rm{2}}}{\rm{ = 2}}\)

Number of integers less than 3 &follow 3 is 2.

So string is 321

\({{\rm{a}}_{\rm{3}}}{\rm{ = 2}}\)

Number of integers less than 4 &follow 4 is 2

So string is 3421

\({{\rm{a}}_{\rm{4}}}{\rm{ = 3}}\)

Number of integers less than 5 & follow 5 is 3

So string is 35421

Thus, required permutation is 35421 .

Therefore, the required permutation of \({\rm{\{ 1,2,3,4,5\} }}\) that corresponds to cantor integer 89 is 35421 .

Find the permutation

(c)

Considering the given information:

The cantor integer 111, set \({\rm{\{ 1,2,3,4,5\} }}\)

Using the following concept:

Cantor expansion is given as.

\({{\rm{a}}_{\rm{1}}}{\rm{(1!) + }}{{\rm{a}}_{\rm{2}}}{\rm{(2!) + }}{{\rm{a}}_{\rm{3}}}{\rm{(3!) + }}{{\rm{a}}_{\rm{4}}}{\rm{(4!) = }}\)Canter integer.

By using cantor expansion.

\(\begin{array}{l}{\rm{ = }}{{\rm{a}}_{\rm{1}}}{\rm{(1!) + }}{{\rm{a}}_{\rm{2}}}{\rm{(2!) + }}{{\rm{a}}_{\rm{3}}}{\rm{(3!) + }}{{\rm{a}}_{\rm{4}}}{\rm{(4!) = 111}}\\{\rm{ = }}{{\rm{a}}_{\rm{1}}}{\rm{(1) + }}{{\rm{a}}_{\rm{2}}}{\rm{(2) + }}{{\rm{a}}_{\rm{3}}}{\rm{(6) + }}{{\rm{a}}_{\rm{4}}}{\rm{(24) = 111}}\\{\rm{ = 1(1) + 2(2) + 2(6) + 3(24) = 111 }}\\{\rm{so, }}{{\rm{a}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{a}}_{\rm{2}}}{\rm{ = 2,}}{{\rm{a}}_{\rm{3}}}{\rm{ = 2,}}{{\rm{a}}_{\rm{4}}}{\rm{ = 4}}\end{array}\)

Since, all of them is non-zero so, 1 is at the right most place. \({{\rm{a}}_{\rm{4}}}{\rm{ = 4}}\)that implies 5 is at the leftmost place. \({a_2} = 1 \Rightarrow 3\)Will be at the left from 1 and 2 is at the left of 3 but 4 is just the immediate left of 3 .

As a result, the permutation required is 52431

Therefore, the permutation required is 52431 .