Question

.1. Write a Python script for compute P50 k=1 P100 j=−2 (k − 2j)

2. It is clear that the cardinality of the Natural numbers is no more than the cardinality of the Rational numbers. Show that Rational numbers have cardinality no greater than the natural numbers (and therefore they have the same cardinality).

3. Prove (by contradiction) that the real numbers are uncountable.

Answer 1

Ans :

As HOMEWORKLIB RULES i suppose to do only one question.

2)

3)

Real numbers are uncountable but they are simultaneously countable and this is mathematically possible too, so to understand this I have an example and then I will prove the theorem that is given by Georg Cantor and commonly known as cantor's theorem which proves that real numbers are uncountable (by contradiction)

example

suppose there is a number line and we spot any two numbers on the number line and then we have to find the numbers between the two.

let us take 0 and 1

and we have to find all the numbers between them, and the numbers between the numbers are infinite like 0.1, 0.2, 0.3, ......

so actually we can count limited real numbers but we can't count all real numbers

Theorem (by contradiction)

lets A be the set and P(A) be the power set of A then |A|<|P(A)| i.e. cardinality of a set is strictly dominated by the cardinality of its power set.

Let A be any non-empty set and P(A) be its power set.

To prove: |A|<|P(A)|

proof: let us define set B B={{x}|x∈A}

let us define a function also f: B →A

f({x}) = x

B ~ A which means |B|=|A| equation 1

B ⊆ P(A)

|B| ≤ |P(A)| equation 2

|A| ≤ |P(A)|

have to prove |A| ≤ |P(A)|

let us assume |A| = |P(A)| (assuming contradictingly)

let g:A → P(A)

x∈ A = g(x) ∈ p(A) equation 3

c = {x ∈ A | x ∉ g(x) } equation 4

from equation 3 g(x) ⊂ A

from equation 4 c ⊂ A

g(x) = c equation 5

now α ∈ A , α ∈ c or α ∉ c

case 1: when α ∈ A then α ∉ c

then α ∉ g(x) from equation 4

α ∉ c as g(α)=c

case 2: when α ∈ A then α ∉ c

α ∉ g(α) as g(α)=c

α ∈ c from equation 4

|A| ≠ |P(A)|

Hence |A| < |P(A)|

Thank you...