Using the following Make a conjecture and prove by induction

Using the following:

Make a conjecture and prove by induction the following:

Definition 1 A set is a collection of objects. The otjects in a set X are called elements of X. Notation 2 A set can be described using set-builder notation. That is, a set can be described by writing P(z) where P(r) is a statement concerning an object r. For erample, N tz: z is a natural number or (2 {r: is a real number and 2 r S Remark 3 Some authors use the notation (rlP(z) in place of tr P(z) Notation 4 The set that contains no elements is called the empty set and is denoted 2. /Note: is NOT the empty set Notation 5 If an o r is an element of a set A, then we write z E A. If r is not an bject element of A, then we write z A. Definition 6 Two sets A and B are equalif and only if every element of A belongs to Band every element of B belongs to A Definition 7 If A and Bare sets such that each element of A is also an element of B, then we say A is a subset of B, and notate this ASB. Remark 8 Some authors use the notation A CBif A is a subset of B. We reserve the notation C or something else later. Proposition 9 Let A and B be sets. Then A Bif and only if ACB and BCA. Definition 10 Let A be a set. The power set of A, denoted POA) or 2 is the set of all subsets of A.

Solution

1. Given that A be a set with n elements, where n is some natural number.

|A| = n, where |A| denotes the number of elements of set A.

Then the power set of A, denoted as P(A), defined as the collection of all subsets of the set A.

P(A) = {S : S is a subset of A}

Then, |P(A)| = 2ⁿ

In other words, P(A) has 2ⁿ elements, where n = |A|

2. We will prove the conjecture by induction method.

Let n = 1, So, A contains only one element. Say A = {a}

Therefore, P(A) = { {}, A}

{} denotes the null set.

Therefore, |P(A)| = 2 = 2¹

So, the conjecture is true for n = 1

Let n = 2, then A = {a, b}

Therefore, P(A) = { {}, {a}, {b}, {a,b}}

So, |P(A)| = 4 = 2²

So, the conjecture is true for n = 2

We suppose by induction that the statement holds for n = k

Now let the set A contain k + 1 elements. We can write A = B U {x}, and consider how to form subsets of A. B is a subset of A such that |B| = k and x is not an element of B.

We take all elements of P(B), and by the inductive hypothesis there are 2^k of these

Then we add the element x to each of these subsets of B, resulting in another 2^k subsets of A.

This exhausts the list of subsets of A, and so the total is 2^k + 2^k = 2(2^k) = 2^{k + 1} elements of the power set of A.

Therefore, the conjecture is true for n = k + 1 when it is true for n = k. Hence, by method of induction, the conjecture is true for all natural number n. (Proved)