Advertisements
Advertisements
Question
Prove that the number of subsets of a set containing n distinct elements is 2n, for all n \[\in\] N .
Advertisements
Solution
\[\text{ Let the given statement be defined as } P\left( n \right): \text {The number of subsets of a set containing n distinct elements } = 2^n ,\text{ for all n } \in N . \]
\[\text{ Step I: For n } = 1, \]
\[\text { LHS = As, the subsets of a set containing only 1 element are: } \phi \text{ and the set itself } . \]
\[\text{ i . e . the number of subsets of a set containing only 1 element } = 2\]
\[\text{ RHS } = 2^1 = 2\]
\[\text{ As, LHS = RHS } \]
\[\text{ So, it is true for n } = 1 . \]
\[\text{ Step II: For n = k } , \]
\[\text{ Let } P\left( k \right): \text{ The number of subsets of a set containing k distinct elements } = 2^k , \text{ be true for some k } \in N . \]
\[\text{ Step III: For n } = k + 1, \]
\[P\left( k + 1 \right): \]
\[\text{ Let } A = \left\{ a_1 , a_2 , a_3 , . . . , a_k , b \right\} \text{ so that A has } \left( k + 1 \right) \text{ elements .} \]
\[\text{ So, the subset of A can be divided into two collections; first contains subsets of A which don't have b in them and } \]
\[ \text{ the second contains subsets of A which do have b in them } . \]
\[i . e . \]
\[\text{ First collection: } \left\{ \right\}, \left\{ a_1 \right\}, \left\{ a_1 , a_2 \right\}, \left\{ a_1 , a_2 , a_3 \right\}, . . . , \left\{ a_1 , a_2 , a_3 , . . . , a_k \right\} \text{ and } \]
\[\text { Second collection } : \left\{ b \right\}, \left\{ a_1 , b \right\}, \left\{ a_1 , a_2 , \right\}, \left\{ a_1 , a_2 , a_3 , b \right\}, . . . , \left\{ a_1 , a_2 , a_3 , . . . , a_k , b \right\}\]
\[\text{ It can be clearly seen that: } \]
\[\text{ The number of subsets of A in first collection = The number of subsets of set with k elements i . e } . \left\{ a_1 , a_2 , a_3 , . . . , a_k \right\} = 2^k \left( \text{ Using step II } \right)\]
\[\text{ Also, it follows that the second collection must have the same number of the subsets as that of the first } = 2^k \]
\[\text{ So, the total number of subsets of } A = 2^k + 2^k = 2 \times 2^k = 2^{k + 1} .\]
Hence, the number of subsets of a set containing n distinct elements is 2n , for all n \[\in\] N .
APPEARS IN
RELATED QUESTIONS
Prove the following by using the principle of mathematical induction for all n ∈ N:
`1^3 + 2^3 + 3^3 + ... + n^3 = ((n(n+1))/2)^2`
Prove the following by using the principle of mathematical induction for all n ∈ N:
Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2
Prove the following by using the principle of mathematical induction for all n ∈ N: x2n – y2n is divisible by x + y.
If P (n) is the statement "n3 + n is divisible by 3", prove that P (3) is true but P (4) is not true.
Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^n} = 1 - \frac{1}{2^n}\]
a + ar + ar2 + ... + arn−1 = \[a\left( \frac{r^n - 1}{r - 1} \right), r \neq 1\]
Prove that 1 + 2 + 22 + ... + 2n = 2n+1 - 1 for all n \[\in\] N .
\[\frac{(2n)!}{2^{2n} (n! )^2} \leq \frac{1}{\sqrt{3n + 1}}\] for all n ∈ N .
\[\text{ A sequence } x_0 , x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_0 = 5 and x_k = 4 + x_{k - 1}\text{ for all natural number k . } \]
\[\text{ Show that } x_n = 5 + 4n \text{ for all n } \in N \text{ using mathematical induction .} \]
Prove by method of induction, for all n ∈ N:
3 + 7 + 11 + ..... + to n terms = n(2n+1)
Prove by method of induction, for all n ∈ N:
12 + 22 + 32 + .... + n2 = `("n"("n" + 1)(2"n" + 1))/6`
Prove by method of induction, for all n ∈ N:
(23n − 1) is divisible by 7
Prove by method of induction, for all n ∈ N:
(24n−1) is divisible by 15
Prove by method of induction, for all n ∈ N:
(cos θ + i sin θ)n = cos (nθ) + i sin (nθ)
Prove by method of induction, for all n ∈ N:
Given that tn+1 = 5tn + 4, t1 = 4, prove that tn = 5n − 1
Answer the following:
Prove by method of induction
`[(3, -4),(1, -1)]^"n" = [(2"n" + 1, -4"n"),("n", -2"n" + 1)], ∀ "n" ∈ "N"`
The distributive law from algebra says that for all real numbers c, a1 and a2, we have c(a1 + a2) = ca1 + ca2.
Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a1, a2, ..., an are any real numbers, then c(a1 + a2 + ... + an) = ca1 + ca2 + ... + can.
Show by the Principle of Mathematical Induction that the sum Sn of the n term of the series 12 + 2 × 22 + 32 + 2 × 42 + 52 + 2 × 62 ... is given by
Sn = `{{:((n(n + 1)^2)/2",", "if n is even"),((n^2(n + 1))/2",", "if n is odd"):}`
Let P(n): “2n < (1 × 2 × 3 × ... × n)”. Then the smallest positive integer for which P(n) is true is ______.
Prove the statement by using the Principle of Mathematical Induction:
32n – 1 is divisible by 8, for all natural numbers n.
Prove the statement by using the Principle of Mathematical Induction:
For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.
Prove the statement by using the Principle of Mathematical Induction:
n2 < 2n for all natural numbers n ≥ 5.
Prove the statement by using the Principle of Mathematical Induction:
1 + 5 + 9 + ... + (4n – 3) = n(2n – 1) for all natural numbers n.
A sequence a1, a2, a3 ... is defined by letting a1 = 3 and ak = 7ak – 1 for all natural numbers k ≥ 2. Show that an = 3.7n–1 for all natural numbers.
If 10n + 3.4n+2 + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is ______.
