Proof:
r = (0, 1, 2, 3, . . . n) Hence the total number of subsets i..
r = (0, 1, 2, 3, . . . n) Hence the total number of subsets i..Proof:
Similarly, ..
Similarly, ..Proof:
..
..Proof:
We shall use PMI to prove tha..
We shall use PMI to prove tha..Proof:
Similarly,..
Similarly,..Proof:
Let p(x) be a polynomial divided by (x-a). Let q(x) be the quotient and R be the remainder. By division algorithm, Dividend = (Divisor x quotient) + Remainder p(x) = q(x) . (x-a) + R Substitute x = a, p(a) = q(a) (a-a) + R p(a) = R (a - a = 0, 0 - q (a) = 0) Hence Remainder = p(..
Proof:
Let A be any set. In order to prove that f A we must show that there is no element of f which is not present in A. And since f contains no element at all, no such element can be found out. Hence f A...
Let A be any set. In order to prove that f A we must show that there is no element of f which is not present in A. And since f contains no element at all, no such element can be found out. Hence f A...Proof:
The null set f is a subset of every set, in particular f A. By hypothesis, A f . The two conditions imply A = f..
Alternative Proof of Binomial Theorem for Positive Integral Index (Combinatorial Method)
We have, (a + b) n = (a + b) (a + b) ....... n times. The terms on the RHS are obtained by taking one letter from each factor and multiplying them together. Choosing 'a' from all the factors, we get the term a n..
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