Solution: According to division algorithm, we have n = 6q + r ; 0 ≤ r < 6 Let, A = n(7n2 + 5) For r = 0, we have A = n(7n2 + 5) = (6q + r) {7(6q + r)2 + 5} = 6q (7.36q2 + 5) = ...

Read More »## Verify that if an integer is simultaneously a square and a cube, then it must be either of the form 7k or 7k + 1

Proof: Let A be an integer. We have to show that the cube and square of A is of the form 7k or 7k + 1. First we show that, a square is either of the form 7k or 7k + 1. According to the division algorithm, we have A ...

Read More »## PDF|SOLUTION OF EXERCISE – 9.2| CLASS – 7

PDF|SOLUTION OF EXERCISE – 9.2| CLASS – 7

Read More »## Ideals | Lattices and Boolean algebra

Definition: A non empty subset I of a lattice L is called an ideal of L if (i) a, b ∊I ⟹a ˅ b ∊ I (ii) a ∊ I, l ∊ L ⟹ a ˄ l ∊ I. Example: Let {1, 2, 5, 10} be a lattice of factors ...

Read More »## Basic properties of congruence and their proved

i) a ≡ b (mod n) Proof: For any integer a, we have 0/n = 0 ⟹ (a – a) / n = 0 ⟹ a – a = 0.n ⟹ a ≡ a (mod n) (proved) ii) If a ≡ b (mod n) then b ≡ a(mod n) ...

Read More »## Congruence

Definition: Let n be a fixed positive integer. Two integers a and b are said to be congruent modulo n, symbolized by a ≡ b (mod n) if divides the difference a – b ; that is, provided that a – b = kn for some integer k. a is ...

Read More »## There are an infinite number of primes of the form 4n + 3.

Lemma: The product of two or more integers of the form 4n + 1 is of the same form. Proof: It is sufficient to consider the product of just two integers. Let us take k = 4n + 1 and k′ = 4m + 1. Multiplying these together, we obtain ...

Read More »## Sublattice and Convex sublattice

Sublattice: Let (L, ˄, ˅) be a lattice. A non empty subset S of L is called a Sublattice of L if S itself is a lattice under same operations ˄ and ˅ in L. Or, A non empty subset S of a lattice L is called a Sublattice of ...

Read More »## Theorem of Euclid’s on prime number on prime number

Theorem of Euclid’s on prime number : There are an infinite number of primes. Proof: Euclid’s proof is by contradiction let p1 = 2, p2 = 3, p3 = 5, p4 = 7, … be the primes in according order and suppose that there is a least prime called pn. ...

Read More »## Correspondence theorem on homomorphism

Let f be a homomorphism of a group G onto a group Gʹ with Kernel K(=Kerf). Let Hʹ be a subgroup of Gʹ and H = {x ∊G | f(x) ∊ Hˊ} = f -1(Hʹ). Then H is a subgroup of G containing K. If Hʹ is a normal subgroup ...

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