devotion & mathematics

    what heppened when   ∞/∞ =1?  " Infinity " , In my earlier post , we have discussed about infinity and its properties. Come to our question " what happened when ∞ / ∞ = 1 ? " Assume   ∞/∞ = 1 that implies ∞ = ∞. Let the ∞ in the L.H.S. is the number of water drops in a riverand the ∞ in the R.H.S. is the  number of  water drops in a sea. It is impossible to count the number of water drops in a sea as  well as in the river.  That why we  consider both having infinite water drops. Now as our  assumption  if ∞ = ∞ then the  number of water drops in a river must equal to the number of  water  drops in a sea it results  the river  and sea must be of equal in size.  Is it true? No. Hence ∞  ≠ ∞ i.e. ∞/∞ ≠ 1. First of all this question may arise only if ∞ is a number.  Now my question " Is ∞ a mumber ? "  Answer is  ∞ is not a number. It is a symbolic representation of n...

                                          Theorem : In a group , the identity element is unique :   proof :               Suppose G is a group and e be the identity element in G.              Now we prove e is only the identity element in G.              In contrary assume e' is another identity element in G.             To prove the identity element is unique, we have to prove e = e'.             Since e is identity in G, we have e a = a = a e ∀ a 𝞊 G.             In particular , since e' 𝞊 G, we have e e' = e' = e' e     .............I             Since e' is identity in G, we have e' a = a = a e' ∀ a 𝞊 G.     ...

Theorem : In a group G, inverse of  any element is unique.    Proof :              Let G be a group with the identity element e.                             Let a 𝞊 G.                         Since G is a group, a has an inverse element say b.                        Now we prove b is only the inverse element of a in G.                         In contrary, assume c is also the inverse of a in G.              To prove a has unique inverse element , we have to prove b = c.                            Since b is inverse of a in G, we have a b = e = b a ............I     ...

  Problem   :   Show that the operation ‘ o ‘ given by aob = a b is a binary operation on the set of natural numbers N. Is this operation associative and commutative in N? Solution :    Consider N,the set of all natural numbers and ‘o’ is operation defined on N such that aob = a b   ∀ a,b 𝟄 N.      Let a,b 𝟄 N.    Now aob = a b                    = a x a x … x b times                    = a natural number                    𝟄 N   ∴ aob 𝟄 N     ∀   a,b 𝟄 N   'o' is a binary operation on N.   Checking for associative law :   Let a,b,c 𝟄 N   Now ao(boc) = aob c       | since ...

  Problem :   Let S be a non-empty set and ‘ o ‘ be an operation on S defined by aob = a for a,b 𝟄 S. Determine whether   o is commutative and associative in S? Solution :   Let S be a non-empty set and ‘ o ‘ be an operation on S defined by aob = a for a,b 𝟄 S. Checking for associative :     Let a,b,c 𝟄 S.    Now (aob)oc = aoc         | since aob = a |                            =   a              | since aoc = a |       ∴ (aob)oc = a   Also   ao(boc) = aob           | since boc = b |                        ...

Basic Definitions :  * Binary Operation  * Algebraic Structure * Quasi- Group or Groupoid * Semi Group * Monoid * Group * Abelian Group Binary Operation :            Let S be a non-empty set . If f : SxS→R is a mapping , then f is called binary                   operation or binary  composition on S.           Thus        If a relation in S is such that every pair ( distinct or equal ) of elements of S  taken in definite  order is associated with a unique element of S then it is called a binary operation in S. Otherwise the relationis not binary operation in S and the relation is simply an operation in S.      (a,b) 𝟄 SxS , ∃ a unique element f(a,b) 𝟄 S. We observe that addition, multiplication, subtraction are binary operations in R and division is not a binary operation in R  why because division by 0 is not defined...