 Hello, dear learners welcome you all to today's audiovisual lectures of unit 2 binary operation of the course abstract algebra and discrete mathematics of the program BA Mathematics Second Semesters. So, today we will discuss unit 2 binary operations. This is the 3rd lecture of the course abstract algebra and discrete mathematics. So, last 2 lecture we have already discussed relations and functions and there are various properties and we also discuss various examples of relations and functions. So, today we will discuss unit 2 binary operations. First we will define what is binary operations then we discuss various properties of binary operations and we also discuss examples related to binary operations and their properties. So, let us discuss this unit 2. So, now we start today's lecture on the topic binary operation. So, in this lecture we will discuss first what is binary operation. So, first we define binary operation then we discuss various properties of binary operation. So, let us define what is binary operations. So, we all know that there are 4 basic aromatic operations addition, subtraction, multiplication and division. So, these 4 binary operations already we know. So, to begin with the concept we consider an example involving the basic aromatic operations addition and multiplication of any 2 natural numbers. Suppose here m and n belongs to natural number m and n any natural any 2 national numbers. So, if we add 2 national numbers the result will be again a national number. So, m plus n belongs to n and similarly their multiplication also belongs to n. So, from these 2 operations addition and multiplication we observed the following results. Here you see at a time exactly elements of n are process they are 2 elements m and n and resulting element is also n element of n here m plus n belongs to n and m multiply n is also belongs to n. Suppose if we consider m is equal to 3 n is equal to 4 then 3 plus 4 belongs to n and 3 into 4 it is 12 also belongs to n. So, their result is also belongs to n. So, now we define binary operations. Here a binary operation star on a non-empty set s is a function from s cross s to s here star we define is a function already we discussed what is a function. So, star is a function star is a function from s into s to s here this is a function and this function is defined as star. So, a binary operation is also a function from s cross s if we consider any 2 element from this set here result will be a star b this also belongs to s. So, here image a b your belongs to s cross s another function star is defined this star a b is also defined as this way star a b. So, star is a binary operation on set s if a star b belongs to s for all a b belongs to s they are if we consider any 2 element from s cross s so result will be a star b also belongs to s. So, then we call that star is a binary operation on set s if we consider any 2 elements from this set s cross s the result will be a star b. So, this will be suppose c element c so c also belongs to s then star will be a binary operation. Here you see star is just a symbol which may be addition multiplication subtraction division it may be matrix addition and matrix multiplication depending on the set which it is defined. So, star is just a symbol and star represent these operations addition subtraction multiplication division matrix addition matrix multiplication. So, here addition and subtraction multiplication or binary operation n if we consider one example suppose addition and subtraction or binary operation on n but subtraction is not a binary operation why why subtraction is not a binary operation. So, addition and multiplication are binary operation on n natural number but subtraction is not binary operation to verify this result we consider one example suppose we consider 5 6 belongs to n cross n suppose from this set n n cross n from this set we consider 1 element 5 6 your subtraction is a binary operation. So, subtraction is a binary operation on n subtraction is not a binary operation on n to verify this result we consider one example 5 6 belongs to n cross n but 5 actually this this is 3 4 3 4 so 3 minus 4 equal to minus 1 so this not belongs to this n so this does not belong to n. So, therefore, subtraction is not binary operation 3 minus 4 is equal to minus 1 so minus 1 does not belong to n so subtraction is not a binary operation. So, therefore, this condition is not held it in in case of subtraction. So, now, we observe this table here in this direction there are number system and in this direction there are operations there are 4 operations addition subtraction multiplication and division and here they are we consider this number system ratio number set of integers ratio number r real number c complex number q subtract this non-zero ratio number this is non-zero real number. So, for addition addition is binary operation on n z q r c but addition is not binary operation on this set non-zero real number and similarly non-zero in case of non-zero real number addition is not binary operation. Similarly, subtraction is not binary operation on n but subtraction is binary operation on z q r c but not binary operation on non-zero ratio number and non-zero real numbers. Similarly, multiplication is binary operation on n z q r c non-zero ratio number also non-zero real number and division is not binary operation on national number not integers yet not rational number not real number not complex number but binary operation on this set of non-zero rational number and set of non-zero real numbers. These are the four binary operations on different number systems. So, in case of some number system these are binary system and some number system some operation or not binary operations. So, here are some examples of binary operations already we discussed in the table. So, addition is binary operation on n subtraction is not binary operation on n multiplication is binary operation on n addition is binary operation on z addition is not binary operation on set s s is equal to z minus 0 this means set of non-zero integers non-zero integers. Substraction is not binary operation is a binary operation on z. So, another example s get a collection of real matrices matrix addition is not binary operation on s. So, we cannot find some of 2 cross 3 matrix and a 3 cross 4 matrix. Similarly, this is another example if m be the collection of all m all m cross n real numbers. So, matrix addition is binary operation on this set. Similarly, m be the all n cross n real matrices the matrix multiplication is a binary operation on this matrix in this set. So, addition n also multiplication is a binary operation on the set of c complex number. Now, there is a formula how can we find number of binary operations on a finite set. So, this is a standard formula if we consider 2 sets of finite sets number of elements of m is m number of elements of b is n the number of functions this number of functions may be binary number of binary version from a cross b is n to the power m. Suppose, number of elements of a set is 3 number of elements of b is 2 the number of binary operations will be is equal to 2 to the power 3 is 8. So, there are 8 binary operations on from a to b like there is a person table or this is also called composition table. If h is the finite set the number of elements of s b is small then we can construct a table known as a personal composition table. So, in the next in a discussion of questions some questions we will discuss a composition table. We give an example of a composition table how can we find whether a operation is binary operation or not with the help of composition. So, now we discuss various properties of a binary operation. So, first properties is permutative property any operation is there define on a non-empty set s is set to be permutative if this condition is satisfied means a star b is equal to b star a for all elements a b belongs to the given set. So, next properties is associative property. So, any binary operation star define on a set s is set to be satisfied by associative property if this condition is satisfied bracket a star b star c is equal to a star b bracket b star c for all elements a b c belongs to the given set s. So, next properties existence of identity element existence of identity element suppose we consider set s as the non-empty set if you put a binary operation star an element e belongs to s is calling identity element for if a star e is equal to a is equal to e star a for all a belongs to the given set s. If this condition is satisfied then e is called the identity element of the set s under the binary operation star identity element of a binary operation if it exists is unique. There is a only one binary identity element in a binary operation on a set here now we consider some examples how can we find identity element of a binary operation on a set. Suppose multiplication we know multiplication is a binary operation on n then there exist one belongs to n such that a into 1 a into 1 is equal to a is equal to 1 into a for any elements belongs to n suppose 3 suppose. So, 3 into 1 is equal to 3 to 1 into 3 for any element this condition is satisfied suppose 10 into 1 is equal to 10 is equal to 1 into 10. So, for any elements of natural number if we consider 1 then this condition is satisfied. So, therefore, 1 is an identity element of a binary operation multiplication on the set n natural number set of natural number. So, 1 is the identity element of for multiplication in n. Similarly, we consider another example 2 3 and 4 here 2 you see addition is binary operation on n. So, there exists 0 belongs to z such that a plus 0 is equal to a plus 0 plus a for all a belongs to z 3 plus 0 is equal to 3 is equal to 0 plus 3. For any set of integers any element from the set of integers. So, this condition is satisfied. So, therefore, 0 will be the identity element for addition in z. So, 0 is also called additive identity. So, addition is binary operation on n, but there does not exist identity element for addition in n because for any element of on the set of integers this condition is not satisfied for addition. Similarly, matrix multiplication is a binary operation on this set m set of all m cross n matrices the set of all m cross matrices the m cross n null matrix 0 is the identity element for matrix addition. So, for matrix multiplication set null matrix is the identity element for matrix addition on this set m set of all m cross n real numbers. So, next property is existence of inverse of an element existence of inverse. So, let S be a non-empty set if you put a binary operation is third and with identity element E star is a binary operation on set S and E directly element then an element B belongs to S is called an inverse element of A under star if this condition is satisfied A star B is equal to E and B star E then B is the inverse of A. If inverse of an element under a binary operation exist is a unique. So, there is a only one inverse element of a binary operation. So, now we discuss examples of a inverse element of a binary operation. So, we know that multiplication is a binary operation on n and one belongs to n is the identity element for multiplication, but no element of a except one possesses inverse under multiplication on n no element of n except one possesses inverse under multiplication of n. Similarly, addition is a binary operation on Z and 0 is the identity element for addition. So, for every A belongs to A there exists minus A belongs to Z such that this condition is satisfied for any element of Z then minus A is the inverse of A under addition. Suppose 3 is a element of Z and minus 3 is also element of Z then 3 plus minus 3 is equal to 0 similarly minus 3 plus 3 is equal to 0. So, this condition is satisfied for for binary version addition for binary version addition on Z. So, therefore minus 3 is the inverse of 3. So, this minus 3 is also called additive inverse because operation is addition. So, learner can also follow these books for better understanding of this chapter. So, these are some reference books. So, with the help of this reference you can understand this concept with the some various examples from this book you can consider. So, now we discuss some questions. So, example 1 here operation star is defined by this A star B is equal to A minus B plus 10 we have to find whether this operation is binary or not. So, here A comma B belongs to Z. So, Z is a set of integers. So, clearly A minus B plus 10 also belongs to Z because subtraction of any of any two numbers also belongs to Z and addition is also binary person on Z. So, therefore A minus B plus 10 is belongs to Z for all A B belongs to Z. So, therefore obviously A star B is equal to A minus B plus 10. So, clearly A minus B plus 10 belongs to Z. So, obviously A star B belongs to Z for any elements of A B belongs to Z. So, star is a binary person on Z. So, another example here operation star is defined by A star B is equal to A B by 5. So, now we find whether star is binary person or not on Q. Q is the set of rational number. So, since A B belongs to Q obviously, their modulation also belongs to Q. So, part of any two rational number is also rational number. So, and A B by A B by 5 also belongs to Q set of rational number. So, here A star B is equal to A B by 5. So, therefore obviously A star B belongs to Q for any elements of A B belongs to Q. So, star is a binary person on Q. You can take any two numbers suppose P 4 belongs to Q obviously 3 into 4, 12 belongs to Q and 12 by 5 also belongs to Q. This way we can prove that this A star B is equal to A B by 5 is a binary person on. Next example 3, there we have to find whether multiplication is a binary person on the set this set minus 1 the set consisting the elements minus 1, 0, 1. So, now we prove this result with the proper composition table or operation table. Here you see the elements of the set we put in horizontally and vertically minus 1, 0, 1 and vertically minus 1, 0, 1. Now here this cell the outcome of this results you see minus 1 into 1 minus 1 into minus 1 is 1 we put here minus 1 to 0, 0 minus 1 into 1 minus 1, 0 into minus 1, 0, 0 into 0, 0, 0 into 1, 0, 1 into minus 1, minus 1, 1 into 0, 0, 1 into 1, 1. You see all these nine elements also belongs to this set each of the nine entries also belongs to E. So, all change all the elements belongs to this set therefore, multiplication is a binary operation on this set. If the set consisting small number of elements then we can apply this table which is called operation table or composition table. Next example, example 4 here star is a binary operation defined by this a star b is equal to a minus b plus 5. So, now we say this operation for commutative whether it is commutative or not we say commutative property of for this result. So, we have this result a star b is equal to a minus b plus 5. So, obviously b star a is also b minus a plus 5 for all a b belongs to this set you see this a star b not is equal to b star a. So, clearly this a star b not is equal to b star a for commutative the result is a star b is equal to b star a for all elements of a b for any set. So, here a star b not is equal to b star a. So, star is not commutative here star is defined by this same result actually same for same example, same result of example 4 the operation is same a star b is equal to a star b is equal to a minus b plus 5. So, now we say associative property for this binary operation. So, for associative property already we define associative property for associative property we have to prove these results. If this result is satisfied then star will be associated. So, first we say 11 side part bracket a star b is equal to bracket a star b star c is equal to this here a star b is defined by a minus b plus 5. So, in place of this we prove this result then star c again for this 2 elements again we apply this results. So, in this case now a will be this element a minus b plus 5 and b will be c. So, accordingly we put this elements in this result. So, a minus bracket a minus b plus 5 minus c plus c. So, a minus b minus c plus 10 after simplification we get this result. Now, we find right inside part. So, a star bracket b star c a star. So, we apply this result here a means b and b means c. Accordingly we put this result b minus c plus 5. So, again we apply this result then we get here a means a and b means this result b minus c plus 5. So, we put here after simplification we get this result. You see the result of 11 side and result of right inside is not equal. So, therefore, bracket a star b star c not is equal to a star bracket b star c. So, the associative result is not satisfied by this binary person. So, let us sum up today's lecture. So, today's lecture we discuss this points a binary operation we define binary operation. So, a binary operation on a set s is a function from s cross s to h. If h is a finite set having m elements the number of binary operation on n is m to the power m square. A binary operation star on a set is commutative if a star b is equal to b star a for all elements a b belongs to s. A binary operation star on a non-empty set s is associative if this condition is satisfied bracket a star b star c is equal to a star bracket b star c for all elements a b c belongs to s. And so, we discuss exist how to find identity element of a binary person. And an element e on a non-empty set s equivit a binary operation star is called identity element if for a binary operation star if a star e is equal to a is equal to e star a for all elements a belongs to s. Similarly, we also discuss how to find inverse element of a binary operation and element b in a non-empty set s equivit a binary operation star is called a inverse element of a if this condition is satisfied a star b is equal to e is equal to b star a. So, this concepts of a binary operation we discuss today's lecture as we define what is a binary operation and we also discuss various properties of binary operations. So, a binary operation is very important for discussing the next sectors unit 3 for group theory. So, this concept is very important. So, to advance the studies of this course in this course we also discuss in block 2 ring field and vector space for that concept also the binary operation is a very important concept for understanding these topics. So, hope this lecture will benefit all of you. So, with this we end today's lecture. Thank you all.