Infinite set

Infinite set :-

A set is said to be an infinite set whose elements cannot be listed if it has an unlimited (i.e. uncountable) by the natural number 1, 2, 3, 4, ………… infinite , for any natural number n is called a infinite set. 

A set which is not finite is called an infinite set.
example of infinite sets:-
set of natural numbers , N={1,2,3,4,.........................}
set of whole numbers  ,W= {0,1,2,3,.........................}
set of real numbers    , R etc

Since we cannot ever finish counting an infinite set, we need a different approach to

thinking about “how many elements” such a set contains. In the early 1900s, Georg

Cantor presented an idea that has clarified thinking about this issue and had enormous

impact on opening new pursuits in logic.

Two (finite or) infinite sets A and B are said to have the same cardinality (equipotent) if

there is a one-to-one, onto function (i.e., a bijection) f: A Æ B.

 Since it is impossible to ascribe an ordinary number to the size of an infinite set, we

don’t try. Instead, the comparative idea above is used. If we can “line the elements up” in

exactly corresponding pairs, we say the two sets have the same cardinality (i.e., size).

So, for instance, the set of natural numbers N = {0, 1, 2, … } has the same cardinality as

the collection of all strings over the alphabet.

 ifnfinite sets that have the same cardinality as N = {0, 1, 2, … } are called countably

infinite. A set that has a larger cardinality than this is called uncountably infinite.

Assertion: there are uncountably infinite sets.

Proof by contradiction:

Let [0,1] denote the interval of all real numbers x, 0≤x≤1 — this set is uncountably

infinite. First of all, note that there is an injection f: N Æ [0,1] defined by f(n) = n/(n+1).

This maps N to a proper subset of [0,1], so [0,1] has cardinality at least as large as N.

To show that the cardinality of [0,1] is actually larger than N, we proceed by

contradiction. Suppose that [0,1] is countably infinite. Suppose that g :N Æ [0,1] is a

bijection, and therefore {g(n) | nŒN} is the set of all real number in the interval [0,1]. We

show that there must be at least one real number missing in this enumeration providing

a contradiction to the existence of g, and hence there is no bijection. Specifically, we

construct the decimal expansion of a number a that is missing from {g(n) | nŒN}. Take

the decimal expansion of a to be .d1d2d3 …, where dn = 5 if the nth digit of g(n) is 6, and

dn = 6 otherwise. Then for each n, the nth digit of a differs from the nth digit of g(n).

Therefore a differs from every number in the enumeration in at least one digit and so

aœ{g(n) | nŒN}, a contradiction since this set was assumed to be all such numbers.

Ref from homepage.

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Finite set

Finite set:-

              In mathematics, a finite set is a  set that has a finite number of elements. s a finite set with five elements. The number of elements of a finite set is a natural number (a non-negative integer) and is called the cardinality of the set. A set that is not finite is called infinite. For example, the set of all positive integers is infinite  :Finite sets are particularly important in combinatorics, the mathematical study of counting. Many arguments involving finite sets rely on the pigeonhole principle, which states that there cannot exist an injective function from a larger finite set to a smaller finite set.

Example. {2,4,6,8} this is a finite set because it have finite number of element....

A= {1,2,3,4,5,6,7,8,9} it is also a finite set....

In other way.

In a finite set the element can be listed if it has a limited i.e. countable by natural number 1, 2, 3, ……… and the process of listing terminates at a certain natural number N.

The number of distinct elements counted in a finite set S is denoted by n(S). The number of elements of a finite set A is called the order or cardinal number of a set A and is symbolically denoted by n(A). Thus, if the set A be that of the English alphabets, then n(A) = 26: For, it contains 26 elements in it. Again if the set A be the vowels of the English alphabets i.e. A = {a, e, i, o, u} then n(A) = 5.

there are more example of finite set  like group of class 12 students 
group of all real number define as R={x:x is a number on real line}
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Linear equation of two variables

Linear equation of two variables

A linear system of two equations with two variables is any system that can be written in the form.
                             ax +  by =p
                             cx + dy =q

where any of the constants can be zero with the exception that each equation must have at least one variable in it.

Also, the system is called linear if the variables are only to the first power, are only in the numerator and there are no products of variables in any of the equations.

Here is an example of a system with numbers.
                                                                 3x - y=7     ................(a)
                                                                2x+2y=3    .................(b)
                                           now (a) x 2 + (b)
                                                             
                                                               6x - 2y =14
                                                               2x + 2y =3
                                                             ___________
                                                             8x          =17               because 2y to 2y is cencle
                                                    therefore     x=17/8
 put this value in (a)
                                                          3 *17/8 -y =7
                                                   =>  51/8 - 7    =y
                                                  =>     51-56/8  =y
                                                  =>           -5/8 = y         

Before we discuss how to solve systems we should first talk about just what a solution to a system of equations is.  A solution to a system of equations is a value of x and a value of y that, when substituted into the equations, satisfies both equations at the same time.

 Solve each of the following systems of equations

(a) x+y=1              ...................1

     2x+y=2             ..................2

solve->

                       for solve this equation  , we can subtract it from 2 to 1

 so we can find

                                      2x+y=2

                                      x+y=1

                                     -   - = -

                                   ________

                                     x=1

put x=1 in equation (1)

so y=0..
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Cyclic group

Cyclic Group->

        A group (G,o) is called cyclic if for a∈G there are exist a∈G is in the form of aⁿ , where n is an integer. 

in mathematical form                          (a)={aⁿ :n∈Z} where a is the generator of G.

Theoram:-  

Every cyclic group is Abelian.

Proof. The elements of cyclic groups are of the form aⁿ

. Commutativity

amounts to proving that aⁿa

 = ajai.

aⁿa^m  =>a^n+m

        ^=>a^{m+n  addition of interger is commutative}

                         ^=>a^maⁿ

therefore
                  

aⁿa^m

=a^maⁿ

 ^{G is a cyclic group.}

                                                                                                              ^{hence proved}

^{Something about cyclic subgroup..}

Cyclic Subgroups

If we pick some element a from a group G then we can consider the subset of all elements of G that are powers ofa. This subset forms a subgroup of G and is called thecyclic subgroup generated by a. If forms a subgroup since it is

• Closed. If you multiply powers of a you end up with powers of a
• Has the identity. a • a^-1 = a⁰ = e
• Has inverses. The inverse of any product of a's is a similar product of a^-1 's.

But this is the long way of proving subgrouphood. Let's use our theorem that says if x and y are in the subset implies that x • y^-1 is in the subset then the subset is a group. This is simple here. If y is a power of a then so is y^-1 and so, therefore, is x • y^-1 .

A few facts about cyclic groups and cyclic subgroups:

1. Cyclic groups are Abelian.
2. All groups of prime order are cyclic.
3. The subgroup of a group G generated by a is the intersection of all subgroups of G containing a
4. All infinite cyclic groups look like the additive group of integers.

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I

Sub group

Subgroup:à

                            Suppose (G,o) be a group and H⊆G where H≠ Ø  .. Then H is called Subgroup of g if

  1.        H is stable for operation “o”.

2. (H,O) is also a Group.

In other words , H is called subgroup of G if it is satisfy all condition of Group…

Complex of a Groupà

Suppose (G,o) be a Group ,then a subset of  H of G is called complex iff    a,b∈H => aob∈H for all a,b ∈H

Proper and improper subgroupà

If a group (G,o) have two subgroup (G,o) and ({e},o) then that subgroup is called improper or trival .. and different from it that is others subgroup is called  proper (non-trival subgroup)…

Some example of subgroup:- let (Z,+) be a set of integer.let (E,+) be a set denoted to even  numbers and (O,+) be set that denoted to odd numbers.set of even numbers is a subgroup of Z ie (E,+) is a subgroup of Z but (O,+) is not a subgroup of Z because it does not satisfies closure law because if 1,3 ∈O then 1+3=4 does not belogs to O.and we can also say (E,+) is a proper subgroup of Z.

* (Z,+) is a subgroup of set of all rational numbers.and it is also a proper subgroup of (Q,+).

* if H={1,-1} and G={1,-1,i,-i} then (H,.) is a subgroup of (G,.).this is because if we take two element of H then we can get a new value of H which is also in set of H. then we can say it is satisfies closure ,associate , identity (ie 1) and inverse .then it is a subgroup of G.

Theoram-

For a subgroup , of a non-empty set H⊆G the necessary and sufficient condition is that

a∈H ,b∈H =>ab^-1∈H   where b^-1 is inverse of b.

proof-

The condition is necessary:à let H be a subgroup of G.

Then to showà a∈H ,b∈H =>ab^-1∈H

We know that a H is a subgroup then it have a inverse of each element i.e. if à a∈H=> a^-1∈H.

And H should be closure with respect to .(dot)

Therefore  a∈H , b∈H => a∈H,b^-1∈H    (inverse low)

                                 => ab^-1∈H          (closure low)  proved I condition…

The condition is sufficientà

Let H be a non-empty set of G s.t. a∈H ,b∈H =>ab^-1∈H……………..(i)

Then to showà H is a subgroup of G.

For this we have to show four exioms.

1 (identity)à a∈H ,a∈H =>aa^-1∈H                from(i)

                                    =>e∈H

Therefor H have a identity element.

II(inverse)à

          Let a is element of H

Then     e∈H ,a∈H =>ea^-1∈H

                           =>a^-1∈H

i.e. H contains inverse of each element.

III(closure)à

          Let a,b∈H then

            a∈H ,b∈H => a∈H ,b^-1∈H

                            =>a (b^-1 )^-1∈H

                            =>ab∈H

                        Therefor H is closure.

IV(associate)à H satisfies associate law because

 for all a,b,c∈H=> a,b,c∈H

                       =>a,b,c∈G

                   =>a(bc)=(ab)c for all a,b,c ∈H

H satisties all condition of group Therefore H is a Subgroup.

 hence proved

Q 2              A non-empty set H⊆G is a subgroup of G iff 

     (i)a∈H ,b∈H =>ab∈H

  (ii) a∈H=>a^-1∈H  where  a^-1 is the inverse of a

 Proof:-

                  Suppose H be a subgroup of G .

To show:- (i)a∈H ,b∈H =>ab∈H

     (ii) a∈H=>a^-1∈H

Since H is a subgroup of G so it is satisfies all condition of group.

                  (i)a∈H ,b∈H =>ab∈H  (closure in H)

                  ⁽ⁱⁱ⁾ a∈H=>a^-1∈H   (inverse in H). first part prove

Conversaly:-

      Let H be a non-empty set of G.and it is satisfies two conditions (i) and (ii)

To show:-  H shall be subgroup.

1. closure:-  a∈H ,b∈H =>ab∈H             . . . . . . from (i)

        H is satisfies closure low.

2.                  H satisfies associate law because

         for all a,b,c∈H  => a,b,c∈H                       H⊆G

                                              =>a,b,c∈G

                                              =>a(bc)=(ab)c for all a,b,c ∈H

3. Identity low  =>     form (ii)   a∈H=>a^-1∈H  

                                 a∈H , a^-1∈H   =>aa^-1∈H       from (i)

     ð  e∈H  

  therefore indentity satisfies identity low.

 4. inverse low:-  a∈H=> a^-1∈H    from(ii)

                       So H have inverse of each element.

H satisfies all condition of group therefore H is a subgroup of G.

 

                                                                                                                                    Hence proved.

Th 3.            For a subgroup , of a non-empty set H⊆G the necessary and sufficient condition   HH^-1⊆H.

Proof:-         necessary condition:-

suppose H be a subgroup of G.

                   Let ab^-1∈HH^-1 be a orbitary element .

                   a∈H , b∈H => a∈H,b^-1∈H    (inverse low in H ,because H is a subgroup)

     =>ab^-1∈H

So, ab^-1∈HH^-1 => ab^-1∈H , for all a,b in H

Therefore HH^-1⊆H….

Suffiecient condition:-

                         Let H be a non-empty set such that HH^-1⊆H.

To show :-        H shall be a subgroup of G.

                         a∈H , b∈H => ab^-1∈HH^-1   (by the definition of HH^-1 )

                                           => ab^-1∈H              (HH^-1⊆H)

i.e.                    a∈H , b∈H => ab^-1∈H   for all a,b in H.

                         therefore H is a Subgroup of G.

                                                                                                                     hence proved……………