Friday 18 August 2017

RELATION : BINARY RELATION AND POSET

RELATION:- In order to express a relation from set A to Set B , we always need a statement which connects the elements of A with the elements of B.
example ;-A ={1,2,3,4} B={2,4,6} and relation is less than than we express it like (1,2) (1,4) (1,6) (2,4) (2,6) (3,4) (4,6). that is 1 is lass than 2 , 1 is less than 4 ...etc

BINARY RELATION - 
let A and B be two non-empty sets . A relation from A to B is a subset of A-B and is denoted by R .R is a relation from A to B such that R ⊆ A-B
R = {(a,b) : a∈A , b∈B and aRb}

example ;-A ={1,2,3,4} B={2,4,6} and relation is less than than we express it like (1,2) (1,4) (1,6) (2,4) (2,6) (3,4) (4,6). that is 1 is lass than 2 , 1 is less than 4 ...etc that is R ⊆ A-B  we can write this like R ⊆ AxB 

memory :- if Set A and B has m and m elements that we write it total number of elements is m.m or squre of m.
if Set A has m and set B has n elements than we write is m.n elements 

INVERSE RELATION:-if R ⊆ AxB is a relation from A to B then inverse relation define as R-1= { (b,a):(a,b)∈A  a∈A , b∈B and aRb}


COMPOSITE RELATION:- Let A , B,C be three non-empty sets  and R be a relation from A to B and S be a relation from B to C st R ⊆ AxB  and S ⊆ BxC then the composite relation of two relations is a relation from A to C define as SoR ={(a,b) :  an element c∈B st (a,c)∈R , (c,b)∈S}

i.e. (a,c)∈R , (c,b)∈S=> (a,b)∈ SoR .

Rusult:- if  R-1 and  S-1 are inverse of the relations R and S respectively then
(RoS)-1 =  S-1 o  R-1

RELATION ON A SET :- Let R be a relation from P to Q . if P =Q then we say that R is a relation on a Set P. i.e R ⊆ PxQ. 

TYPES OF BINARY RELATIONS:-
1. Reflexive:- Let A be a non- empty set and R be a binary relation in A i.e. R ⊆ AxA then the relation R is called relexive relation if every element of A is R related to itself.
i.e. for all (a,a)∈R

2. Symmetric :-

If R is a relation in the set A then R is called symmetric relation if a is related to a then a is related to b.

i.e. (a,b)∈R => (b,a)∈R.


3. Anty-symmetric Relation :- 

If R is a relation in the set A , then B is called anty-symmetric if (a,b)∈R (b,a)∈R. => a=b

4. Transitive Relation :- 

If R is a relation in the set A then R is called transitive if a is related to b and b related to c then a is related to c i.e.

(a,b)∈R (b,c)∈R =>(a,c)∈R.


Partial Order set (poset):- A relation R on a set A is said to be Partial order set if it is Reflexice , Antisymmtric and Transitive. And if it satisfies another axiom trichnotomy low then it is called total ordering relation. trichnotomy law is any two element of set is comparble.
compleltely defination of Partial order set is :-  A set A with a partial ordering relationon A is called partial order set and its denoted by( A,)

for Example:- If N be the set of partial integers then prove that relation ≤ is a partial order set on N.
solve:- 
1. Reflexivity:- for each a∈N . since a is less or equal to itself then we can say a≤a. i.e. aRa.
2. Anti-symmtry:- let a,b∈N  such that a≤b , b≤a so no element cannot be less or equal each other. therefore a=b.
3.Transitivity:- let a,b,c∈N st a≤b , b≤c then a≤c.
and hence it is a partial order set. hence proved.

Maximal element:- let ( A,) be a partially order set .An element a in A is called maximal element if 

Minimal element:- let ( A,) be a partially order set .An element p in A is called maximal element if x≤ p => m=x.

Least Upper Bound:-
Let S is a subset of R. And aR. Then a is called least upper bound of S if it is satisfied two axioms.
1) x≤a for all x∈S.
2) there are exist a Upper Bound t such that a≤t.
I.e. lub is less or equal all other upper bounds.


Greatest lower Bound:-

Let S is a subset of R. And a∈R. Then a is called least upper bound of S if it is satisfied two axioms.
1) a≤x for all x∈S.
2) there are exist a lower Bound w such that w≤a.
I.e. glb is greater or equal all other lower bounds. 


LATTICE:- A partially ordered set ( L,) is said to be lattice if evety two elements in the set L has a unique least upper bound and a unique greatest lower bound or its inf or sup exist.

DUAL LATTICE:- let ( L,) be a partial order set for any two element a,b∈L the converse of the relation "≤ " denoted by "≥" defined as  b≥a <=>a≤b  . in other words ( L,)  is a lattice then ( L,) also a lattice.
continue..

Monday 14 August 2017

Vector space

Vector space:-
Let V be a set with operation + and let F be a field with the operations + and . (dot) . an algebric expression ((V,+),(F,+.), .) with the internal and external operations is called vector space if it is  satisfies following axioms.
1. (V,+) be an abelian group.
2. (F,+,.)  Be closer with respect to dot.
3. a(æ+ß) = aæ+aß for all a,b is in F and æ,ß in V.
In other words we can express it as below..
we can describe this
1. (V,+) be an abelian group i,e, it satisfies 5 charactiristic closure associate identity inverse and commutative /
1  α,β in V then α+β in V.
2  α,β,γ in V then (α+β)+γ= α+(β+γ)
3  α in V , then there are exist 0 st α+0 = α
4  if α in V then -α in V st α+(α) in V.
5  α,β in V then α+β =β+α
2. (F,+,.)  Be closer with respect to dot. i,e,

a in F and α in V st aα in V.

* set of real no is a vector space .




Thursday 3 August 2017

Useful formula Geometry formulla (उपयोगी सूत्र )

Geometry formula


(α в ¢)²= α² в² ¢² 2(αв в¢ ¢α)
1. (α в)²= α² 2αв в²
2. (α в)²= (α-в)² 4αв b
3. (α-в)²= α²-2αв в²
4. (α-в)²= f(α в)²-4αв
5. α²   в²= (α в)² - 2αв.
6. α²   в²= (α-в)²   2αв.
7. α²-в² =(α   в)(α - в)
8. 2(α²   в²) = (α  в)²   (α - в)²
9. 4αв = (α   в)² -(α-в)²
10. αв ={(α в)/2}²-{(α-в)/2}²
11. (α   в   ¢)² = α²   в²   ¢²   2(αв   в¢   ¢α)
12. (α   в)³ = α³   3α²в   3αв²   в³
13. (α   в)³ = α³   в³   3αв(α   в)
14. (α-в)³=α³-3α²в 3αв²-в³
15. α³   в³ = (α   в) (α² -αв   в²)
16. α³   в³ = (α  в)³ -3αв(α  в)
17. α³ -в³ = (α -в) (α²   αв   в²)
18. α³ -в³ = (α-в)³   3αв(α-в)
ѕιη0° =0
ѕιη30° = 1/2
ѕιη45° = 1/√2
ѕιη60° = √3/2
ѕιη90° = 1
¢σѕ ιѕ σρρσѕιтє σƒ ѕιη
тαη0° = 0
тαη30° = 1/√3
тαη45° = 1
тαη60° = √3
тαη90° = ∞
¢σт ιѕ σρρσѕιтє σƒ тαη
ѕє¢0° = 1
ѕє¢30° = 2/√3
ѕє¢45° = √2
ѕє¢60° = 2
ѕє¢90° = ∞
¢σѕє¢ ιѕ σρρσѕιтє σƒ ѕє¢
2ѕιηα¢σѕв=ѕιη(α в) ѕιη(α-в)
2¢σѕαѕιηв=ѕιη(α в)-ѕιη(α-в)
2¢σѕα¢σѕв=¢σѕ(α в) ¢σѕ(α-в)
2ѕιηαѕιηв=¢σѕ(α-в)-¢σѕ(α в)
ѕιη(α в)=ѕιηα ¢σѕв  ¢σѕα ѕιηв.
» ¢σѕ(α в)=¢σѕα ¢σѕв - ѕιηα ѕιηв.
» ѕιη(α-в)=ѕιηα¢σѕв-¢σѕαѕιηв.
» ¢σѕ(α-в)=¢σѕα¢σѕв ѕιηαѕιηв.
» тαη(α в)= (тαηα   тαηв)/ (1−тαηαтαηв)
» тαη(α−в)= (тαηα − тαηв) / (1  тαηαтαηв)
» ¢σт(α в)= (¢σтα¢σтв −1) / (¢σтα   ¢σтв)
» ¢σт(α−в)= (¢σтα¢σтв   1) / (¢σтв− ¢σтα)
» ѕιη(α в)=ѕιηα ¢σѕв  ¢σѕα ѕιηв.
» ¢σѕ(α в)=¢σѕα ¢σѕв  ѕιηα ѕιηв.
» ѕιη(α-в)=ѕιηα¢σѕв-¢σѕαѕιηв.
» ¢σѕ(α-в)=¢σѕα¢σѕв ѕιηαѕιηв.
» тαη(α в)= (тαηα   тαηв)/ (1−тαηαтαηв)
» тαη(α−в)= (тαηα − тαηв) / (1  тαηαтαηв)
» ¢σт(α в)= (¢σтα¢σтв −1) / (¢σтα   ¢σтв)
» ¢σт(α−в)= (¢σтα¢σтв   1) / (¢σтв− ¢σтα)
α/ѕιηα = в/ѕιηв = ¢/ѕιη¢
» α = в ¢σѕ¢   ¢ ¢σѕв
» в = α ¢σѕ¢   ¢ ¢σѕα
» ¢ = α ¢σѕв   в ¢σѕα
» ¢σѕα = (в²   ¢²− α²) / 2в¢
» ¢σѕв = (¢²   α²− в²) / 2¢α
» ¢σѕ¢ = (α²   в²− ¢²) / 2¢α
» Δ = αв¢/4я
» ѕιηΘ = 0 тнєη,Θ = ηΠ
» ѕιηΘ = 1 тнєη,Θ = (4η   1)Π/2
» ѕιηΘ =−1 тнєη,Θ = (4η− 1)Π/2
» ѕιηΘ = ѕιηα тнєη,Θ = ηΠ (−1)^ηα

1. ѕιη2α = 2ѕιηα¢σѕα
2. ¢σѕ2α = ¢σѕ²α − ѕιη²α
3. ¢σѕ2α = 2¢σѕ²α − 1
4. ¢σѕ2α = 1 − ѕιη²α
5. 2ѕιη²α = 1 − ¢σѕ2α
6. 1   ѕιη2α = (ѕιηα   ¢σѕα)²
7. 1 − ѕιη2α = (ѕιηα − ¢σѕα)²
8. тαη2α = 2тαηα / (1 − тαη²α)
9. ѕιη2α = 2тαηα / (1   тαη²α)
10. ¢σѕ2α = (1 − тαη²α) / (1   тαη²α)
11. 4ѕιη³α = 3ѕιηα − ѕιη3α
12. 4¢σѕ³α = 3¢σѕα   ¢σѕ3α

» ѕιη²Θ ¢σѕ²Θ=1
» ѕє¢²Θ-тαη²Θ=1
» ¢σѕє¢²Θ-¢σт²Θ=1
» ѕιηΘ=1/¢σѕє¢Θ
» ¢σѕє¢Θ=1/ѕιηΘ
» ¢σѕΘ=1/ѕє¢Θ
» ѕє¢Θ=1/¢σѕΘ
» тαηΘ=1/¢σтΘ
» ¢σтΘ=1/тαηΘ
» тαηΘ=ѕιηΘ/¢σѕΘ continue...


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Wednesday 2 August 2017

Maths formulla (rectangle) in hindi

आयत के  फ़ॉर्मूलास : √ल ० का वर्ग  + चौ ० का वर्ग
आयत का विकर्ण =
आयत का क्षेत्रफल = ल *चौ
आयत का परिमिति या  परिमाप = २ (ल +चौ )
आयत के  क्षेत्रफल में वृद्धि = a+b+ab/100
आयत के  क्षेत्रफल में कमी  = a-b-ab/100
एक आयताकार खेत की ल ० a  और चौ b तथा इसके चारो ओर x चौ का रास्ता हो तो रस्ते का क्षेत्रफल           =       2x (a+b+2x)
एक आयताकार खेत की ल ० a  और चौ b तथा इसके अंदर  चारो ओर x चौ का रास्ता हो तो रस्ते का क्षेत्रफल           =       2x (a+b-2x)



Area
वर्ग का परिमाप = 4A
वर्ग का क्छेत्रफल = A *A
Area of square = a2
Dimension of square= 4a    where a is side of square.

Circle

वृत  के  क्षेत्रफल = π*r2
वृत की परिधि = 2πr
Area of circle = πr2
 cirumference of  circle = 2πr

पाइथागोरस प्रमेय - किसी समकोण त्रिभुज में कर्ण का वर्ग अन्य दो भुजाओ के वर्गों के योग के वरावर होता है।
A2+B2=C2 where A and B आधार और लंब और C कर्ण है।