What is Cartesian Product ?
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Friends iss article mein aaj me aapko bataane waala hoon ki Cartesian Product kya hota hai. Iss article ko poora padhne ke baad aap jaan paayenge ki kinhi diye gaye do(two) sets ka Cartesian Product kaise nikaal sakte hain aur agar hame do(two) sets ka Cartesian Product diya hua hai to hum unn dono sets ko pata kar sakte hain! So friends...
Let's Begin...
"Given two non-empty sets P and Q. The cartesian product P✕Q is the set of all ordered pairs of elements from P and Q, i.e., P✕Q = {(p, q): p ∈ P, q ∈ Q}".
Matlab agar P aur Q do(two) non-empty set hai to unka Cartesian Product P✕Q wo set hota hai jisme wo sabhi ordered pair aate hain jinka pehla member P mein se ho aur doosra member Q mein se ho.
[Things to know:
1. Ordered pair do(two) objects ya members ka ek pair hota hai jinke likhne ka order(क्रम) fixed/certain hota hai. Ordered pair mein do(two) members hote hain. Wo dono members kuch bhi ho sakte hain jaise ki numbers, English alphabets ke letters, names of people etc. Unn dono members ko comma ( , ) ka use karke alag-alag (separate) karke likha jaata hai. Unn dono members ko small bracket ke andar close karke likha jaata hai. For example,
(2, -3) ⇒ It is an ordered pair.
2. Ordered pair ko likhne ka order(क्रम) fixed hota hai. Member ke order(क्रम) change karne par ordered pair bhi change ho jaata hai. For example ye ordered pair (4, 2) aur ye ordered pair (2, 4) alag (different) hain.
(4, 2) ≠ (2, 4)
3. Koi bhi do(two) ordered pair tab he equal kahe jaate hain jab unka pehla member aur doosra member alag-alag (separately) equal ho, i.e.,
(a, b) = (c, d) ⇔ a = c and b = d
4. Agar P ek empty set hai ya phir Q ek empty set hai ya phir P aur Q dono hi empty set hain to P✕Q bhi ek empty set hoga. Isi tarah se Q✕P bhi ek empty set hoga.
5. P aur Q non-empty set ho aur P ek infinite set hai ya phir Q ek infinite set hai ya phir P aur Q dono hi infinite set hain to P✕Q bhi ek infinite set hoga. Isi tarah se Q✕P bhi ek infinite set hoga.
6. Agar set P mein x elements ho aur set Q mein y elements ho to set P✕Q mein x✕y elements honge, i.e., n(P) = x and n(Q) = y then n(P✕Q) = xy].
Chaliye Cartesian Product ko aur sahi tarike se samajhne ke liye ek example dekh lete hain.
If P = {1, 2} and Q = {a, b, c} then find P✕Q. And Q✕P also.
Ab P✕Q ko find karne ke liye aapko same to same wahi karna hai jaisa maine likha hai aur bataaya hai:
(i) Sabse pahle ' P✕Q ' likhiye.
P✕Q
(ii) Phir uske baad (after that) ' = ' likhiye.
P✕Q =
(iii) Phir uske baad open flower bracket matlab ' { ' likhiye.
P✕Q = {
(iv) Ab hame ordered pair banana hai; iske liye set P ka sabse pehla element lijiye (kyuki P✕Q mein sabse pehle P likha hai isliye hum ordered pair mein set P ka hi element sabse pehle rakhenge). Ab iske saath set Q ka element comma ( , ) ka use karke likh dijiye. Ye bann gaya ek ordered pair aur isko small bracket ke andar close karke likh lijiye. Isi tarah P ke pehle element ke sath set Q ka doosra element lekar aap ek aur ordered pair bana sakte hain. Isi tarah se apko set P ke pehle element ke saath set Q ka teesra(third) element lena hoga jisse ek aur ordered pair bann jaayega. Isi tarah se set P ke pehle element ke saath set Q ka ek-ek element lekar ordered pair aapko banaate jaana hai aur tab tak banaate jaana hai jab tak ki set Q ke sabhi elements khatam naa ho jaaye.
Agar aap set P ke pehle element ke saath set Q ke elements ek-ek karke lete jaa rahe hain aur ordered pair banaate jaa rahe hain and then aapne set P ke pehle element ke saath set Q ke sabhi elements ek-ek karke le liye, tab aapko set P ke doosre element ke saath set Q ke sabhi elements ek-ek karke lete jaana hai aur ordered pair banaate jaana hai. Isi tarah se agar set P ka doosra element ke saath set Q ke sabhi elements aapne le liye hain to ab set P ke teesre(third) element ke saath set Q ke sabhi elements ek-ek karke lete jaana hai aur ordered pair banaate jaana hai aur isi tarah se ye process aage badhaate jaana hai jab tak ki hame sabhi possible ordered pairs naa mil jaayein.
P✕Q = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)
(vi) Sabhi ordered pairs likh lene ke baad closed flower bracket matlab ' } ' likhna hai.
P✕Q = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}
That's it! Hamne P✕Q find kar liya!
Isi procedure ka use karke hum Q✕P bhi find kar sakte hain-
Q✕P = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}
Agar iss question ko formal way mein kiya jaaye to ye kuch aisa hoga-
P✕Q = {1, 2, 3} ✕ {a, b}
= {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}
Isi tarah se,
Q✕P = {a, b} ✕ {1, 2, 3}
= {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}
Chaliye kuch aur examples dekh lete hain:
Ex.1. If G = {7, 8} and H = {5, 4, 2}, find G✕H and H✕G.
Solution: G = {7, 8}
H = {5, 4, 2}
G✕H = {7, 8}✕{5, 4, 2}
= {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H✕G = {5, 4, 2} ✕ {7, 8}
= {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
Ex.2. If A = {1, 2, 3, 4, 5} and B = {-1, -2, -3} then find A✕B, B✕A and B✕B.
Solution: A = {1, 2, 3, 4, 5}
B = {-1, -2, -3}
A✕B = {1, 2, 3, 4, 5} ✕ {-1, -2, -3}
= {(1, -1), (1, -2), (1, -3), (2, -1), (2, -2), (2, -3), (3, -1), (3, -2), (3, -3), (4, -1), (4, -2), (4, -3), (5, -1), (5, -2), (5, -3)}
B✕A = {-1, -2, -3} ✕ {1, 2, 3, 4, 5}
= {(-1, 1), (-1, 2), (-1, 3), (-1, 4), (-1, 5), (-2, 1), (-2, 2), (-2, 3), (-2, 4), (-2, 5), (-3, 1), (-3, 2), (-3, 3), (-3, 4), (-3, 5)}
B✕B = {-1, -2, -3} ✕ {-1, -2, -3}
= {(-1, -1), (-1, -2), (-1, -3), (-2, -1), (-2, -2), (-2, -3), (-3, -1), (-3, -2), (-3, -3)}
Ex.3. If A = {1, 2}, B = {a, b} and C = {m, n} then find A✕B✕C.
Solution: A = {1, 2}
B = {a, b}
C = {m, n}
A✕B✕C = {1, 2} ✕ {a, b} ✕ {m, n}
= {(1, a, m), (1, a, n), (1, b, m), (1, b, n), (2, a, m), (2, a, n), (2, b, m), (2, b, n)}
Inn ordered triplets ko kaise banana hai ye aapko iss diagram ko dekhkar samajh mein aa jaayega-
[Note: If A, B, C are non-empty and finite sets then A✕B✕C = {(a, b, c) : a ∈ A, b ∈ B and c ∈ C} here (a, b, c) is called ordered triplet].
Ex.4. P = {1, 2}. Form the set P✕P✕P.
Solution: P = {1, 2}
P✕P✕P = {1, 2} ✕ {1, 2} ✕ {1, 2}
= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}
Ex.5. Let A = {1, 2, 3} and B = {3, 4} and C = {4, 5, 6}. Find A✕(B⋂C) and (A✕B)⋂(A✕C). Are the sets A✕(B⋂C) and (A✕B)⋂(A✕C) equal?
Solution: A = {1, 2, 3}
B = {3, 4}
C = {4, 5, 6}
B⋂C = {3, 4} ⋂ {4, 5, 6}
= {4}
A✕B = {1, 2, 3} ✕ {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
A✕C = {1, 2, 3} ✕ {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
So, A✕(B⋂C) = {1, 2, 3} ✕ {4}
= {(1, 4), (2, 4), (3, 4)}
(A✕B)⋂(A✕C) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ⋂ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
= {(1, 4), (2, 4), (3, 4)}
Yes, the sets A✕(B⋂C) and (A✕B)⋂(A✕C) are equal.
[Note: ' ⋂ ' ye sign hai intersection ka. Agar aap intersection ke baare mein detail mein jaanna chahte hain to iss blue link par click kijiye >>>
Intersection of Sets].
Read Also:
Ex.6. If (x + 1, y - 2) = (3, 1), find the value of x and y.
Solution: Ordered pairs are equal so the corresponding elements will also be equal.
So, x + 1 = 3
x = 3 - 1
x = 2
and
y - 2 = 1
y = 1 + 2
y = 3
Ex.7. If A✕B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution: A✕B mein sabse pehle likha hai A aur phir B likha hai iska matlab ye hai ki inn sabhi ordered pairs - (a, x), (a, y), (b, x) aur (b, y), mein pehle elements/members set A ke hain aur doosre elements/members set B ke hain.
So, A = {a, a, b, b}
= {a, b}
B = {x, x, y, y}
= {x, y}
I hope ki mera ye article aap ko pasand aaya hoga. Agar aap ko mera ye article pasand aaya ho to comment karke hame bata sakte hain. Aapke dwaara kiye gaye comment se hame iss tarah ke post/article ko likne ke liye motivation milta hai.
THANKS FOR READING THIS BLOG.
Let's Begin...
What is Cartesian Product?
"Given two non-empty sets P and Q. The cartesian product P✕Q is the set of all ordered pairs of elements from P and Q, i.e., P✕Q = {(p, q): p ∈ P, q ∈ Q}".
Matlab agar P aur Q do(two) non-empty set hai to unka Cartesian Product P✕Q wo set hota hai jisme wo sabhi ordered pair aate hain jinka pehla member P mein se ho aur doosra member Q mein se ho.
[Things to know:
1. Ordered pair do(two) objects ya members ka ek pair hota hai jinke likhne ka order(क्रम) fixed/certain hota hai. Ordered pair mein do(two) members hote hain. Wo dono members kuch bhi ho sakte hain jaise ki numbers, English alphabets ke letters, names of people etc. Unn dono members ko comma ( , ) ka use karke alag-alag (separate) karke likha jaata hai. Unn dono members ko small bracket ke andar close karke likha jaata hai. For example,
(2, -3) ⇒ It is an ordered pair.
2. Ordered pair ko likhne ka order(क्रम) fixed hota hai. Member ke order(क्रम) change karne par ordered pair bhi change ho jaata hai. For example ye ordered pair (4, 2) aur ye ordered pair (2, 4) alag (different) hain.
(4, 2) ≠ (2, 4)
3. Koi bhi do(two) ordered pair tab he equal kahe jaate hain jab unka pehla member aur doosra member alag-alag (separately) equal ho, i.e.,
(a, b) = (c, d) ⇔ a = c and b = d
4. Agar P ek empty set hai ya phir Q ek empty set hai ya phir P aur Q dono hi empty set hain to P✕Q bhi ek empty set hoga. Isi tarah se Q✕P bhi ek empty set hoga.
5. P aur Q non-empty set ho aur P ek infinite set hai ya phir Q ek infinite set hai ya phir P aur Q dono hi infinite set hain to P✕Q bhi ek infinite set hoga. Isi tarah se Q✕P bhi ek infinite set hoga.
6. Agar set P mein x elements ho aur set Q mein y elements ho to set P✕Q mein x✕y elements honge, i.e., n(P) = x and n(Q) = y then n(P✕Q) = xy].
Chaliye Cartesian Product ko aur sahi tarike se samajhne ke liye ek example dekh lete hain.
If P = {1, 2} and Q = {a, b, c} then find P✕Q. And Q✕P also.
Ab P✕Q ko find karne ke liye aapko same to same wahi karna hai jaisa maine likha hai aur bataaya hai:
(i) Sabse pahle ' P✕Q ' likhiye.
P✕Q
(ii) Phir uske baad (after that) ' = ' likhiye.
P✕Q =
(iii) Phir uske baad open flower bracket matlab ' { ' likhiye.
P✕Q = {
(iv) Ab hame ordered pair banana hai; iske liye set P ka sabse pehla element lijiye (kyuki P✕Q mein sabse pehle P likha hai isliye hum ordered pair mein set P ka hi element sabse pehle rakhenge). Ab iske saath set Q ka element comma ( , ) ka use karke likh dijiye. Ye bann gaya ek ordered pair aur isko small bracket ke andar close karke likh lijiye. Isi tarah P ke pehle element ke sath set Q ka doosra element lekar aap ek aur ordered pair bana sakte hain. Isi tarah se apko set P ke pehle element ke saath set Q ka teesra(third) element lena hoga jisse ek aur ordered pair bann jaayega. Isi tarah se set P ke pehle element ke saath set Q ka ek-ek element lekar ordered pair aapko banaate jaana hai aur tab tak banaate jaana hai jab tak ki set Q ke sabhi elements khatam naa ho jaaye.
Agar aap set P ke pehle element ke saath set Q ke elements ek-ek karke lete jaa rahe hain aur ordered pair banaate jaa rahe hain and then aapne set P ke pehle element ke saath set Q ke sabhi elements ek-ek karke le liye, tab aapko set P ke doosre element ke saath set Q ke sabhi elements ek-ek karke lete jaana hai aur ordered pair banaate jaana hai. Isi tarah se agar set P ka doosra element ke saath set Q ke sabhi elements aapne le liye hain to ab set P ke teesre(third) element ke saath set Q ke sabhi elements ek-ek karke lete jaana hai aur ordered pair banaate jaana hai aur isi tarah se ye process aage badhaate jaana hai jab tak ki hame sabhi possible ordered pairs naa mil jaayein.
P✕Q = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)
(vi) Sabhi ordered pairs likh lene ke baad closed flower bracket matlab ' } ' likhna hai.
P✕Q = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}
That's it! Hamne P✕Q find kar liya!
Isi procedure ka use karke hum Q✕P bhi find kar sakte hain-
Q✕P = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}
Agar iss question ko formal way mein kiya jaaye to ye kuch aisa hoga-
P✕Q = {1, 2, 3} ✕ {a, b}
= {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}
Isi tarah se,
Q✕P = {a, b} ✕ {1, 2, 3}
= {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}
Chaliye kuch aur examples dekh lete hain:
Ex.1. If G = {7, 8} and H = {5, 4, 2}, find G✕H and H✕G.
Solution: G = {7, 8}
H = {5, 4, 2}
G✕H = {7, 8}✕{5, 4, 2}
= {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H✕G = {5, 4, 2} ✕ {7, 8}
= {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
Ex.2. If A = {1, 2, 3, 4, 5} and B = {-1, -2, -3} then find A✕B, B✕A and B✕B.
Solution: A = {1, 2, 3, 4, 5}
B = {-1, -2, -3}
A✕B = {1, 2, 3, 4, 5} ✕ {-1, -2, -3}
= {(1, -1), (1, -2), (1, -3), (2, -1), (2, -2), (2, -3), (3, -1), (3, -2), (3, -3), (4, -1), (4, -2), (4, -3), (5, -1), (5, -2), (5, -3)}
B✕A = {-1, -2, -3} ✕ {1, 2, 3, 4, 5}
= {(-1, 1), (-1, 2), (-1, 3), (-1, 4), (-1, 5), (-2, 1), (-2, 2), (-2, 3), (-2, 4), (-2, 5), (-3, 1), (-3, 2), (-3, 3), (-3, 4), (-3, 5)}
B✕B = {-1, -2, -3} ✕ {-1, -2, -3}
= {(-1, -1), (-1, -2), (-1, -3), (-2, -1), (-2, -2), (-2, -3), (-3, -1), (-3, -2), (-3, -3)}
Ex.3. If A = {1, 2}, B = {a, b} and C = {m, n} then find A✕B✕C.
Solution: A = {1, 2}
B = {a, b}
C = {m, n}
A✕B✕C = {1, 2} ✕ {a, b} ✕ {m, n}
= {(1, a, m), (1, a, n), (1, b, m), (1, b, n), (2, a, m), (2, a, n), (2, b, m), (2, b, n)}
Inn ordered triplets ko kaise banana hai ye aapko iss diagram ko dekhkar samajh mein aa jaayega-
[Note: If A, B, C are non-empty and finite sets then A✕B✕C = {(a, b, c) : a ∈ A, b ∈ B and c ∈ C} here (a, b, c) is called ordered triplet].
Ex.4. P = {1, 2}. Form the set P✕P✕P.
Solution: P = {1, 2}
P✕P✕P = {1, 2} ✕ {1, 2} ✕ {1, 2}
= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}
Ex.5. Let A = {1, 2, 3} and B = {3, 4} and C = {4, 5, 6}. Find A✕(B⋂C) and (A✕B)⋂(A✕C). Are the sets A✕(B⋂C) and (A✕B)⋂(A✕C) equal?
Solution: A = {1, 2, 3}
B = {3, 4}
C = {4, 5, 6}
B⋂C = {3, 4} ⋂ {4, 5, 6}
= {4}
A✕B = {1, 2, 3} ✕ {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
A✕C = {1, 2, 3} ✕ {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
So, A✕(B⋂C) = {1, 2, 3} ✕ {4}
= {(1, 4), (2, 4), (3, 4)}
(A✕B)⋂(A✕C) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} ⋂ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
= {(1, 4), (2, 4), (3, 4)}
Yes, the sets A✕(B⋂C) and (A✕B)⋂(A✕C) are equal.
[Note: ' ⋂ ' ye sign hai intersection ka. Agar aap intersection ke baare mein detail mein jaanna chahte hain to iss blue link par click kijiye >>>
Intersection of Sets].
Read Also:
- What is Set ?
- Types of Set
- Venn Diagrams
- Intersection of Sets
- Difference of Sets
- Symmetric Difference of Sets
- What are Intervals ?
Ex.6. If (x + 1, y - 2) = (3, 1), find the value of x and y.
Solution: Ordered pairs are equal so the corresponding elements will also be equal.
So, x + 1 = 3
x = 3 - 1
x = 2
and
y - 2 = 1
y = 1 + 2
y = 3
Ex.7. If A✕B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution: A✕B mein sabse pehle likha hai A aur phir B likha hai iska matlab ye hai ki inn sabhi ordered pairs - (a, x), (a, y), (b, x) aur (b, y), mein pehle elements/members set A ke hain aur doosre elements/members set B ke hain.
So, A = {a, a, b, b}
= {a, b}
B = {x, x, y, y}
= {x, y}
I hope ki mera ye article aap ko pasand aaya hoga. Agar aap ko mera ye article pasand aaya ho to comment karke hame bata sakte hain. Aapke dwaara kiye gaye comment se hame iss tarah ke post/article ko likne ke liye motivation milta hai.
THANKS FOR READING THIS BLOG.
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