wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Prove that : (i) (AB)×C=(A×C)(B×C)

(ii) (AB)×C=(A×C)(B×C).

Open in App
Solution

(i) Let (a, b) be an arbitary element of (AB)×C. Then,

(a,b)ϵ(AB)×C

aϵAB and bϵC [By defination]

(aϵA) and bϵC) or (a \epsilon A)\) and bϵC)

(a,b)ϵA×C or (a, b) ϵB×C

(a,b)ϵ(A×C)(B×C)

(a,b)ϵ(AB)×C

(a,b)ϵ(A×C)(B×C)

(AB)×C(A×C)(B×C) ....(i)

Again, let (x, y) be an arbitary element of (A×C)(B×C). Then,

(x,y)ϵ(A×C)(B×C)

(x,y)ϵA×C or (x, y) \epsilon B \times C\)

xϵA and yϵ C or xϵB and yϵC

(xϵAB) and yϵC

(x,y)ϵ(AB)×C

(x,y)ϵ(A×C)(B×C)

(x,y)ϵ(AB)×C

(A×C)(B×C)(AB)×C ....(i)

Using equation (i) and equation (ii), we get (AB)×C=(A×C)(B×C)

Hence proved.

(ii) Let (a, b) be an arbitary element of (AB)×C. Then,

(a,b)ϵ(AB)×C

aϵAB and bϵC

(aϵA and aϵB) and bϵC

[By defination]

(aϵA and bϵC) and (aϵB and bϵC)

(a,b)ϵA×C and (a,b)ϵB×C

(a,b)ϵ(A×C)(B×C)

(a,b)ϵ(AC)×C

(a,b)ϵ(A×C)(B×C)

(AC)×C(A×C)(B×C) ....(i)

Let (x, y) be an arbitary element of (A×C)(B×C). Then,

(x,y)ϵ(A×C)(B×C)

(x,y)ϵA×C and (x,y)ϵB×C

[By defination]

(xϵA and yϵC) and (xϵB and yϵC)

(xϵA and xϵB) and yϵC

xϵAB and yϵC

(x,y)ϵ(AB)×C

(A×C)(B×C)(AB)×C ....(ii)

Using equation (i) and equation (ii), we get (AB)×C=(A×C)(B×C)

Hence proved.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Cartesian Product
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon