# Set Theory, Calendars, Clocks and Binomial Theorem

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## Bijective Functions

If set A and set B are bijective and set C and set D are bijective too, State whether there exist a bijection between AC + BD or not
1. Yes
2. No
3. Data insufficient
4. Cannot be determined

Choice A. Yes

## Detailed Solution

A/Q the functions f:A → B and g:C → D are bijections
Then g-1 must exist .
Then for a function h∈ Ac we may define a function T: AC → BBD by T(h) = f o h o g-1 . That is , for d∈D, T(h)(d) = f(h(g-1(d)))
Since g-1:D→C, the expression g-1(d) must exist
Now,
As h:C → A and g-1(d)∈C then the expression h(g-1(d)) must exist
Again, As h(g-1(d)) ∈ A and f:A → B, the expression f(h(g-1(d))) must exist
Now it only remains to prove that R(h) = f o h o g-1 is a bijection.
To do so, we need to simply provide an inverse.
Now T o R(h) = f o (f-1 o h o g) o g-1
= (f o f-1) o h o (g o g-1 )
= idB o h o idD
= h Therefore R:h → f-1 oh o g exists and is an inverse to T
Hence there exists a bijection between AC + BD

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