Consider a function f:Y_n→X_n^2 For any (a,b) in Y_n, let f((a,b))=f(a,b)=an+b.

Will Underwood 2022-09-26 Answered
Consider a function f : Y n X n 2
For any ( a , b ) Y n , let f ( ( a , b ) ) = f ( a , b ) = a n + b . Prove that f is a bijection and find its inverse f 1 .
How to find an inverse of this multivariable function. I was already able to prove that f ( a , b ) = a n + b is a bijection, but not certain how to find its specific inverse. It would be important to note that Y n = X n × X n where X n = { 0 , 1 , 2 , . . . , n 1 }.
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Answers (1)

Ronan Rollins
Answered 2022-09-27 Author has 9 answers
injectivity: Let f ( a 1 , b 1 ) = f ( a 2 , b 2 ).
Then ( a 1 a 2 ) n = b 1 b 2 .
If a 1 a 2 , | L H S | > n > | R H S | , which is absurd. Therefore, L H S = R H S = 0.
surjectivity: Let c X n 2 . By Division Algorithm (𝑛 is the divisor), there exists unique quotient a N { 0 } and remainder b X n such that c = a n + b.
Check that a X n . It's obvious that a 0. If a n, c a n n 2 , so c X n 2 , which is absurd.
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