q4 3. Consider the set of all possible polynomials which can be formed which hav
ID: 3589094 • Letter: Q
Question
q4
Explanation / Answer
Please see the below answer.
Answer:3)
The set P of all polynomials having rational (integer) coefficients represented as P = n-1 Pn,
where Pn = {c n n + c n-1xn-1 + · · · + c 1x + c0 : cn Q {0},c n-1, c n-2, . . . , c0 Q} (n N) as the set of all polynomials of degree n with a coefficient of rational (integer) number.
The set of polynomials Pn has equal cardinality as (Q {0}) × Qn.
As the set (Q {0}) × Qn is considered as countable as a cartesian product of a finite number of sets which are countable ,then it is concluded that P = n-1 Pn is countable as a countable union of countable sets.
Answer:4)
If there is a bijection between the set of natural numbers N, and the S is set of natural numbers where S N.
So by bijection from N to SxN. There is an injection f:N->SxN as f(n)=(n,1).
There has to be an injection g:NxN->N g(m,n)=(2^m)(3^n).
According to Cantor-Bernstein-Schroeder theorem there is a bijection
h:NxN->N.So if by surjection k:NxN->N by k(m,n)=m/n. Then t=kh(-1):N->N
is a surjection.
As the union of two countable sets is countable that there is a bijection
g:N x N->N.
A nonempty set N is countable if and only if there exists a surjection N --> N as S-- > N
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