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seen Feb 10 '13 at 17:49

I have a BS in Math with 9-12 certification, and an MEd in Curriculum and Instruction with an emphasis in Mathematics Education. I have just returned to school to start work on an MA in mathematics...almost 9 years after finishing my undergrad. Needless to say, the noggin is a little rusty, especially with proofs!


Sep
24
awarded  Autobiographer
Feb
9
awarded  Supporter
Feb
9
comment Exhibit a bijection between $\mathbb{N}$ and all odd $\mathbb{Z}$ greater than 13.
Deinitely would not have thought of it this way. I need to let it sit for a bit, then come back to it. Thanks!
Feb
8
comment Exhibit a bijection between $\mathbb{N}$ and all odd $\mathbb{Z}$ greater than 13.
Start with Let $f(m) \in \mathbb{Z}$ such that $f(m)=2n+1$ or should it be $2n+13$ in this case? The hardest part for me seems to be starting a proof. Once I get it rolling, I'm usually okay.
Feb
8
comment Exhibit a bijection between $\mathbb{N}$ and all odd $\mathbb{Z}$ greater than 13.
I meant $2n + 13$. Shoot!
Feb
8
revised Exhibit a bijection between $\mathbb{N}$ and all odd $\mathbb{Z}$ greater than 13.
meant 2n + 13 !
Feb
8
comment Exhibit a bijection between $\mathbb{N}$ and all odd $\mathbb{Z}$ greater than 13.
Do you mean odds represented as $2n+1$, and work backwards somehow?
Feb
8
asked Exhibit a bijection between $\mathbb{N}$ and all odd $\mathbb{Z}$ greater than 13.
Feb
8
answered Which Mathematical Analysis I Book or Textbook Is The Best?
Feb
3
revised Showing inverse composed with function is $x$ for all $x$ in the domain.
improved formatting, offered questionable answer
Feb
3
awarded  Editor
Feb
3
revised Showing inverse composed with function is $x$ for all $x$ in the domain.
improved formatting
Feb
3
awarded  Student
Feb
3
comment Showing inverse composed with function is $x$ for all $x$ in the domain.
(Again, I apologize, I'm trying to learn how to code this, and don't have a clue!). Inverse: If f mapping A onto B is a bijection of A onto B, then g:={(b, a) element of BxA: (a,b) element of f} is a function of B into A.
Feb
3
asked Showing inverse composed with function is $x$ for all $x$ in the domain.