Classification of mathematical objects

[Office_Shredder]

1) $m-1\geq1 \Rightarrow m\geq2 \Rightarrow m\in S$ contradiction because $m-1\leq m$ and m is the smallest element of S.

2) (i)$m$ is of the form $2k$ $\Rightarrow$ $m-1$ is an odd,

or (ii) $m$ is of the form $2k-1$ $\Rightarrow$ $m-1$ is an even, but we said that m-1 is not a natural number , contradiction.

Both cases show that there is no such $m\in S$

I don't get this at all. $m-1\notin S$ means that $m-$ can be written as $2k$ or $2k+1$. You need to use that.

 

[Vanadium 50]

<sigh>
Writing random statements hoping you will accidentally prove this is not a good strategy.

<double sigh>
If m is the smallest number that is neither even nor odd, is m-2 even? Is it odd? Is it neither?

 

[trees and plants]

I don't get this at all. $m-1\notin S$ means that $m-$ can be written as $2k$ or ##2k+1#. You need to use that.

I am sorry for the mistakes before.
a) $m-1\notin S$
(i) if $m-1=2k$ $\Rightarrow$ $m=2k+1$ $m$ is odd contradiction because we said that $m\in S$

(ii) if $m-1=2k-1$ $\Rightarrow$ $m=2k-2$ $\Rightarrow$ $m$ is even, contradiction because we said that $m\in S$

(b) if it is neither of these two forms then it is a contradiction because then $m-1\leq m$ but we know that $m$ is the smallest element of $S$.

 

[trees and plants]

Is this effort correct?

 

[trees and plants]

So, i had to suppose that there is a set $S$ that contains the rest of the natural numbers and prove that it is empty.

Thank you for your help and time.

 

[Office_Shredder]

If m-1=2k-1 then m=2k, not 2k-2. But now what you have looks a lot more like a correct proof.

I think you are still not handling the m-1 not a natural number case. That just means m is 1 or 0 depending on your definition of natural numbers. You need to confirm that is either even or odd also.

 

[Vanadium 50]

I think you are still not handling the m-1 not a natural number case.

This proof requires careful examination of the edge cases. Also, as originally posed (by you) it was a statement about integers, not natural numbers, and that will make zero an edge case as well.

 

[trees and plants]

This proof requires careful examination of the edge cases. Also, as originally posed (by you) it was a statement about integers, not natural numbers, and that will make zero an edge case as well.

Is the proof for integers analogous to the one i made? Is there anything else needed?

 

[Vanadium 50]

Is the proof for integers analogous to the one i made?

You tell me. You won't get very far asking other people to prove things. You want to get better, you need to work yourself.