Classification of mathematical objects

[Vanadium 50]

What should my plan be?

If I tell you that, it's not your proof, is it? It's mine.

What plans have you made and why did you discard them? And if you really don't have a clue on where to start, maybe we need to step back and try a simpler proof.

 

[trees and plants]

If I tell you that, it's not your proof, is it? It's mine.

What plans have you made and why did you discard them? And if you really don't have a clue on where to start, maybe we need to step back and try a simpler proof.

1. Let us suppose there is a natural number that is neither an odd nor an even number. 2. Try the cases of $k=1$ for $n=2k$ for evens and then $n=2k-1$ for odds and then supposing it works for $k=n$ we prove for $k=n+1$ 3. Try by using successors but i do not know how to do it. These are my plans.

 

[Vanadium 50]

That's not a plan. A plan explains how the process will work. It has a beginning, a middle and an end. Your plan is similar to (but not as good as) a plan to drive from Rome to Paris: "Drive out of Rome. Drive on various streets until I bump into Paris." The driving plan is better because it at least states that it's random luck that this plan is counting on.

An example of a plan - and I wouldn't use this plan - would be:

1. Start with the set of natural numbers
2. Remove all the even numbers
3. Remove all the odd numbers
4. Show that the resultant set is empty. QED.

See how it has a beginning, a middle and (most importantly) an end? That's a plan that one might execute. (Although I would not use this plan for this problem)

 

[trees and plants]

That's not a plan. A plan explains how the process will work. It has a beginning, a middle and an end. Your plan is similar to (but not as good as) a plan to drive from Rome to Paris: "Drive out of Rome. Drive on various streets until I bump into Paris." The driving plan is better because it at least states that it's random luck that this plan is counting on.

An example of a plan - and I wouldn't use this plan - would be:

1. Start with the set of natural numbers
2. Remove all the even numbers
3. Remove all the odd numbers
4. Show that the resultant set is empty. QED.

See how it has a beginning, a middle and (most importantly) an end? That's a plan that one might execute. (Although I would not use this plan for this problem)

1. $N=\{1,2,3,...\}$

2.mathematical induction for the even numbers which means $2$ is even, and supposing that $2k$ is even we show that $2(k+1)$ is even

3. mathematical induction for the odd numbers but now we have $2k-1$

4. suppose that an element exists that belongs to the$N$ and is neither an even nor an odd and showing that this leads to contradiction, which means it is empty.

I get stuck at the step number 4.What should i do?

 

[Office_Shredder]

I think the proof by contradiction is a fine approach. Let S be the set of natural numbers that are not of the form 2k or 2k+1. Let m be the smallest element of S. There are two possibilities:
1.) m-1 is a natural number. If it is what can you say about it?
2.) m-1 is not a natural number. If it's not, what can you say about m?

 

[Vanadium 50]

1. Extra close brace or missing open brace

You might want to fix this.

Let m be the smallest element of S.

This is a very, very common approach. Learn it. Love it.

I get stuck at the step number 4.What should i do?

Probably ask us to do the rest for you. Why should this be any different?

You are still confusing a plan for a proof with a proof itself. Until you stop doing this, you will be unable to prove anything that requires anything beyond what you can immediately see. That means you be unable to prove anything that is non-obvious.

@Office_Shredder has written down a plan. His plan is:

• We're going to do this by contradiction
• If the set of numbers that is neither even nor odd is non-empty, it has a smallest member.
• Look at properties of this smallest member and show that they lead to a contradiction.

That is a good plan. It is also not a proof. It is a plan for a proof.

 

[trees and plants]

I think the proof by contradiction is a fine approach. Let S be the set of natural numbers that are not of the form 2k or 2k+1. Let m be the smallest element of S. There are two possibilities:
1.) m-1 is a natural number. If it is what can you say about it?
2.) m-1 is not a natural number. If it's not, what can you say about m?

Should i use infimums and supremums and lower bounds or upper bounds?

 

[Vanadium 50]

I don't understand why you are ignoring the advice that @Office_Shredder gave you.

 

[trees and plants]

I think the proof by contradiction is a fine approach. Let S be the set of natural numbers that are not of the form 2k or 2k+1. Let m be the smallest element of S. There are two possibilities:
1.) m-1 is a natural number. If it is what can you say about it?
2.) m-1 is not a natural number. If it's not, what can you say about m?

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 $m\in S$

 

[trees and plants]

I want to say that i knew about the method of telling $S$ has its smallest element, I had read in other proofs of other theorems but i did not know about continuing with $m-1$ and reaching a contradiction.

Thank you.

 

[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.