Classification of mathematical objects

[jbriggs444]

2(2e-2f)=1=>4(e-f)=1

but this is a contradiction because e and f must be odd integers and

WHY must e and f be odd integers. You've asserted that they are but you have not proved it.

You'll also want to prove that there is no natural number that, when multiplied by 8, yields 1.

But if you could prove that, then you could likely prove more easily that there is no natural number that, when multiplied by 4 yields 1.

But if you could prove that, then you could likely prove more easily that there is no natural number that, when multiplied by 2 yields 1.

But if you could do that then all these equations would have been wasted.

 

[trees and plants]

WHY must e and f be odd integers. You've asserted that they are but you have not proved it.

If they are a) e=2k and f=2l-1 or b) e=2k and f=2l., where k,l are natural numbers then it still is a contradiction.

 

[jbriggs444]

If they are a) e=2k and f=2l-1 or b) e=2k and f=2l., where k,l are natural numbers then it still is a contradiction.

So you are retracting the claim that e and f are odd. That's fine.

Now prove the contradiction. Proof by personal incredulity does not count.

If you had a definition for multiplication by 2, a definition for addition and some axioms for the behavior of the successor function, that might be a place to start.

 

[trees and plants]

a) (2k-1)-(2l)=2(k-l)-1 so

4(e-f)=1=>4((2(k-l)-1)=1 ,4 times an integer can not equal 1 contradiction

b) (2k)-(2l)=2(k-l) so

4(e-f)=1=>4((2(k-l))=1 =>8(k-l)=1 ,contradiction

for k,l natural numbers a) is the case where e is odd and f is even, b) is the case where e is even and f is even. May i ask something if this is correct do you know a similar proof of the statement like the one i probably made?

 

[jbriggs444]

a) (2k-1)-(2l)=2(k-l)-1 so

4(e-f)=1=>4((2(k-l)-1)=1 ,4 times an integer can not equal 1 contradiction

b) (2k)-(2l)=2(k-l) so

4(e-f)=1=>4((2(k-l))=1 =>8(k-l)=1 ,contradiction

for k,l natural numbers a) is the case where e is odd and f is even, b) is the case where e is even and f is even. May i ask something if this is correct do you know a similar proof of the statement like the one i probably made?

You say that 4 times an integer can not equal 1. But saying it is not the same as proving it. Prove it.

 

[trees and plants]

Ok, we have for 4n=1, where n is a natural number. 4x1=1 is wrong, 4x2=1 is wrong, so supposing that 4n=1 is also wrong, we will prove that 4(n+1)=1 is also wrong. 4(n+1)=1=>4n+4=1=>4n=-3 which is wrong because n is a natural number.

 

[jbriggs444]

Ok, we have for 4n=1, where n is a natural number. 4x1=1 is wrong, 4x2=1 is wrong, so supposing that 4n=1 is also wrong, we will prove that 4(n+1)=1 is also wrong. 4(n+1)=1=>4n+4=1=>4n=-3 which is wrong because n is a natural number.

How are you proposing to prove that 4(n+1)=1 is also wrong? At some point you are going to need to refer to the axioms.

It can be proven. But not the way you are going about it.

 

[mathwonk]

@pbuk: my apologies for a rash post. I have deleted (hopefully) the offending parts.

 

[Vanadium 50]

Lordy, this is going downhill fast.

If you want a proof, you need a plan. That's the very second thing you need. (The first is the set of axioms and definitions you are using.) An example of a plan would be:

I will prove this for natural numbers, and then extend the proof to all integers. I will prove it for natural numbers by induction, showing that if it is true for the set {1,2...n} it is true for the set {1,2...n+1}. I will extend the proof to natural numbers by showing that if a is even, so is -a, and if a is odd, so is -a. I will complete the proof by showing 0 is even.

Now you have a plan you can execute. It is not necessarily a good plan (in fact, there is a logical hole in it - can you find it?) but it is a plan. From the plan, you can then attempt a proof.

 

[trees and plants]

I think i found a proof that shows that n where n is natural number can be only odd or even. So, $n=2k$, k is a natural number or $n=2k-1$ are the only options we want for n.

Let us suppose that a number n that is in the set of natural numbers is neither odd nor even, then these hold: $n\neq2k$ and $n\neq2k-1$ $\Rightarrow$ ($n>2k$, $n>2k-1$) (1) or ($n<2k$ (i),$n>2k-1$(ii))(2) hold there are two other cases also which have similar proofs.

Let us take (1), if we subtract on both sides those inequalities, we get $0>1$ which is a contradiction, in (2), we multiply the second inequality both sides by $-1$ so we get $-n>-2k+1$(3) adding (i) and (3) we get $0<-1$ which is a contradiction.

There is left no other choice for a natural number than to be even or odd.

 

[Mark44]

Let us suppose that a number n that is in the set of natural numbers is neither odd nor even, then these hold: $n\neq2k$ and $n\neq2k-1$ $\Rightarrow$ ($n>2k$, $n>2k-1$) (1) or ($n<2k$ (i),$n>2k-1$(ii))(2) hold there are two other cases also which have similar proofs.

For your case 1, if n > 2k, necessarily n > 2k - 1.
For case 2, if n < 2k, does it necessarily follow that n > 2k - 1? For example, if k = 4, you have n < 8 and n > 7. Is that possible for natural numbers n?

 

[trees and plants]

For your case 1, if n > 2k, necessarily n > 2k - 1.
For case 2, if n < 2k, does it necessarily follow that n > 2k - 1? For example, if k = 4, you have n < 8 and n > 7. Is that possible for natural numbers n?

I made this mistake, sorry for that. So is what i wrote correct without that case?

 

[Vanadium 50]

It's good to see you regain interest after four months. But you seem to have missed an important point: If you want a proof, first you need a plan. Don't just start writing statements down.

 

[Office_Shredder]

I would suggest trying to do a proof by induction. Suppose n is the smallest number that cannot be written as 2k or 2k+1. What can you do with it?

 

[Mark44]

Inequality 1: n > 2k and n > 2k - 1

Let us take (1), if we subtract on both sides those inequalities, we get 0>1 which is a contradiction

If we subtract what from both sides of the inequalities? Without details, it is "arm waving" to conclude that 0 > 1.
As already noted, your inequality 2 is flawed, since there are no natural number n for which n < 2k and n > 2k - 1.

 

[trees and plants]

I would suggest trying to do a proof by induction. Suppose n is the smallest number that cannot be written as 2k or 2k+1. What can you do with it?

We could say that l is the minimum of a set $L=\{m: m\neq2k, m\neq2k-1, m\in N\}$, so we want to prove that $L=\emptyset$. We want to reach a contradiction? Is the proof you want like this? i have not finished this proof yet.

 

[trees and plants]

I think the way i am trying to prove it is a wrong one.

 

[Vanadium 50]

I think the way i am trying to prove it is a wrong one.

Start with a plan. Office Shredder mentioned induction. Can you form a plan around that? You mentioned contradiction. Can you form a plan around that?

 

[Vanadium 50]

If we subtract what from both sides of the inequalities?

Well, he seems to have lost interest and wandered off again. But he seems to have subtracted one inequality from another. Why he thinks this is valid is unclear. 10 > 9 and 6 > 3, but when you subtract them you get 4 > 6, which should have given him pause.

 

[trees and plants]

Start with a plan. Office Shredder mentioned induction. Can you form a plan around that? You mentioned contradiction. Can you form a plan around that?

What should my plan be?How should i start? Apart from knowing the Peano axioms what else should i know and use?

 

[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.