MHB Do we have to check the second case?

  • Thread starter Thread starter evinda
  • Start date Start date
AI Thread Summary
The discussion centers on the necessity of checking a second case in a proof regarding the immediate successor of a natural number, n'. It is established that if the relationship \( n < m \) is defined as \( n \in m \), then the second case can be deemed unnecessary if this relationship is previously proven. Participants explore how to prove this relationship, referencing definitions from lecture notes that clarify the order of natural numbers. The conclusion reached is that the defined relationship allows for the omission of the second case in the proof. The conversation emphasizes the importance of definitions in mathematical proofs.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hi! (Smile)

Proposition:
The natural number $n'$ is the immediate successor of $n$, i.e. there is no natural number $m$ such that $n<m \wedge m<n'$.

Proof:
We assume that there is a $m$ such that $n<m \wedge m<n'$. Then $n \subset m$ and $m \subset n \cup \{n\}$. 1st case: $n \in m$. Then $n \cup \{n\} \subset m$ and so $n'=m$, contradiction.

2nd case: $n \notin m$. Then since $m \subset n \cup \{n\}$ we have that $m \subset n$. Since $n \subset m$ we have $n=m$, contradiction.
Do we have to check the second case knowing that $n<m \leftrightarrow n \in m$ ? (Thinking)
 
Physics news on Phys.org
evinda said:
Do we have to check the second case knowing that $n<m \leftrightarrow n \in m$ ?
No, if $n<m \leftrightarrow n \in m$ is proved before the current proposition, then the second case is unnecessary.
 
Evgeny.Makarov said:
No, if $n<m \leftrightarrow n \in m$ is proved before the current proposition, then the second case is unnecessary.

It hasn't been proven, it is referred in a proposition. (Worried)
How could we prove it? (Thinking)
 
evinda said:
It hasn't been proven, it is referred in a proposition.
What does this mean: "it is referred in a proposition"?

evinda said:
How could we prove it?
You should start with a definition of $<$. Isn't this statement proved in your lecture notes?
 
According to my notes:

Definition: For any natural numbers $m,n$ we define:
$m<n \leftrightarrow m \in n (\leftrightarrow \langle m,n \rangle \in \epsilon_{\omega})$

$(m \leq n \leftrightarrow m \in n \lor m=n)$​
where $\epsilon_{\omega}=\{ \langle m,n \rangle\in \omega^2: m \in n \}$.$\epsilon_{\omega}$ defines an order on $\omega$ and is symbolized with $<$.
The non-strict order of $\omega$ that corresponds to the order $\epsilon_{\omega}$ is symbolized with $\leq$, so:

$$n \leq m \leftrightarrow n \in m \lor n=m$$
$$(n<m \leftrightarrow n \in m)$$

Proposition:

  • For any natural numbers $m,n$: $m \leq n \leftrightarrow m \subset n$
  • For any natural numbers $m,n$: $m<n \leftrightarrow m \subsetneq n (\leftrightarrow m \in n)$
 
Last edited:
Evgeny.Makarov said:
No, if $n<m \leftrightarrow n \in m$ is proved before the current proposition, then the second case is unnecessary.
evinda said:
How could we prove it?

evinda said:
According to my notes:

Definition: For any natural numbers $m,n$ we define:
$m<n \leftrightarrow m \in n$​
Then you answered your own question.
 
So we can use the fact that $m<n \leftrightarrow m \in n$ without proving it since we defined it like that, right? (Smile)
 
Yes.
 
Nice... Thanks a lot! (Yes)
 

Similar threads

Replies
6
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
5
Views
1K
Replies
19
Views
3K
Replies
30
Views
5K
Back
Top