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Peano axioms: prove that there is no natural number between n and sucessor of n

Proving every positive natural number has unique predecessorIs every natural number even or odd?Prove $n<m++leq n++iff m=n$ from Peano AxiomsHow can I prove this proposition from Peano Axioms?Prove that the quotient of a nonzero rational number and an irrational number is irrationalPeano and induction according to Schaum's OutlineProve that the system $(P, S, 0)$ satisfy Peano Axioms.Peano axioms-Mathematical Induction$x_1$ and $x_2$ roots of $f$ with $f'(x_1) gt 0$ and $f'(x_2) gt 0$. Existence of another root between $(x_1,x_2)$.“No infinite descending membership chains”, Peano Axioms, Axiom of Regularity

1

$begingroup$

Let $m in mathbbN$. Show that $nexists n in mathbbN$ such that $m < n < s(m)$, where $s$ is the successor function.

---

Here's my proof using only the Peano Axioms I was introduced. I'd appreciate someone to check my work.

Let's prove it by contradiction. Suppose that $exists n in mathbbN$ such that (1) $m < n$ and (2) $ n < s(m)$.

From the definition of order we have that:

(1) implies that $exists c in mathbbN setminus 0$ such that $m + c = n$.

(2) imples that $exists c' in mathbbN setminus 0$ such that $n + c' = s(m)$.

Therefore
$$
beginalign*
n + c' + c &= s(m) + c\
&= s(m + c)\
&= s(n)\
endalign*
$$

Since $c' neq 0$ and $c neq 0$, we know that $c' + c neq 0$ and therefore $exists k in mathbbN$ such that $s(k) = c' + c$. Hence:

$$
beginalign*
n + s(k) &= s(n)\
s(n + k) &= s(n) rightarrow n + k = n rightarrow k = 0
endalign*
$$

Since we've concluded that $k=0$, we have that $s(k) = s(0) = 1 = c + c'$, where follows that $c$ or $c'$ needs to be equals to $0$, and that is a contradiction.

---

I also have one more question... After defining what a function is, can't we conclude that $s : mathbbN setminus 0 rightarrow mathbbN$ is actually $s(n) = n + 1$ ?

If so, using that the proof would've been more immediate...

Any kind of comments and critics are highly appreciated! Thank you!

real-analysis proof-verification peano-axioms

share|cite|improve this question

asked Mar 30 at 5:19

Bruno ReisBruno Reis

1,039418

$endgroup$

add a comment | 

1

$begingroup$

Let $m in mathbbN$. Show that $nexists n in mathbbN$ such that $m < n < s(m)$, where $s$ is the successor function.

---

Here's my proof using only the Peano Axioms I was introduced. I'd appreciate someone to check my work.

Let's prove it by contradiction. Suppose that $exists n in mathbbN$ such that (1) $m < n$ and (2) $ n < s(m)$.

From the definition of order we have that:

(1) implies that $exists c in mathbbN setminus 0$ such that $m + c = n$.

(2) imples that $exists c' in mathbbN setminus 0$ such that $n + c' = s(m)$.

Therefore
$$
beginalign*
n + c' + c &= s(m) + c\
&= s(m + c)\
&= s(n)\
endalign*
$$

Since $c' neq 0$ and $c neq 0$, we know that $c' + c neq 0$ and therefore $exists k in mathbbN$ such that $s(k) = c' + c$. Hence:

$$
beginalign*
n + s(k) &= s(n)\
s(n + k) &= s(n) rightarrow n + k = n rightarrow k = 0
endalign*
$$

Since we've concluded that $k=0$, we have that $s(k) = s(0) = 1 = c + c'$, where follows that $c$ or $c'$ needs to be equals to $0$, and that is a contradiction.

---

I also have one more question... After defining what a function is, can't we conclude that $s : mathbbN setminus 0 rightarrow mathbbN$ is actually $s(n) = n + 1$ ?

If so, using that the proof would've been more immediate...

Any kind of comments and critics are highly appreciated! Thank you!

real-analysis proof-verification peano-axioms

share|cite|improve this question

asked Mar 30 at 5:19

Bruno ReisBruno Reis

1,039418

$endgroup$

add a comment | 

$begingroup$

Let $m in mathbbN$. Show that $nexists n in mathbbN$ such that $m < n < s(m)$, where $s$ is the successor function.

---

Here's my proof using only the Peano Axioms I was introduced. I'd appreciate someone to check my work.

Let's prove it by contradiction. Suppose that $exists n in mathbbN$ such that (1) $m < n$ and (2) $ n < s(m)$.

From the definition of order we have that:

(1) implies that $exists c in mathbbN setminus 0$ such that $m + c = n$.

(2) imples that $exists c' in mathbbN setminus 0$ such that $n + c' = s(m)$.

Therefore
$$
beginalign*
n + c' + c &= s(m) + c\
&= s(m + c)\
&= s(n)\
endalign*
$$

Since $c' neq 0$ and $c neq 0$, we know that $c' + c neq 0$ and therefore $exists k in mathbbN$ such that $s(k) = c' + c$. Hence:

$$
beginalign*
n + s(k) &= s(n)\
s(n + k) &= s(n) rightarrow n + k = n rightarrow k = 0
endalign*
$$

Since we've concluded that $k=0$, we have that $s(k) = s(0) = 1 = c + c'$, where follows that $c$ or $c'$ needs to be equals to $0$, and that is a contradiction.

---

I also have one more question... After defining what a function is, can't we conclude that $s : mathbbN setminus 0 rightarrow mathbbN$ is actually $s(n) = n + 1$ ?

If so, using that the proof would've been more immediate...

Any kind of comments and critics are highly appreciated! Thank you!

real-analysis proof-verification peano-axioms

share|cite|improve this question

asked Mar 30 at 5:19

Bruno ReisBruno Reis

1,039418

$endgroup$

share|cite|improve this question

asked Mar 30 at 5:19

Bruno ReisBruno Reis

1,039418

share|cite|improve this question

asked Mar 30 at 5:19

Bruno ReisBruno Reis

1,039418

asked Mar 30 at 5:19

Bruno ReisBruno Reis

1,039418

asked Mar 30 at 5:19

Bruno ReisBruno Reis

1,039418

1 Answer
1

active

oldest

votes

1

$begingroup$

Your proof doesn't work as it is, because from "$n+k = n$" you cannot get "$k = 0$" just like that. You need induction in order to prove this cancellation fact. Either that or you just use induction to directly prove the original desired theorem.

share|cite|improve this answer

answered Mar 31 at 10:46

user21820user21820

40.1k544162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That ancellation fact was previously proved using induction.
  $endgroup$
  – Bruno Reis
  Mar 31 at 12:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BrunoReis: Then that's fine, but you should have stated it, otherwise people like me will guess that you're cancelling without justification. It's a common mistake since many students don't stick strictly to the deductive rules and given axioms. =)
  $endgroup$
  – user21820
  Mar 31 at 15:24
  

  
  
  
  

add a comment | 

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1

$begingroup$

Your proof doesn't work as it is, because from "$n+k = n$" you cannot get "$k = 0$" just like that. You need induction in order to prove this cancellation fact. Either that or you just use induction to directly prove the original desired theorem.

share|cite|improve this answer

answered Mar 31 at 10:46

user21820user21820

40.1k544162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That ancellation fact was previously proved using induction.
  $endgroup$
  – Bruno Reis
  Mar 31 at 12:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BrunoReis: Then that's fine, but you should have stated it, otherwise people like me will guess that you're cancelling without justification. It's a common mistake since many students don't stick strictly to the deductive rules and given axioms. =)
  $endgroup$
  – user21820
  Mar 31 at 15:24
  

  
  
  
  

add a comment | 

1

$begingroup$

Your proof doesn't work as it is, because from "$n+k = n$" you cannot get "$k = 0$" just like that. You need induction in order to prove this cancellation fact. Either that or you just use induction to directly prove the original desired theorem.

share|cite|improve this answer

answered Mar 31 at 10:46

user21820user21820

40.1k544162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That ancellation fact was previously proved using induction.
  $endgroup$
  – Bruno Reis
  Mar 31 at 12:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BrunoReis: Then that's fine, but you should have stated it, otherwise people like me will guess that you're cancelling without justification. It's a common mistake since many students don't stick strictly to the deductive rules and given axioms. =)
  $endgroup$
  – user21820
  Mar 31 at 15:24
  

  
  
  
  

add a comment | 

$begingroup$

Your proof doesn't work as it is, because from "$n+k = n$" you cannot get "$k = 0$" just like that. You need induction in order to prove this cancellation fact. Either that or you just use induction to directly prove the original desired theorem.

share|cite|improve this answer

answered Mar 31 at 10:46

user21820user21820

40.1k544162

$endgroup$

share|cite|improve this answer

answered Mar 31 at 10:46

user21820user21820

40.1k544162

answered Mar 31 at 10:46

user21820user21820

40.1k544162

answered Mar 31 at 10:46

user21820user21820

40.1k544162