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A question on Goldrei Theorem 3.13 : prove that if $n>m$ and $a>0$ then $acdot n>acdot m$
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This question is about the proof of Theorem 3.13 b) in Goldreis' "Classic Set Theory":
For all natural numbers $n,m,a$, if $a>0$ and $n>m$ then $acdot n>acdot m quad(1.)$
It has previously been established that if $m<n$ then $m+a<n+a quad (2.)$.
$n^+$ denotes the "successor" $S(n)$ of any natural number $n$ (n+1 if you will).
Goldrei writes that for all $n leq m$ the statement $(*)$ holds vacuously, so "...that the smallest $n$ for which there are anything significant to prove, namely because $m<n$, is $m^+$.". He proceeds with proving $(1.)$ when $n=m^+$.
This is where my first question comes from, because I don't think proving this "base case" is necessary, precisely because $(1.)$ holds vacuously for all $nleq m$. Even if these cases are true vacuously they are still true, right? So we could just use $n=0$ as the base case.
Then he aims to prove $(1.)$ by induction with the induction hypothesis that $(1.)$ holds for $n > m$ in the following way:
$acdot n^+ = (acdot n) + a > (acdot n) + 0quad(2.) = acdot n > acdot m$ (by induction hypothesis)
My issue here is that the last step using the induction hypothesis assumes that $n > m$ but the only thing we know from $n^+ > m$ is that $ngeq m$. Of course this is not a big problem since it is sufficient that $acdot n = a cdot m$ for $acdot n^+ > a cdot n = a cdot m$
So the big question for me is was the base case with $n=m^+$ even necessary?
elementary-set-theory induction natural-numbers
asked 2 hours ago
DancingIceCream
112
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This question is about the proof of Theorem 3.13 b) in Goldreis' "Classic Set Theory":
For all natural numbers $n,m,a$, if $a>0$ and $n>m$ then $acdot n>acdot m quad(1.)$
It has previously been established that if $m<n$ then $m+a<n+a quad (2.)$.
$n^+$ denotes the "successor" $S(n)$ of any natural number $n$ (n+1 if you will).
Goldrei writes that for all $n leq m$ the statement $(*)$ holds vacuously, so "...that the smallest $n$ for which there are anything significant to prove, namely because $m<n$, is $m^+$.". He proceeds with proving $(1.)$ when $n=m^+$.
This is where my first question comes from, because I don't think proving this "base case" is necessary, precisely because $(1.)$ holds vacuously for all $nleq m$. Even if these cases are true vacuously they are still true, right? So we could just use $n=0$ as the base case.
Then he aims to prove $(1.)$ by induction with the induction hypothesis that $(1.)$ holds for $n > m$ in the following way:
$acdot n^+ = (acdot n) + a > (acdot n) + 0quad(2.) = acdot n > acdot m$ (by induction hypothesis)
My issue here is that the last step using the induction hypothesis assumes that $n > m$ but the only thing we know from $n^+ > m$ is that $ngeq m$. Of course this is not a big problem since it is sufficient that $acdot n = a cdot m$ for $acdot n^+ > a cdot n = a cdot m$
So the big question for me is was the base case with $n=m^+$ even necessary?
elementary-set-theory induction natural-numbers
asked 2 hours ago
DancingIceCream
112
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is about the proof of Theorem 3.13 b) in Goldreis' "Classic Set Theory":
For all natural numbers $n,m,a$, if $a>0$ and $n>m$ then $acdot n>acdot m quad(1.)$
It has previously been established that if $m<n$ then $m+a<n+a quad (2.)$.
$n^+$ denotes the "successor" $S(n)$ of any natural number $n$ (n+1 if you will).
Goldrei writes that for all $n leq m$ the statement $(*)$ holds vacuously, so "...that the smallest $n$ for which there are anything significant to prove, namely because $m<n$, is $m^+$.". He proceeds with proving $(1.)$ when $n=m^+$.
This is where my first question comes from, because I don't think proving this "base case" is necessary, precisely because $(1.)$ holds vacuously for all $nleq m$. Even if these cases are true vacuously they are still true, right? So we could just use $n=0$ as the base case.
Then he aims to prove $(1.)$ by induction with the induction hypothesis that $(1.)$ holds for $n > m$ in the following way:
$acdot n^+ = (acdot n) + a > (acdot n) + 0quad(2.) = acdot n > acdot m$ (by induction hypothesis)
My issue here is that the last step using the induction hypothesis assumes that $n > m$ but the only thing we know from $n^+ > m$ is that $ngeq m$. Of course this is not a big problem since it is sufficient that $acdot n = a cdot m$ for $acdot n^+ > a cdot n = a cdot m$
So the big question for me is was the base case with $n=m^+$ even necessary?
elementary-set-theory induction natural-numbers
asked 2 hours ago
DancingIceCream
112
This question is about the proof of Theorem 3.13 b) in Goldreis' "Classic Set Theory":
For all natural numbers $n,m,a$, if $a>0$ and $n>m$ then $acdot n>acdot m quad(1.)$
It has previously been established that if $m<n$ then $m+a<n+a quad (2.)$.
$n^+$ denotes the "successor" $S(n)$ of any natural number $n$ (n+1 if you will).
Goldrei writes that for all $n leq m$ the statement $(*)$ holds vacuously, so "...that the smallest $n$ for which there are anything significant to prove, namely because $m<n$, is $m^+$.". He proceeds with proving $(1.)$ when $n=m^+$.
This is where my first question comes from, because I don't think proving this "base case" is necessary, precisely because $(1.)$ holds vacuously for all $nleq m$. Even if these cases are true vacuously they are still true, right? So we could just use $n=0$ as the base case.
Then he aims to prove $(1.)$ by induction with the induction hypothesis that $(1.)$ holds for $n > m$ in the following way:
$acdot n^+ = (acdot n) + a > (acdot n) + 0quad(2.) = acdot n > acdot m$ (by induction hypothesis)
My issue here is that the last step using the induction hypothesis assumes that $n > m$ but the only thing we know from $n^+ > m$ is that $ngeq m$. Of course this is not a big problem since it is sufficient that $acdot n = a cdot m$ for $acdot n^+ > a cdot n = a cdot m$
So the big question for me is was the base case with $n=m^+$ even necessary?
elementary-set-theory induction natural-numbers
asked 2 hours ago
DancingIceCream
112
edited 1 hour ago
asked 2 hours ago
DancingIceCream
112
asked 2 hours ago
DancingIceCream
112
asked 2 hours ago
DancingIceCream
112
112
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