Sum/Product of two natural numbers is a natural number
$begingroup$
I wanted to prove that the sum and the product of two natural numbers is a natural number. Intuitively it's clear to my why that is true, however I couldn't prove it.
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , 1leq n$ .
So this is the information we have so far. But I am having difficulty proving that the sum/product of two natural numbers is a natural number. Can someone give me a clue how to prove this?
I am trying to prove this with the definition of a Natural numbers, but it's not working so far...
Thank you.
set-theory natural-numbers
edited Feb 1 '15 at 11:38
Martin Sleziak
44.7k10119272
asked Oct 9 '14 at 13:29
Charles CarmichaelCharles Carmichael
30917
$endgroup$
|
show 5 more comments
$begingroup$
I wanted to prove that the sum and the product of two natural numbers is a natural number. Intuitively it's clear to my why that is true, however I couldn't prove it.
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , 1leq n$ .
So this is the information we have so far. But I am having difficulty proving that the sum/product of two natural numbers is a natural number. Can someone give me a clue how to prove this?
I am trying to prove this with the definition of a Natural numbers, but it's not working so far...
Thank you.
set-theory natural-numbers
edited Feb 1 '15 at 11:38
Martin Sleziak
44.7k10119272
asked Oct 9 '14 at 13:29
Charles CarmichaelCharles Carmichael
30917
$endgroup$
1
$begingroup$
First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
$endgroup$
– Daniel Fischer♦
Oct 9 '14 at 13:33
$begingroup$
Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
$endgroup$
– Charles Carmichael
Oct 9 '14 at 13:57
$begingroup$
inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
$endgroup$
– Charles Carmichael
Oct 9 '14 at 14:01
$begingroup$
@DavidC You need to tag him, otherwise he won't get the notification of your comment.
$endgroup$
– Vincenzo Oliva
Oct 11 '14 at 9:24
1
$begingroup$
I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
$endgroup$
– Martin Sleziak
Feb 1 '15 at 11:40
|
show 5 more comments
$begingroup$
I wanted to prove that the sum and the product of two natural numbers is a natural number. Intuitively it's clear to my why that is true, however I couldn't prove it.
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , 1leq n$ .
So this is the information we have so far. But I am having difficulty proving that the sum/product of two natural numbers is a natural number. Can someone give me a clue how to prove this?
I am trying to prove this with the definition of a Natural numbers, but it's not working so far...
Thank you.
set-theory natural-numbers
edited Feb 1 '15 at 11:38
Martin Sleziak
44.7k10119272
asked Oct 9 '14 at 13:29
Charles CarmichaelCharles Carmichael
30917
$endgroup$
I wanted to prove that the sum and the product of two natural numbers is a natural number. Intuitively it's clear to my why that is true, however I couldn't prove it.
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , 1leq n$ .
So this is the information we have so far. But I am having difficulty proving that the sum/product of two natural numbers is a natural number. Can someone give me a clue how to prove this?
I am trying to prove this with the definition of a Natural numbers, but it's not working so far...
Thank you.
set-theory natural-numbers
set-theory natural-numbers
edited Feb 1 '15 at 11:38
Martin Sleziak
44.7k10119272
asked Oct 9 '14 at 13:29
Charles CarmichaelCharles Carmichael
30917
edited Feb 1 '15 at 11:38
Martin Sleziak
44.7k10119272
asked Oct 9 '14 at 13:29
Charles CarmichaelCharles Carmichael
30917
edited Feb 1 '15 at 11:38
Martin Sleziak
44.7k10119272
edited Feb 1 '15 at 11:38
Martin Sleziak
44.7k10119272
edited Feb 1 '15 at 11:38
Martin Sleziak
44.7k10119272
44.7k10119272
asked Oct 9 '14 at 13:29
Charles CarmichaelCharles Carmichael
30917
asked Oct 9 '14 at 13:29
Charles CarmichaelCharles Carmichael
30917
asked Oct 9 '14 at 13:29
Charles CarmichaelCharles Carmichael
30917
30917
1
$begingroup$
First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
$endgroup$
– Daniel Fischer♦
Oct 9 '14 at 13:33
$begingroup$
Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
$endgroup$
– Charles Carmichael
Oct 9 '14 at 13:57
$begingroup$
inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
$endgroup$
– Charles Carmichael
Oct 9 '14 at 14:01
$begingroup$
@DavidC You need to tag him, otherwise he won't get the notification of your comment.
$endgroup$
– Vincenzo Oliva
Oct 11 '14 at 9:24
1
$begingroup$
I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
$endgroup$
– Martin Sleziak
Feb 1 '15 at 11:40
|
show 5 more comments
1
$begingroup$
First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
$endgroup$
– Daniel Fischer♦
Oct 9 '14 at 13:33
$begingroup$
Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
$endgroup$
– Charles Carmichael
Oct 9 '14 at 13:57
$begingroup$
inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
$endgroup$
– Charles Carmichael
Oct 9 '14 at 14:01
$begingroup$
@DavidC You need to tag him, otherwise he won't get the notification of your comment.
$endgroup$
– Vincenzo Oliva
Oct 11 '14 at 9:24
1
$begingroup$
I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
$endgroup$
– Martin Sleziak
Feb 1 '15 at 11:40
1
1
$begingroup$
First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
$endgroup$
– Daniel Fischer♦
Oct 9 '14 at 13:33
$begingroup$
First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
$endgroup$
– Daniel Fischer♦
Oct 9 '14 at 13:33
$begingroup$
Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
$endgroup$
– Charles Carmichael
Oct 9 '14 at 13:57
$begingroup$
Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
$endgroup$
– Charles Carmichael
Oct 9 '14 at 13:57
$begingroup$
inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
$endgroup$
– Charles Carmichael
Oct 9 '14 at 14:01
$begingroup$
inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
$endgroup$
– Charles Carmichael
Oct 9 '14 at 14:01
$begingroup$
@DavidC You need to tag him, otherwise he won't get the notification of your comment.
$endgroup$
– Vincenzo Oliva
Oct 11 '14 at 9:24
$begingroup$
@DavidC You need to tag him, otherwise he won't get the notification of your comment.
$endgroup$
– Vincenzo Oliva
Oct 11 '14 at 9:24
1
1
$begingroup$
I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
$endgroup$
– Martin Sleziak
Feb 1 '15 at 11:40
$begingroup$
I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
$endgroup$
– Martin Sleziak
Feb 1 '15 at 11:40
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
How are sum/product operations in $mathbb{N}$ defined in your context?
Let $n,minmathbb{N}$.
Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
$$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$
Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:
Addition (or "sum"):
- If $m=0$, then $n+ m := m$
- If $mneq 0$, then $n+ m := s(n+l)$
Multiplication (or "product"):
- If $m=0$, then $ncdot m = 0$
- If $mneq 0$, then $ncdot m = ncdot l + n$
Thus, by this definition and using induction, the result of both sum and product are also natural numbers.
answered Dec 5 '18 at 18:27
Dr PotatoDr Potato
375
$endgroup$
add a comment |
$begingroup$
The OP stated
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.
I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.
So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).
With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.
With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.
Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).
We conclude this by defining addition.
If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:
Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.
The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.
answered Jan 9 at 11:05
CopyPasteItCopyPasteIt
4,1781628
$endgroup$
$begingroup$
Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
$endgroup$
– Asaf Karagila♦
Jan 9 at 11:09
add a comment |
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I'm going to do something I rarely do, namely defining $Bbb N$ to include $0$ as an element. It makes the treatment herein a little easier or at least a little more Peano-conventional; feel free to adjust it, as an exercise, to start $Bbb N$ at $1$.
From $$0inBbb N,,forall ninBbb N (SninBbb N),,a+0=a,,a+Sb=S(a+b),,atimes 0=a,,atimes Sb=atimes b+a$$and $$varphi(0)landforall n(varphi(n)tovarphi(n+1))toforall ninBbb N(varphi(n))$$we'll prove $forall a,,binBbb N(a+binBbb N)$, and as a separate theorem $forall a,,binBbb N(atimes binBbb N)$. For both proofs we'll induct on $b$.
First, addition. The case $b=0$ is trivial because $a+0=ainBbb N$. Now we just need the inductive step; if it works when $b=k$ then $a+kinBbb Nimplies a+Sk=S(a+k)inBbb N$.
Next, multiplication; the addition result is actually needed in the inductive step. Again, $b=0$ is trivial because $atimes 0=0inBbb N$. Now the inductive step, assuming $atimes k$ exists; then $atimes Sk=(atimes k)+a$ is the sum of two elements of $Bbb N$, completing the proof by the above result.
You may also wish as an exercise to use this result about multiplication to in turn prove $a^binBbb N$ using $a^0=1,,a^{Sb}=a^btimes a$.
answered Feb 8 at 20:17
J.G.J.G.
27.9k22843
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
How are sum/product operations in $mathbb{N}$ defined in your context?
Let $n,minmathbb{N}$.
Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
$$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$
Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:
Addition (or "sum"):
- If $m=0$, then $n+ m := m$
- If $mneq 0$, then $n+ m := s(n+l)$
Multiplication (or "product"):
- If $m=0$, then $ncdot m = 0$
- If $mneq 0$, then $ncdot m = ncdot l + n$
Thus, by this definition and using induction, the result of both sum and product are also natural numbers.
answered Dec 5 '18 at 18:27
Dr PotatoDr Potato
375
$endgroup$
add a comment |
$begingroup$
How are sum/product operations in $mathbb{N}$ defined in your context?
Let $n,minmathbb{N}$.
Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
$$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$
Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:
Addition (or "sum"):
- If $m=0$, then $n+ m := m$
- If $mneq 0$, then $n+ m := s(n+l)$
Multiplication (or "product"):
- If $m=0$, then $ncdot m = 0$
- If $mneq 0$, then $ncdot m = ncdot l + n$
Thus, by this definition and using induction, the result of both sum and product are also natural numbers.
answered Dec 5 '18 at 18:27
Dr PotatoDr Potato
375
$endgroup$
add a comment |
$begingroup$
How are sum/product operations in $mathbb{N}$ defined in your context?
Let $n,minmathbb{N}$.
Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
$$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$
Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:
Addition (or "sum"):
- If $m=0$, then $n+ m := m$
- If $mneq 0$, then $n+ m := s(n+l)$
Multiplication (or "product"):
- If $m=0$, then $ncdot m = 0$
- If $mneq 0$, then $ncdot m = ncdot l + n$
Thus, by this definition and using induction, the result of both sum and product are also natural numbers.
answered Dec 5 '18 at 18:27
Dr PotatoDr Potato
375
$endgroup$
How are sum/product operations in $mathbb{N}$ defined in your context?
Let $n,minmathbb{N}$.
Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
$$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$
Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:
Addition (or "sum"):
- If $m=0$, then $n+ m := m$
- If $mneq 0$, then $n+ m := s(n+l)$
Multiplication (or "product"):
- If $m=0$, then $ncdot m = 0$
- If $mneq 0$, then $ncdot m = ncdot l + n$
Thus, by this definition and using induction, the result of both sum and product are also natural numbers.
answered Dec 5 '18 at 18:27
Dr PotatoDr Potato
375
edited Dec 6 '18 at 17:06
answered Dec 5 '18 at 18:27
Dr PotatoDr Potato
375
answered Dec 5 '18 at 18:27
Dr PotatoDr Potato
375
answered Dec 5 '18 at 18:27
Dr PotatoDr Potato
375
375
add a comment |
add a comment |
$begingroup$
The OP stated
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.
I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.
So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).
With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.
With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.
Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).
We conclude this by defining addition.
If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:
Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.
The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.
answered Jan 9 at 11:05
CopyPasteItCopyPasteIt
4,1781628
$endgroup$
$begingroup$
Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
$endgroup$
– Asaf Karagila♦
Jan 9 at 11:09
add a comment |
$begingroup$
The OP stated
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.
I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.
So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).
With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.
With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.
Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).
We conclude this by defining addition.
If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:
Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.
The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.
answered Jan 9 at 11:05
CopyPasteItCopyPasteIt
4,1781628
$endgroup$
$begingroup$
Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
$endgroup$
– Asaf Karagila♦
Jan 9 at 11:09
add a comment |
$begingroup$
The OP stated
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.
I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.
So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).
With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.
With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.
Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).
We conclude this by defining addition.
If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:
Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.
The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.
answered Jan 9 at 11:05
CopyPasteItCopyPasteIt
4,1781628
$endgroup$
The OP stated
So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.
I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.
So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).
With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.
With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.
Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).
We conclude this by defining addition.
If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:
Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.
The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.
answered Jan 9 at 11:05
CopyPasteItCopyPasteIt
4,1781628
answered Jan 9 at 11:05
CopyPasteItCopyPasteIt
4,1781628
answered Jan 9 at 11:05
CopyPasteItCopyPasteIt
4,1781628
answered Jan 9 at 11:05
CopyPasteItCopyPasteIt
4,1781628
4,1781628
$begingroup$
Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
$endgroup$
– Asaf Karagila♦
Jan 9 at 11:09
add a comment |
$begingroup$
Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
$endgroup$
– Asaf Karagila♦
Jan 9 at 11:09
$begingroup$
Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
$endgroup$
– Asaf Karagila♦
Jan 9 at 11:09
$begingroup$
Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
$endgroup$
– Asaf Karagila♦
Jan 9 at 11:09
add a comment |
$begingroup$
I'm going to do something I rarely do, namely defining $Bbb N$ to include $0$ as an element. It makes the treatment herein a little easier or at least a little more Peano-conventional; feel free to adjust it, as an exercise, to start $Bbb N$ at $1$.
From $$0inBbb N,,forall ninBbb N (SninBbb N),,a+0=a,,a+Sb=S(a+b),,atimes 0=a,,atimes Sb=atimes b+a$$and $$varphi(0)landforall n(varphi(n)tovarphi(n+1))toforall ninBbb N(varphi(n))$$we'll prove $forall a,,binBbb N(a+binBbb N)$, and as a separate theorem $forall a,,binBbb N(atimes binBbb N)$. For both proofs we'll induct on $b$.
First, addition. The case $b=0$ is trivial because $a+0=ainBbb N$. Now we just need the inductive step; if it works when $b=k$ then $a+kinBbb Nimplies a+Sk=S(a+k)inBbb N$.
Next, multiplication; the addition result is actually needed in the inductive step. Again, $b=0$ is trivial because $atimes 0=0inBbb N$. Now the inductive step, assuming $atimes k$ exists; then $atimes Sk=(atimes k)+a$ is the sum of two elements of $Bbb N$, completing the proof by the above result.
You may also wish as an exercise to use this result about multiplication to in turn prove $a^binBbb N$ using $a^0=1,,a^{Sb}=a^btimes a$.
answered Feb 8 at 20:17
J.G.J.G.
27.9k22843
$endgroup$
add a comment |
$begingroup$
I'm going to do something I rarely do, namely defining $Bbb N$ to include $0$ as an element. It makes the treatment herein a little easier or at least a little more Peano-conventional; feel free to adjust it, as an exercise, to start $Bbb N$ at $1$.
From $$0inBbb N,,forall ninBbb N (SninBbb N),,a+0=a,,a+Sb=S(a+b),,atimes 0=a,,atimes Sb=atimes b+a$$and $$varphi(0)landforall n(varphi(n)tovarphi(n+1))toforall ninBbb N(varphi(n))$$we'll prove $forall a,,binBbb N(a+binBbb N)$, and as a separate theorem $forall a,,binBbb N(atimes binBbb N)$. For both proofs we'll induct on $b$.
First, addition. The case $b=0$ is trivial because $a+0=ainBbb N$. Now we just need the inductive step; if it works when $b=k$ then $a+kinBbb Nimplies a+Sk=S(a+k)inBbb N$.
Next, multiplication; the addition result is actually needed in the inductive step. Again, $b=0$ is trivial because $atimes 0=0inBbb N$. Now the inductive step, assuming $atimes k$ exists; then $atimes Sk=(atimes k)+a$ is the sum of two elements of $Bbb N$, completing the proof by the above result.
You may also wish as an exercise to use this result about multiplication to in turn prove $a^binBbb N$ using $a^0=1,,a^{Sb}=a^btimes a$.
answered Feb 8 at 20:17
J.G.J.G.
27.9k22843
$endgroup$
add a comment |
$begingroup$
I'm going to do something I rarely do, namely defining $Bbb N$ to include $0$ as an element. It makes the treatment herein a little easier or at least a little more Peano-conventional; feel free to adjust it, as an exercise, to start $Bbb N$ at $1$.
From $$0inBbb N,,forall ninBbb N (SninBbb N),,a+0=a,,a+Sb=S(a+b),,atimes 0=a,,atimes Sb=atimes b+a$$and $$varphi(0)landforall n(varphi(n)tovarphi(n+1))toforall ninBbb N(varphi(n))$$we'll prove $forall a,,binBbb N(a+binBbb N)$, and as a separate theorem $forall a,,binBbb N(atimes binBbb N)$. For both proofs we'll induct on $b$.
First, addition. The case $b=0$ is trivial because $a+0=ainBbb N$. Now we just need the inductive step; if it works when $b=k$ then $a+kinBbb Nimplies a+Sk=S(a+k)inBbb N$.
Next, multiplication; the addition result is actually needed in the inductive step. Again, $b=0$ is trivial because $atimes 0=0inBbb N$. Now the inductive step, assuming $atimes k$ exists; then $atimes Sk=(atimes k)+a$ is the sum of two elements of $Bbb N$, completing the proof by the above result.
You may also wish as an exercise to use this result about multiplication to in turn prove $a^binBbb N$ using $a^0=1,,a^{Sb}=a^btimes a$.
answered Feb 8 at 20:17
J.G.J.G.
27.9k22843
$endgroup$
I'm going to do something I rarely do, namely defining $Bbb N$ to include $0$ as an element. It makes the treatment herein a little easier or at least a little more Peano-conventional; feel free to adjust it, as an exercise, to start $Bbb N$ at $1$.
From $$0inBbb N,,forall ninBbb N (SninBbb N),,a+0=a,,a+Sb=S(a+b),,atimes 0=a,,atimes Sb=atimes b+a$$and $$varphi(0)landforall n(varphi(n)tovarphi(n+1))toforall ninBbb N(varphi(n))$$we'll prove $forall a,,binBbb N(a+binBbb N)$, and as a separate theorem $forall a,,binBbb N(atimes binBbb N)$. For both proofs we'll induct on $b$.
First, addition. The case $b=0$ is trivial because $a+0=ainBbb N$. Now we just need the inductive step; if it works when $b=k$ then $a+kinBbb Nimplies a+Sk=S(a+k)inBbb N$.
Next, multiplication; the addition result is actually needed in the inductive step. Again, $b=0$ is trivial because $atimes 0=0inBbb N$. Now the inductive step, assuming $atimes k$ exists; then $atimes Sk=(atimes k)+a$ is the sum of two elements of $Bbb N$, completing the proof by the above result.
You may also wish as an exercise to use this result about multiplication to in turn prove $a^binBbb N$ using $a^0=1,,a^{Sb}=a^btimes a$.
answered Feb 8 at 20:17
J.G.J.G.
27.9k22843
answered Feb 8 at 20:17
J.G.J.G.
27.9k22843
answered Feb 8 at 20:17
J.G.J.G.
27.9k22843
answered Feb 8 at 20:17
J.G.J.G.
27.9k22843
27.9k22843
add a comment |
add a comment |
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$begingroup$
First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
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– Daniel Fischer♦
Oct 9 '14 at 13:33
$begingroup$
Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
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– Charles Carmichael
Oct 9 '14 at 13:57
$begingroup$
inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
$endgroup$
– Charles Carmichael
Oct 9 '14 at 14:01
$begingroup$
@DavidC You need to tag him, otherwise he won't get the notification of your comment.
$endgroup$
– Vincenzo Oliva
Oct 11 '14 at 9:24
1
$begingroup$
I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
$endgroup$
– Martin Sleziak
Feb 1 '15 at 11:40