Mixing
Division Algorithm proof
$begingroup$
May someone tell me if there is anything wrong with my proof? And what can I do to improve it, please?
So the theorem is
Let a,b$in$ $mathbb{N}$ with b$>$0. Then $exists$ q,r$in$ $mathbb{N}$ : $a=qb+r$ where $0 leq r < b$
Now, I'm only considering the case where $b<a$.
Proof: Let $a,binmathbb{N}$ such that $a>b$. Assume that for $1,2,3,dots,a-1$, the result holds. Now consider three cases:
1) a-b=b and so setting q=1 and r=0 gives the desired result.
2) a-b$<$b and so setting q=0 and r=a-b gives the desired result.
3)$a-b>b$ and since a-b$geq$ 1 it follows by the induction hypothesis that there exists t and r such that a-b=bt+r and so setting q=t+1 gives the desired result.
abstract-algebra number-theory proof-verification
asked Jan 27 at 13:41
mathsssislifemathsssislife
448
$endgroup$
|
show 10 more comments
$begingroup$
May someone tell me if there is anything wrong with my proof? And what can I do to improve it, please?
So the theorem is
Let a,b$in$ $mathbb{N}$ with b$>$0. Then $exists$ q,r$in$ $mathbb{N}$ : $a=qb+r$ where $0 leq r < b$
Now, I'm only considering the case where $b<a$.
Proof: Let $a,binmathbb{N}$ such that $a>b$. Assume that for $1,2,3,dots,a-1$, the result holds. Now consider three cases:
1) a-b=b and so setting q=1 and r=0 gives the desired result.
2) a-b$<$b and so setting q=0 and r=a-b gives the desired result.
3)$a-b>b$ and since a-b$geq$ 1 it follows by the induction hypothesis that there exists t and r such that a-b=bt+r and so setting q=t+1 gives the desired result.
abstract-algebra number-theory proof-verification
asked Jan 27 at 13:41
mathsssislifemathsssislife
448
$endgroup$
$begingroup$
Need $b>0$ in your version.
$endgroup$
– coffeemath
Jan 27 at 13:45
$begingroup$
Yes, clearly that is the case. Notice, I said 0$leq$ r $<$ b.
$endgroup$
– mathsssislife
Jan 27 at 13:46
2
$begingroup$
It is not clearly the case, as you have only stated that $bneq0$. In fact the statement you are proving is false; you claim the existence of some $rgeq0$ with $0leq r<b$, for any nonzero $binBbb{Z}$.
$endgroup$
– Servaes
Jan 27 at 13:46
1
$begingroup$
No, that's the thing trying to prove.
$endgroup$
– coffeemath
Jan 27 at 13:47
1
$begingroup$
I don't understand why you're reversing my MathJax edits, making the formulas completely wrong as far as math typesetting is concerned.
$endgroup$
– egreg
Jan 27 at 14:18
|
show 10 more comments
$begingroup$
May someone tell me if there is anything wrong with my proof? And what can I do to improve it, please?
So the theorem is
Let a,b$in$ $mathbb{N}$ with b$>$0. Then $exists$ q,r$in$ $mathbb{N}$ : $a=qb+r$ where $0 leq r < b$
Now, I'm only considering the case where $b<a$.
Proof: Let $a,binmathbb{N}$ such that $a>b$. Assume that for $1,2,3,dots,a-1$, the result holds. Now consider three cases:
1) a-b=b and so setting q=1 and r=0 gives the desired result.
2) a-b$<$b and so setting q=0 and r=a-b gives the desired result.
3)$a-b>b$ and since a-b$geq$ 1 it follows by the induction hypothesis that there exists t and r such that a-b=bt+r and so setting q=t+1 gives the desired result.
abstract-algebra number-theory proof-verification
asked Jan 27 at 13:41
mathsssislifemathsssislife
448
$endgroup$
May someone tell me if there is anything wrong with my proof? And what can I do to improve it, please?
So the theorem is
Let a,b$in$ $mathbb{N}$ with b$>$0. Then $exists$ q,r$in$ $mathbb{N}$ : $a=qb+r$ where $0 leq r < b$
Now, I'm only considering the case where $b<a$.
Proof: Let $a,binmathbb{N}$ such that $a>b$. Assume that for $1,2,3,dots,a-1$, the result holds. Now consider three cases:
1) a-b=b and so setting q=1 and r=0 gives the desired result.
2) a-b$<$b and so setting q=0 and r=a-b gives the desired result.
3)$a-b>b$ and since a-b$geq$ 1 it follows by the induction hypothesis that there exists t and r such that a-b=bt+r and so setting q=t+1 gives the desired result.
abstract-algebra number-theory proof-verification
abstract-algebra number-theory proof-verification
asked Jan 27 at 13:41
mathsssislifemathsssislife
448
asked Jan 27 at 13:41
mathsssislifemathsssislife
448
edited Jan 27 at 14:16
mathsssislife
asked Jan 27 at 13:41
mathsssislifemathsssislife
448
asked Jan 27 at 13:41
mathsssislifemathsssislife
448
asked Jan 27 at 13:41
mathsssislifemathsssislife
448
448
$begingroup$
Need $b>0$ in your version.
$endgroup$
– coffeemath
Jan 27 at 13:45
$begingroup$
Yes, clearly that is the case. Notice, I said 0$leq$ r $<$ b.
$endgroup$
– mathsssislife
Jan 27 at 13:46
2
$begingroup$
It is not clearly the case, as you have only stated that $bneq0$. In fact the statement you are proving is false; you claim the existence of some $rgeq0$ with $0leq r<b$, for any nonzero $binBbb{Z}$.
$endgroup$
– Servaes
Jan 27 at 13:46
1
$begingroup$
No, that's the thing trying to prove.
$endgroup$
– coffeemath
Jan 27 at 13:47
1
$begingroup$
I don't understand why you're reversing my MathJax edits, making the formulas completely wrong as far as math typesetting is concerned.
$endgroup$
– egreg
Jan 27 at 14:18
|
show 10 more comments
$begingroup$
Need $b>0$ in your version.
$endgroup$
– coffeemath
Jan 27 at 13:45
$begingroup$
Yes, clearly that is the case. Notice, I said 0$leq$ r $<$ b.
$endgroup$
– mathsssislife
Jan 27 at 13:46
2
$begingroup$
It is not clearly the case, as you have only stated that $bneq0$. In fact the statement you are proving is false; you claim the existence of some $rgeq0$ with $0leq r<b$, for any nonzero $binBbb{Z}$.
$endgroup$
– Servaes
Jan 27 at 13:46
1
$begingroup$
No, that's the thing trying to prove.
$endgroup$
– coffeemath
Jan 27 at 13:47
1
$begingroup$
I don't understand why you're reversing my MathJax edits, making the formulas completely wrong as far as math typesetting is concerned.
$endgroup$
– egreg
Jan 27 at 14:18
$begingroup$
Need $b>0$ in your version.
$endgroup$
– coffeemath
Jan 27 at 13:45
$begingroup$
Need $b>0$ in your version.
$endgroup$
– coffeemath
Jan 27 at 13:45
$begingroup$
Yes, clearly that is the case. Notice, I said 0$leq$ r $<$ b.
$endgroup$
– mathsssislife
Jan 27 at 13:46
$begingroup$
Yes, clearly that is the case. Notice, I said 0$leq$ r $<$ b.
$endgroup$
– mathsssislife
Jan 27 at 13:46
2
2
$begingroup$
It is not clearly the case, as you have only stated that $bneq0$. In fact the statement you are proving is false; you claim the existence of some $rgeq0$ with $0leq r<b$, for any nonzero $binBbb{Z}$.
$endgroup$
– Servaes
Jan 27 at 13:46
$begingroup$
It is not clearly the case, as you have only stated that $bneq0$. In fact the statement you are proving is false; you claim the existence of some $rgeq0$ with $0leq r<b$, for any nonzero $binBbb{Z}$.
$endgroup$
– Servaes
Jan 27 at 13:46
1
1
$begingroup$
No, that's the thing trying to prove.
$endgroup$
– coffeemath
Jan 27 at 13:47
$begingroup$
No, that's the thing trying to prove.
$endgroup$
– coffeemath
Jan 27 at 13:47
1
1
$begingroup$
I don't understand why you're reversing my MathJax edits, making the formulas completely wrong as far as math typesetting is concerned.
$endgroup$
– egreg
Jan 27 at 14:18
$begingroup$
I don't understand why you're reversing my MathJax edits, making the formulas completely wrong as far as math typesetting is concerned.
$endgroup$
– egreg
Jan 27 at 14:18
|
show 10 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The strong induction hypothesis is
For every $a'$, with $b<a'<a$, there exist $t$ and $r$ such that $a'=bt+r$, with $0le r<b$.
You have $a-b<a$, but you don't know whether $a-b>b$. However the case $a-ble b$ is easily taken care of: in case $a-b<b$ you can set $t=0$ and $r=a-b$; in case $a-b=b$ then take $t=1$ and $r=0$. If $a-b>b$, then the strong induction hypothesis provides $t$ and $r$ with $a-b=bt+r$ and $0le r<b$. Then you can set $q=t+1$.
On the other hand, you can simply avoid the assumption that $a>b$. The strong induction hypothesis then can be
For every $a'<a$, there exist $q$ and $r$ such that $a'=bt+r$, with $a'=bt+r$ and $0le r<b$.
If $ale b$, you can do as done before. If $a>b$, then $a-b<a$ and you can use the induction hypothesis.
answered Jan 27 at 14:08
egregegreg
185k1486206
$endgroup$
$begingroup$
What about now? May you please look at my edit?
$endgroup$
– mathsssislife
Jan 27 at 14:18
add a comment |
$begingroup$
This doesn't answer your questions about the proof you gave, but you might want to note that there's a different approach that doesn't use induction explcitly - it seems simpler to me, or at least cleaner:
Define $$R={r=a-qb: qinBbb N, rge0}.$$Taking $q=0$ shows that $R$ is nonempty. Any nonempty set of non-negative integers has a smallest element; let $r_0$ be the smallest element of $R$.
You're done if you can show that $r_0<b$. But it's easy to show this:
If $r_0ge b$ then $r=r_0-bin R$.
Since $r_0-b<r_0$ this shows that if $r_0ge b$ then $r_0$ is not the smallest element of $R$, contradiction.
Edit: I've been asked why I would ocnsider that set. Hmm...
Define $r$ to be a function of $q$; $$r(q)=a-qb.$$We need to show
There exists $q$ suuch that $0le r(q)<b$.
How might we try to find such a $q$? Well, it's pretty clear that if we first restrict attention to $q$ such that $r(q)ge0$, then if there does exist a $q$ with $r(q)<b$, the smallest possible $r(q)$ must work. So we choose $q$ to minimize $r(q)$ (again, subject to the restriction that $r(q)ge0$.
And now letting $R$ be that set and taking the smallest element of $R$ is the obvious way to show formally that there is a $q$ that minimizes $r(q)$.
Note The statement that every non-empty subset of $Bbb N$ has a smallest element says that $Bbb N$ is "well-ordered". The fact that $Bbb N$ is well ordered is equivalent to the validity of proof by induction - any proof by induction has an equivalent version using well-ordering and vice-versa. Sometimes one comes out seeming simpler, sometimes the other.
Comment: The tags seem to indicate you're taking a course in abstract algebra. It's going to happen a lot that proofs of things involve defining a certain set and deducing properties of it. (If $G$ is a finite group and $ain G$ then the order of $a$ divides the order of $G$: Let $S$ be the subgroup of $G$ generated by $a$...)
answered Jan 27 at 16:22
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
But why would you consider that set?
$endgroup$
– mathsssislife
Jan 27 at 16:39
$begingroup$
@mathsssislife The edit may or may not make this clear...
$endgroup$
– David C. Ullrich
Jan 27 at 17:13
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
The strong induction hypothesis is
For every $a'$, with $b<a'<a$, there exist $t$ and $r$ such that $a'=bt+r$, with $0le r<b$.
You have $a-b<a$, but you don't know whether $a-b>b$. However the case $a-ble b$ is easily taken care of: in case $a-b<b$ you can set $t=0$ and $r=a-b$; in case $a-b=b$ then take $t=1$ and $r=0$. If $a-b>b$, then the strong induction hypothesis provides $t$ and $r$ with $a-b=bt+r$ and $0le r<b$. Then you can set $q=t+1$.
On the other hand, you can simply avoid the assumption that $a>b$. The strong induction hypothesis then can be
For every $a'<a$, there exist $q$ and $r$ such that $a'=bt+r$, with $a'=bt+r$ and $0le r<b$.
If $ale b$, you can do as done before. If $a>b$, then $a-b<a$ and you can use the induction hypothesis.
answered Jan 27 at 14:08
egregegreg
185k1486206
$endgroup$
$begingroup$
What about now? May you please look at my edit?
$endgroup$
– mathsssislife
Jan 27 at 14:18
add a comment |
$begingroup$
The strong induction hypothesis is
For every $a'$, with $b<a'<a$, there exist $t$ and $r$ such that $a'=bt+r$, with $0le r<b$.
You have $a-b<a$, but you don't know whether $a-b>b$. However the case $a-ble b$ is easily taken care of: in case $a-b<b$ you can set $t=0$ and $r=a-b$; in case $a-b=b$ then take $t=1$ and $r=0$. If $a-b>b$, then the strong induction hypothesis provides $t$ and $r$ with $a-b=bt+r$ and $0le r<b$. Then you can set $q=t+1$.
On the other hand, you can simply avoid the assumption that $a>b$. The strong induction hypothesis then can be
For every $a'<a$, there exist $q$ and $r$ such that $a'=bt+r$, with $a'=bt+r$ and $0le r<b$.
If $ale b$, you can do as done before. If $a>b$, then $a-b<a$ and you can use the induction hypothesis.
answered Jan 27 at 14:08
egregegreg
185k1486206
$endgroup$
$begingroup$
What about now? May you please look at my edit?
$endgroup$
– mathsssislife
Jan 27 at 14:18
add a comment |
$begingroup$
The strong induction hypothesis is
For every $a'$, with $b<a'<a$, there exist $t$ and $r$ such that $a'=bt+r$, with $0le r<b$.
You have $a-b<a$, but you don't know whether $a-b>b$. However the case $a-ble b$ is easily taken care of: in case $a-b<b$ you can set $t=0$ and $r=a-b$; in case $a-b=b$ then take $t=1$ and $r=0$. If $a-b>b$, then the strong induction hypothesis provides $t$ and $r$ with $a-b=bt+r$ and $0le r<b$. Then you can set $q=t+1$.
On the other hand, you can simply avoid the assumption that $a>b$. The strong induction hypothesis then can be
For every $a'<a$, there exist $q$ and $r$ such that $a'=bt+r$, with $a'=bt+r$ and $0le r<b$.
If $ale b$, you can do as done before. If $a>b$, then $a-b<a$ and you can use the induction hypothesis.
answered Jan 27 at 14:08
egregegreg
185k1486206
$endgroup$
The strong induction hypothesis is
For every $a'$, with $b<a'<a$, there exist $t$ and $r$ such that $a'=bt+r$, with $0le r<b$.
You have $a-b<a$, but you don't know whether $a-b>b$. However the case $a-ble b$ is easily taken care of: in case $a-b<b$ you can set $t=0$ and $r=a-b$; in case $a-b=b$ then take $t=1$ and $r=0$. If $a-b>b$, then the strong induction hypothesis provides $t$ and $r$ with $a-b=bt+r$ and $0le r<b$. Then you can set $q=t+1$.
On the other hand, you can simply avoid the assumption that $a>b$. The strong induction hypothesis then can be
For every $a'<a$, there exist $q$ and $r$ such that $a'=bt+r$, with $a'=bt+r$ and $0le r<b$.
If $ale b$, you can do as done before. If $a>b$, then $a-b<a$ and you can use the induction hypothesis.
answered Jan 27 at 14:08
egregegreg
185k1486206
edited Jan 27 at 14:14
answered Jan 27 at 14:08
egregegreg
185k1486206
answered Jan 27 at 14:08
egregegreg
185k1486206
answered Jan 27 at 14:08
egregegreg
185k1486206
185k1486206
$begingroup$
What about now? May you please look at my edit?
$endgroup$
– mathsssislife
Jan 27 at 14:18
add a comment |
$begingroup$
What about now? May you please look at my edit?
$endgroup$
– mathsssislife
Jan 27 at 14:18
$begingroup$
What about now? May you please look at my edit?
$endgroup$
– mathsssislife
Jan 27 at 14:18
$begingroup$
What about now? May you please look at my edit?
$endgroup$
– mathsssislife
Jan 27 at 14:18
add a comment |
$begingroup$
This doesn't answer your questions about the proof you gave, but you might want to note that there's a different approach that doesn't use induction explcitly - it seems simpler to me, or at least cleaner:
Define $$R={r=a-qb: qinBbb N, rge0}.$$Taking $q=0$ shows that $R$ is nonempty. Any nonempty set of non-negative integers has a smallest element; let $r_0$ be the smallest element of $R$.
You're done if you can show that $r_0<b$. But it's easy to show this:
If $r_0ge b$ then $r=r_0-bin R$.
Since $r_0-b<r_0$ this shows that if $r_0ge b$ then $r_0$ is not the smallest element of $R$, contradiction.
Edit: I've been asked why I would ocnsider that set. Hmm...
Define $r$ to be a function of $q$; $$r(q)=a-qb.$$We need to show
There exists $q$ suuch that $0le r(q)<b$.
How might we try to find such a $q$? Well, it's pretty clear that if we first restrict attention to $q$ such that $r(q)ge0$, then if there does exist a $q$ with $r(q)<b$, the smallest possible $r(q)$ must work. So we choose $q$ to minimize $r(q)$ (again, subject to the restriction that $r(q)ge0$.
And now letting $R$ be that set and taking the smallest element of $R$ is the obvious way to show formally that there is a $q$ that minimizes $r(q)$.
Note The statement that every non-empty subset of $Bbb N$ has a smallest element says that $Bbb N$ is "well-ordered". The fact that $Bbb N$ is well ordered is equivalent to the validity of proof by induction - any proof by induction has an equivalent version using well-ordering and vice-versa. Sometimes one comes out seeming simpler, sometimes the other.
Comment: The tags seem to indicate you're taking a course in abstract algebra. It's going to happen a lot that proofs of things involve defining a certain set and deducing properties of it. (If $G$ is a finite group and $ain G$ then the order of $a$ divides the order of $G$: Let $S$ be the subgroup of $G$ generated by $a$...)
answered Jan 27 at 16:22
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
But why would you consider that set?
$endgroup$
– mathsssislife
Jan 27 at 16:39
$begingroup$
@mathsssislife The edit may or may not make this clear...
$endgroup$
– David C. Ullrich
Jan 27 at 17:13
add a comment |
$begingroup$
This doesn't answer your questions about the proof you gave, but you might want to note that there's a different approach that doesn't use induction explcitly - it seems simpler to me, or at least cleaner:
Define $$R={r=a-qb: qinBbb N, rge0}.$$Taking $q=0$ shows that $R$ is nonempty. Any nonempty set of non-negative integers has a smallest element; let $r_0$ be the smallest element of $R$.
You're done if you can show that $r_0<b$. But it's easy to show this:
If $r_0ge b$ then $r=r_0-bin R$.
Since $r_0-b<r_0$ this shows that if $r_0ge b$ then $r_0$ is not the smallest element of $R$, contradiction.
Edit: I've been asked why I would ocnsider that set. Hmm...
Define $r$ to be a function of $q$; $$r(q)=a-qb.$$We need to show
There exists $q$ suuch that $0le r(q)<b$.
How might we try to find such a $q$? Well, it's pretty clear that if we first restrict attention to $q$ such that $r(q)ge0$, then if there does exist a $q$ with $r(q)<b$, the smallest possible $r(q)$ must work. So we choose $q$ to minimize $r(q)$ (again, subject to the restriction that $r(q)ge0$.
And now letting $R$ be that set and taking the smallest element of $R$ is the obvious way to show formally that there is a $q$ that minimizes $r(q)$.
Note The statement that every non-empty subset of $Bbb N$ has a smallest element says that $Bbb N$ is "well-ordered". The fact that $Bbb N$ is well ordered is equivalent to the validity of proof by induction - any proof by induction has an equivalent version using well-ordering and vice-versa. Sometimes one comes out seeming simpler, sometimes the other.
Comment: The tags seem to indicate you're taking a course in abstract algebra. It's going to happen a lot that proofs of things involve defining a certain set and deducing properties of it. (If $G$ is a finite group and $ain G$ then the order of $a$ divides the order of $G$: Let $S$ be the subgroup of $G$ generated by $a$...)
answered Jan 27 at 16:22
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
But why would you consider that set?
$endgroup$
– mathsssislife
Jan 27 at 16:39
$begingroup$
@mathsssislife The edit may or may not make this clear...
$endgroup$
– David C. Ullrich
Jan 27 at 17:13
add a comment |
$begingroup$
This doesn't answer your questions about the proof you gave, but you might want to note that there's a different approach that doesn't use induction explcitly - it seems simpler to me, or at least cleaner:
Define $$R={r=a-qb: qinBbb N, rge0}.$$Taking $q=0$ shows that $R$ is nonempty. Any nonempty set of non-negative integers has a smallest element; let $r_0$ be the smallest element of $R$.
You're done if you can show that $r_0<b$. But it's easy to show this:
If $r_0ge b$ then $r=r_0-bin R$.
Since $r_0-b<r_0$ this shows that if $r_0ge b$ then $r_0$ is not the smallest element of $R$, contradiction.
Edit: I've been asked why I would ocnsider that set. Hmm...
Define $r$ to be a function of $q$; $$r(q)=a-qb.$$We need to show
There exists $q$ suuch that $0le r(q)<b$.
How might we try to find such a $q$? Well, it's pretty clear that if we first restrict attention to $q$ such that $r(q)ge0$, then if there does exist a $q$ with $r(q)<b$, the smallest possible $r(q)$ must work. So we choose $q$ to minimize $r(q)$ (again, subject to the restriction that $r(q)ge0$.
And now letting $R$ be that set and taking the smallest element of $R$ is the obvious way to show formally that there is a $q$ that minimizes $r(q)$.
Note The statement that every non-empty subset of $Bbb N$ has a smallest element says that $Bbb N$ is "well-ordered". The fact that $Bbb N$ is well ordered is equivalent to the validity of proof by induction - any proof by induction has an equivalent version using well-ordering and vice-versa. Sometimes one comes out seeming simpler, sometimes the other.
Comment: The tags seem to indicate you're taking a course in abstract algebra. It's going to happen a lot that proofs of things involve defining a certain set and deducing properties of it. (If $G$ is a finite group and $ain G$ then the order of $a$ divides the order of $G$: Let $S$ be the subgroup of $G$ generated by $a$...)
answered Jan 27 at 16:22
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
This doesn't answer your questions about the proof you gave, but you might want to note that there's a different approach that doesn't use induction explcitly - it seems simpler to me, or at least cleaner:
Define $$R={r=a-qb: qinBbb N, rge0}.$$Taking $q=0$ shows that $R$ is nonempty. Any nonempty set of non-negative integers has a smallest element; let $r_0$ be the smallest element of $R$.
You're done if you can show that $r_0<b$. But it's easy to show this:
If $r_0ge b$ then $r=r_0-bin R$.
Since $r_0-b<r_0$ this shows that if $r_0ge b$ then $r_0$ is not the smallest element of $R$, contradiction.
Edit: I've been asked why I would ocnsider that set. Hmm...
Define $r$ to be a function of $q$; $$r(q)=a-qb.$$We need to show
There exists $q$ suuch that $0le r(q)<b$.
How might we try to find such a $q$? Well, it's pretty clear that if we first restrict attention to $q$ such that $r(q)ge0$, then if there does exist a $q$ with $r(q)<b$, the smallest possible $r(q)$ must work. So we choose $q$ to minimize $r(q)$ (again, subject to the restriction that $r(q)ge0$.
And now letting $R$ be that set and taking the smallest element of $R$ is the obvious way to show formally that there is a $q$ that minimizes $r(q)$.
Note The statement that every non-empty subset of $Bbb N$ has a smallest element says that $Bbb N$ is "well-ordered". The fact that $Bbb N$ is well ordered is equivalent to the validity of proof by induction - any proof by induction has an equivalent version using well-ordering and vice-versa. Sometimes one comes out seeming simpler, sometimes the other.
Comment: The tags seem to indicate you're taking a course in abstract algebra. It's going to happen a lot that proofs of things involve defining a certain set and deducing properties of it. (If $G$ is a finite group and $ain G$ then the order of $a$ divides the order of $G$: Let $S$ be the subgroup of $G$ generated by $a$...)
answered Jan 27 at 16:22
David C. UllrichDavid C. Ullrich
61.6k43995
edited Jan 27 at 17:04
answered Jan 27 at 16:22
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 27 at 16:22
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 27 at 16:22
David C. UllrichDavid C. Ullrich
61.6k43995
61.6k43995
$begingroup$
But why would you consider that set?
$endgroup$
– mathsssislife
Jan 27 at 16:39
$begingroup$
@mathsssislife The edit may or may not make this clear...
$endgroup$
– David C. Ullrich
Jan 27 at 17:13
add a comment |
$begingroup$
But why would you consider that set?
$endgroup$
– mathsssislife
Jan 27 at 16:39
$begingroup$
@mathsssislife The edit may or may not make this clear...
$endgroup$
– David C. Ullrich
Jan 27 at 17:13
$begingroup$
But why would you consider that set?
$endgroup$
– mathsssislife
Jan 27 at 16:39
$begingroup$
But why would you consider that set?
$endgroup$
– mathsssislife
Jan 27 at 16:39
$begingroup$
@mathsssislife The edit may or may not make this clear...
$endgroup$
– David C. Ullrich
Jan 27 at 17:13
$begingroup$
@mathsssislife The edit may or may not make this clear...
$endgroup$
– David C. Ullrich
Jan 27 at 17:13
add a comment |
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$begingroup$
Need $b>0$ in your version.
$endgroup$
– coffeemath
Jan 27 at 13:45
$begingroup$
Yes, clearly that is the case. Notice, I said 0$leq$ r $<$ b.
$endgroup$
– mathsssislife
Jan 27 at 13:46
2
$begingroup$
It is not clearly the case, as you have only stated that $bneq0$. In fact the statement you are proving is false; you claim the existence of some $rgeq0$ with $0leq r<b$, for any nonzero $binBbb{Z}$.
$endgroup$
– Servaes
Jan 27 at 13:46
1
$begingroup$
No, that's the thing trying to prove.
$endgroup$
– coffeemath
Jan 27 at 13:47
1
$begingroup$
I don't understand why you're reversing my MathJax edits, making the formulas completely wrong as far as math typesetting is concerned.
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– egreg
Jan 27 at 14:18