Mixing
Considering sets in proofs
Clash Royale CLAN TAG#URR8PPP
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In order to show that $forall xgt0, exists n_0in mathbb N: n_0(n_0+1)le xlt (n_0+1)(n_0+2)$, the proof in my textbook considers the set $A=nin mathbb N :n(n+1)le x$ and then goes to show that $A$ has a maximal element $n_0$, which completes the proof.
Now, I want to know what has lead to this kind of argument, because I began to see it very frequently and even some of my classmates use this method often. But to me it doesn't seem to be very useful and I certainly wouldn't be able to tell if it's a good way to approach a certain proof.
I understand that such motivations usually can't be explained easily, in that case, I'm asking for other problems you know of, that use a similar approach.
Thanks in advance, and I hope I'm not asking much.
proof-writing
asked 21 mins ago
FuzzyPixelz
312214
add a comment |Â
up vote
1
down vote
favorite
In order to show that $forall xgt0, exists n_0in mathbb N: n_0(n_0+1)le xlt (n_0+1)(n_0+2)$, the proof in my textbook considers the set $A=nin mathbb N :n(n+1)le x$ and then goes to show that $A$ has a maximal element $n_0$, which completes the proof.
Now, I want to know what has lead to this kind of argument, because I began to see it very frequently and even some of my classmates use this method often. But to me it doesn't seem to be very useful and I certainly wouldn't be able to tell if it's a good way to approach a certain proof.
I understand that such motivations usually can't be explained easily, in that case, I'm asking for other problems you know of, that use a similar approach.
Thanks in advance, and I hope I'm not asking much.
proof-writing
asked 21 mins ago
FuzzyPixelz
312214
1
The basic idea is that you're "sandwiching" $x$ between two integers of the form $n_0(n_0 + 1) leq x < (n_0 + 1)(n_0 + 2)$. Notice how if you replace $n_0$ by $n_0 + 1$ in the left expression you get the one on the far right. $A$ will have a maximal element since all of the elements in $A$ are less than $x$ and the largest element could be at most $left lfloor x right rfloor$. We also note that by construction of $mathbbN$ that $n_0 in mathbbN$ implies the same for $n_0 + 1$. It's not a formal argument, but I think it has at least most of the pieces necessary to follow along.
– Eevee Trainer
11 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In order to show that $forall xgt0, exists n_0in mathbb N: n_0(n_0+1)le xlt (n_0+1)(n_0+2)$, the proof in my textbook considers the set $A=nin mathbb N :n(n+1)le x$ and then goes to show that $A$ has a maximal element $n_0$, which completes the proof.
Now, I want to know what has lead to this kind of argument, because I began to see it very frequently and even some of my classmates use this method often. But to me it doesn't seem to be very useful and I certainly wouldn't be able to tell if it's a good way to approach a certain proof.
I understand that such motivations usually can't be explained easily, in that case, I'm asking for other problems you know of, that use a similar approach.
Thanks in advance, and I hope I'm not asking much.
proof-writing
asked 21 mins ago
FuzzyPixelz
312214
In order to show that $forall xgt0, exists n_0in mathbb N: n_0(n_0+1)le xlt (n_0+1)(n_0+2)$, the proof in my textbook considers the set $A=nin mathbb N :n(n+1)le x$ and then goes to show that $A$ has a maximal element $n_0$, which completes the proof.
Now, I want to know what has lead to this kind of argument, because I began to see it very frequently and even some of my classmates use this method often. But to me it doesn't seem to be very useful and I certainly wouldn't be able to tell if it's a good way to approach a certain proof.
I understand that such motivations usually can't be explained easily, in that case, I'm asking for other problems you know of, that use a similar approach.
Thanks in advance, and I hope I'm not asking much.
proof-writing
proof-writing
asked 21 mins ago
FuzzyPixelz
312214
asked 21 mins ago
FuzzyPixelz
312214
asked 21 mins ago
FuzzyPixelz
312214
asked 21 mins ago
FuzzyPixelz
312214
asked 21 mins ago
FuzzyPixelz
312214
312214
1
The basic idea is that you're "sandwiching" $x$ between two integers of the form $n_0(n_0 + 1) leq x < (n_0 + 1)(n_0 + 2)$. Notice how if you replace $n_0$ by $n_0 + 1$ in the left expression you get the one on the far right. $A$ will have a maximal element since all of the elements in $A$ are less than $x$ and the largest element could be at most $left lfloor x right rfloor$. We also note that by construction of $mathbbN$ that $n_0 in mathbbN$ implies the same for $n_0 + 1$. It's not a formal argument, but I think it has at least most of the pieces necessary to follow along.
– Eevee Trainer
11 mins ago
add a comment |Â
1
The basic idea is that you're "sandwiching" $x$ between two integers of the form $n_0(n_0 + 1) leq x < (n_0 + 1)(n_0 + 2)$. Notice how if you replace $n_0$ by $n_0 + 1$ in the left expression you get the one on the far right. $A$ will have a maximal element since all of the elements in $A$ are less than $x$ and the largest element could be at most $left lfloor x right rfloor$. We also note that by construction of $mathbbN$ that $n_0 in mathbbN$ implies the same for $n_0 + 1$. It's not a formal argument, but I think it has at least most of the pieces necessary to follow along.
– Eevee Trainer
11 mins ago
1
1
The basic idea is that you're "sandwiching" $x$ between two integers of the form $n_0(n_0 + 1) leq x < (n_0 + 1)(n_0 + 2)$. Notice how if you replace $n_0$ by $n_0 + 1$ in the left expression you get the one on the far right. $A$ will have a maximal element since all of the elements in $A$ are less than $x$ and the largest element could be at most $left lfloor x right rfloor$. We also note that by construction of $mathbbN$ that $n_0 in mathbbN$ implies the same for $n_0 + 1$. It's not a formal argument, but I think it has at least most of the pieces necessary to follow along.
– Eevee Trainer
11 mins ago
The basic idea is that you're "sandwiching" $x$ between two integers of the form $n_0(n_0 + 1) leq x < (n_0 + 1)(n_0 + 2)$. Notice how if you replace $n_0$ by $n_0 + 1$ in the left expression you get the one on the far right. $A$ will have a maximal element since all of the elements in $A$ are less than $x$ and the largest element could be at most $left lfloor x right rfloor$. We also note that by construction of $mathbbN$ that $n_0 in mathbbN$ implies the same for $n_0 + 1$. It's not a formal argument, but I think it has at least most of the pieces necessary to follow along.
– Eevee Trainer
11 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Everyone has their own way of proving things, and that's OK. The statement you made can be shown to be true in different ways, and what counts is that you prove it, not how you prove it. The way you learn how to prove it is that you prove many many other statements as well, and then you get used to it. Repetitio mater studiorum est.
However, if you want a path that leads to the particular proof, in this case, my thoughts would proceed as follows:
- I look at the statement. OK, it's saying that I can squeeze any positive real $x$ between two numbers. OK, let's imagine the $x$ as some point on the positive real line.
- Hmm, the two numbers, $n_0(n_0+1)$ and $(n_0+1)(n_0+2)$ are both integers.
- Not only are they integers, they are two integers from a monotonically increasing sequence of integers, $a_n = n(n+1)$.
- So... this sequence of integers, it's really a series of points on the real line. Let's imagine them as crosses. (yes, really, I do that. Don't judge) The order I draw them in is left to right.
- So what I now have is the statement that there exist two crosses so that the previous cross is to the left of $x$, and the next one is to the right. Well... sure they do! I just need to find the last cross on the left of $x$, and the next one must be on the right of it (otherwise, the previous one wasn't the last one!).
- OK, what I just said in point 5 can be said formally as "I need to find the maximum value of $n$ such that $a_n < x$. This can be rewritten formally as finding a maximum element from some set.
Once this train of thought concludes, I go down to really writing down the proof, and the proof "starts" with introducing the set. Sure, the proof does, but the thought process that lead me to the proof started long before.
answered 10 mins ago
5xum
86.7k388156
This has been very helpful, thank you.
– FuzzyPixelz
4 mins ago
add a comment |Â
up vote
1
down vote
For instance, let $a,b$ be natural numbers with $1leq bleq a$. Then consider the set
$A=a-qbmid a-qbgeq 0$.
Since the set of natural numbers is well-ordered, every nonempty subset $A$ of natural numbers has a least element, here $r = a-qb$ in $A$. It is then clear that $a=qb+r$ and $0leq r<b$ is the remainder.
answered 13 mins ago
Wuestenfux
2,126139
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Everyone has their own way of proving things, and that's OK. The statement you made can be shown to be true in different ways, and what counts is that you prove it, not how you prove it. The way you learn how to prove it is that you prove many many other statements as well, and then you get used to it. Repetitio mater studiorum est.
However, if you want a path that leads to the particular proof, in this case, my thoughts would proceed as follows:
- I look at the statement. OK, it's saying that I can squeeze any positive real $x$ between two numbers. OK, let's imagine the $x$ as some point on the positive real line.
- Hmm, the two numbers, $n_0(n_0+1)$ and $(n_0+1)(n_0+2)$ are both integers.
- Not only are they integers, they are two integers from a monotonically increasing sequence of integers, $a_n = n(n+1)$.
- So... this sequence of integers, it's really a series of points on the real line. Let's imagine them as crosses. (yes, really, I do that. Don't judge) The order I draw them in is left to right.
- So what I now have is the statement that there exist two crosses so that the previous cross is to the left of $x$, and the next one is to the right. Well... sure they do! I just need to find the last cross on the left of $x$, and the next one must be on the right of it (otherwise, the previous one wasn't the last one!).
- OK, what I just said in point 5 can be said formally as "I need to find the maximum value of $n$ such that $a_n < x$. This can be rewritten formally as finding a maximum element from some set.
Once this train of thought concludes, I go down to really writing down the proof, and the proof "starts" with introducing the set. Sure, the proof does, but the thought process that lead me to the proof started long before.
answered 10 mins ago
5xum
86.7k388156
This has been very helpful, thank you.
– FuzzyPixelz
4 mins ago
add a comment |Â
up vote
3
down vote
accepted
Everyone has their own way of proving things, and that's OK. The statement you made can be shown to be true in different ways, and what counts is that you prove it, not how you prove it. The way you learn how to prove it is that you prove many many other statements as well, and then you get used to it. Repetitio mater studiorum est.
However, if you want a path that leads to the particular proof, in this case, my thoughts would proceed as follows:
- I look at the statement. OK, it's saying that I can squeeze any positive real $x$ between two numbers. OK, let's imagine the $x$ as some point on the positive real line.
- Hmm, the two numbers, $n_0(n_0+1)$ and $(n_0+1)(n_0+2)$ are both integers.
- Not only are they integers, they are two integers from a monotonically increasing sequence of integers, $a_n = n(n+1)$.
- So... this sequence of integers, it's really a series of points on the real line. Let's imagine them as crosses. (yes, really, I do that. Don't judge) The order I draw them in is left to right.
- So what I now have is the statement that there exist two crosses so that the previous cross is to the left of $x$, and the next one is to the right. Well... sure they do! I just need to find the last cross on the left of $x$, and the next one must be on the right of it (otherwise, the previous one wasn't the last one!).
- OK, what I just said in point 5 can be said formally as "I need to find the maximum value of $n$ such that $a_n < x$. This can be rewritten formally as finding a maximum element from some set.
Once this train of thought concludes, I go down to really writing down the proof, and the proof "starts" with introducing the set. Sure, the proof does, but the thought process that lead me to the proof started long before.
answered 10 mins ago
5xum
86.7k388156
This has been very helpful, thank you.
– FuzzyPixelz
4 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Everyone has their own way of proving things, and that's OK. The statement you made can be shown to be true in different ways, and what counts is that you prove it, not how you prove it. The way you learn how to prove it is that you prove many many other statements as well, and then you get used to it. Repetitio mater studiorum est.
However, if you want a path that leads to the particular proof, in this case, my thoughts would proceed as follows:
- I look at the statement. OK, it's saying that I can squeeze any positive real $x$ between two numbers. OK, let's imagine the $x$ as some point on the positive real line.
- Hmm, the two numbers, $n_0(n_0+1)$ and $(n_0+1)(n_0+2)$ are both integers.
- Not only are they integers, they are two integers from a monotonically increasing sequence of integers, $a_n = n(n+1)$.
- So... this sequence of integers, it's really a series of points on the real line. Let's imagine them as crosses. (yes, really, I do that. Don't judge) The order I draw them in is left to right.
- So what I now have is the statement that there exist two crosses so that the previous cross is to the left of $x$, and the next one is to the right. Well... sure they do! I just need to find the last cross on the left of $x$, and the next one must be on the right of it (otherwise, the previous one wasn't the last one!).
- OK, what I just said in point 5 can be said formally as "I need to find the maximum value of $n$ such that $a_n < x$. This can be rewritten formally as finding a maximum element from some set.
Once this train of thought concludes, I go down to really writing down the proof, and the proof "starts" with introducing the set. Sure, the proof does, but the thought process that lead me to the proof started long before.
answered 10 mins ago
5xum
86.7k388156
Everyone has their own way of proving things, and that's OK. The statement you made can be shown to be true in different ways, and what counts is that you prove it, not how you prove it. The way you learn how to prove it is that you prove many many other statements as well, and then you get used to it. Repetitio mater studiorum est.
However, if you want a path that leads to the particular proof, in this case, my thoughts would proceed as follows:
- I look at the statement. OK, it's saying that I can squeeze any positive real $x$ between two numbers. OK, let's imagine the $x$ as some point on the positive real line.
- Hmm, the two numbers, $n_0(n_0+1)$ and $(n_0+1)(n_0+2)$ are both integers.
- Not only are they integers, they are two integers from a monotonically increasing sequence of integers, $a_n = n(n+1)$.
- So... this sequence of integers, it's really a series of points on the real line. Let's imagine them as crosses. (yes, really, I do that. Don't judge) The order I draw them in is left to right.
- So what I now have is the statement that there exist two crosses so that the previous cross is to the left of $x$, and the next one is to the right. Well... sure they do! I just need to find the last cross on the left of $x$, and the next one must be on the right of it (otherwise, the previous one wasn't the last one!).
- OK, what I just said in point 5 can be said formally as "I need to find the maximum value of $n$ such that $a_n < x$. This can be rewritten formally as finding a maximum element from some set.
Once this train of thought concludes, I go down to really writing down the proof, and the proof "starts" with introducing the set. Sure, the proof does, but the thought process that lead me to the proof started long before.
answered 10 mins ago
5xum
86.7k388156
answered 10 mins ago
5xum
86.7k388156
answered 10 mins ago
5xum
86.7k388156
answered 10 mins ago
5xum
86.7k388156
86.7k388156
This has been very helpful, thank you.
– FuzzyPixelz
4 mins ago
add a comment |Â
This has been very helpful, thank you.
– FuzzyPixelz
4 mins ago
This has been very helpful, thank you.
– FuzzyPixelz
4 mins ago
This has been very helpful, thank you.
– FuzzyPixelz
4 mins ago
add a comment |Â
up vote
1
down vote
For instance, let $a,b$ be natural numbers with $1leq bleq a$. Then consider the set
$A=a-qbmid a-qbgeq 0$.
Since the set of natural numbers is well-ordered, every nonempty subset $A$ of natural numbers has a least element, here $r = a-qb$ in $A$. It is then clear that $a=qb+r$ and $0leq r<b$ is the remainder.
answered 13 mins ago
Wuestenfux
2,126139
add a comment |Â
up vote
1
down vote
For instance, let $a,b$ be natural numbers with $1leq bleq a$. Then consider the set
$A=a-qbmid a-qbgeq 0$.
Since the set of natural numbers is well-ordered, every nonempty subset $A$ of natural numbers has a least element, here $r = a-qb$ in $A$. It is then clear that $a=qb+r$ and $0leq r<b$ is the remainder.
answered 13 mins ago
Wuestenfux
2,126139
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For instance, let $a,b$ be natural numbers with $1leq bleq a$. Then consider the set
$A=a-qbmid a-qbgeq 0$.
Since the set of natural numbers is well-ordered, every nonempty subset $A$ of natural numbers has a least element, here $r = a-qb$ in $A$. It is then clear that $a=qb+r$ and $0leq r<b$ is the remainder.
answered 13 mins ago
Wuestenfux
2,126139
For instance, let $a,b$ be natural numbers with $1leq bleq a$. Then consider the set
$A=a-qbmid a-qbgeq 0$.
Since the set of natural numbers is well-ordered, every nonempty subset $A$ of natural numbers has a least element, here $r = a-qb$ in $A$. It is then clear that $a=qb+r$ and $0leq r<b$ is the remainder.
answered 13 mins ago
Wuestenfux
2,126139
answered 13 mins ago
Wuestenfux
2,126139
answered 13 mins ago
Wuestenfux
2,126139
answered 13 mins ago
Wuestenfux
2,126139
2,126139
add a comment |Â
add a comment |Â
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1
The basic idea is that you're "sandwiching" $x$ between two integers of the form $n_0(n_0 + 1) leq x < (n_0 + 1)(n_0 + 2)$. Notice how if you replace $n_0$ by $n_0 + 1$ in the left expression you get the one on the far right. $A$ will have a maximal element since all of the elements in $A$ are less than $x$ and the largest element could be at most $left lfloor x right rfloor$. We also note that by construction of $mathbbN$ that $n_0 in mathbbN$ implies the same for $n_0 + 1$. It's not a formal argument, but I think it has at least most of the pieces necessary to follow along.
– Eevee Trainer
11 mins ago