Mixing
Linear order on the set N^N
$begingroup$
Give the example of the Linear order on the set: $mathbb N^mathbb N$
I know that $mathbb N^mathbb N$ is a function from $mathbb N$ to $mathbb N$ and it's a string, but I'm unable to give an example.
discrete-mathematics logic relations
asked Jan 20 at 14:20
KarolKarol
205
$endgroup$
|
show 6 more comments
$begingroup$
Give the example of the Linear order on the set: $mathbb N^mathbb N$
I know that $mathbb N^mathbb N$ is a function from $mathbb N$ to $mathbb N$ and it's a string, but I'm unable to give an example.
discrete-mathematics logic relations
asked Jan 20 at 14:20
KarolKarol
205
$endgroup$
$begingroup$
Hint : There is an order on the alphabet $a,b,c,d,...,x,y,z$, and hence there is an order on words made out of $a,b,c,d,...,x,y,z$ which is reflected in the English dictionary, for example. Think about using this approach on your set : there is a natural order on $mathbb N$, and it should lead to an ordering on $mathbb N^mathbb N$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:29
$begingroup$
@астонвіллаолофмэллбэрг, thanks for answering but haven't learned about the alphabet. Any other ideas?
$endgroup$
– Karol
Jan 20 at 14:38
$begingroup$
I think that he just means the letters that we write with, not a mathematical term. Do you know how to find a word in a dictionary based on your knowledge of the ABC?
$endgroup$
– badjohn
Jan 20 at 14:40
$begingroup$
What I am saying is this : for example if you have two strings $a_1a_2a_3...$ and $b_1b_2b_3...$ of natural numbers, which correspond to two elements of $mathbb N^mathbb N$. You first compare $a_1$ and $b_1$ : if $a_1<b_1$ then the first string is smaller than the second string. If $a_1 > b_1$ then the first string is larger than the second string. If $a_1 = b_1$, then move to comparing $a_2$ and $b_2$ (repeat the above steps with $a_1,b_1$ replaced by $a_2,b_2$). If these are equal then move to $a_3$ and $b_3$. I hope you see how to compare the strings from here.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:43
$begingroup$
@badjohn, so, this set is a set of natural numbers strings, tell me if I'm wrong. So those would be sets of natural numbers, so how exactly do I use the alphabet there?
$endgroup$
– Karol
Jan 20 at 14:43
|
show 6 more comments
$begingroup$
Give the example of the Linear order on the set: $mathbb N^mathbb N$
I know that $mathbb N^mathbb N$ is a function from $mathbb N$ to $mathbb N$ and it's a string, but I'm unable to give an example.
discrete-mathematics logic relations
asked Jan 20 at 14:20
KarolKarol
205
$endgroup$
Give the example of the Linear order on the set: $mathbb N^mathbb N$
I know that $mathbb N^mathbb N$ is a function from $mathbb N$ to $mathbb N$ and it's a string, but I'm unable to give an example.
discrete-mathematics logic relations
discrete-mathematics logic relations
asked Jan 20 at 14:20
KarolKarol
205
asked Jan 20 at 14:20
KarolKarol
205
asked Jan 20 at 14:20
KarolKarol
205
asked Jan 20 at 14:20
KarolKarol
205
asked Jan 20 at 14:20
KarolKarol
205
205
$begingroup$
Hint : There is an order on the alphabet $a,b,c,d,...,x,y,z$, and hence there is an order on words made out of $a,b,c,d,...,x,y,z$ which is reflected in the English dictionary, for example. Think about using this approach on your set : there is a natural order on $mathbb N$, and it should lead to an ordering on $mathbb N^mathbb N$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:29
$begingroup$
@астонвіллаолофмэллбэрг, thanks for answering but haven't learned about the alphabet. Any other ideas?
$endgroup$
– Karol
Jan 20 at 14:38
$begingroup$
I think that he just means the letters that we write with, not a mathematical term. Do you know how to find a word in a dictionary based on your knowledge of the ABC?
$endgroup$
– badjohn
Jan 20 at 14:40
$begingroup$
What I am saying is this : for example if you have two strings $a_1a_2a_3...$ and $b_1b_2b_3...$ of natural numbers, which correspond to two elements of $mathbb N^mathbb N$. You first compare $a_1$ and $b_1$ : if $a_1<b_1$ then the first string is smaller than the second string. If $a_1 > b_1$ then the first string is larger than the second string. If $a_1 = b_1$, then move to comparing $a_2$ and $b_2$ (repeat the above steps with $a_1,b_1$ replaced by $a_2,b_2$). If these are equal then move to $a_3$ and $b_3$. I hope you see how to compare the strings from here.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:43
$begingroup$
@badjohn, so, this set is a set of natural numbers strings, tell me if I'm wrong. So those would be sets of natural numbers, so how exactly do I use the alphabet there?
$endgroup$
– Karol
Jan 20 at 14:43
|
show 6 more comments
$begingroup$
Hint : There is an order on the alphabet $a,b,c,d,...,x,y,z$, and hence there is an order on words made out of $a,b,c,d,...,x,y,z$ which is reflected in the English dictionary, for example. Think about using this approach on your set : there is a natural order on $mathbb N$, and it should lead to an ordering on $mathbb N^mathbb N$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:29
$begingroup$
@астонвіллаолофмэллбэрг, thanks for answering but haven't learned about the alphabet. Any other ideas?
$endgroup$
– Karol
Jan 20 at 14:38
$begingroup$
I think that he just means the letters that we write with, not a mathematical term. Do you know how to find a word in a dictionary based on your knowledge of the ABC?
$endgroup$
– badjohn
Jan 20 at 14:40
$begingroup$
What I am saying is this : for example if you have two strings $a_1a_2a_3...$ and $b_1b_2b_3...$ of natural numbers, which correspond to two elements of $mathbb N^mathbb N$. You first compare $a_1$ and $b_1$ : if $a_1<b_1$ then the first string is smaller than the second string. If $a_1 > b_1$ then the first string is larger than the second string. If $a_1 = b_1$, then move to comparing $a_2$ and $b_2$ (repeat the above steps with $a_1,b_1$ replaced by $a_2,b_2$). If these are equal then move to $a_3$ and $b_3$. I hope you see how to compare the strings from here.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:43
$begingroup$
@badjohn, so, this set is a set of natural numbers strings, tell me if I'm wrong. So those would be sets of natural numbers, so how exactly do I use the alphabet there?
$endgroup$
– Karol
Jan 20 at 14:43
$begingroup$
Hint : There is an order on the alphabet $a,b,c,d,...,x,y,z$, and hence there is an order on words made out of $a,b,c,d,...,x,y,z$ which is reflected in the English dictionary, for example. Think about using this approach on your set : there is a natural order on $mathbb N$, and it should lead to an ordering on $mathbb N^mathbb N$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:29
$begingroup$
Hint : There is an order on the alphabet $a,b,c,d,...,x,y,z$, and hence there is an order on words made out of $a,b,c,d,...,x,y,z$ which is reflected in the English dictionary, for example. Think about using this approach on your set : there is a natural order on $mathbb N$, and it should lead to an ordering on $mathbb N^mathbb N$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:29
$begingroup$
@астонвіллаолофмэллбэрг, thanks for answering but haven't learned about the alphabet. Any other ideas?
$endgroup$
– Karol
Jan 20 at 14:38
$begingroup$
@астонвіллаолофмэллбэрг, thanks for answering but haven't learned about the alphabet. Any other ideas?
$endgroup$
– Karol
Jan 20 at 14:38
$begingroup$
I think that he just means the letters that we write with, not a mathematical term. Do you know how to find a word in a dictionary based on your knowledge of the ABC?
$endgroup$
– badjohn
Jan 20 at 14:40
$begingroup$
I think that he just means the letters that we write with, not a mathematical term. Do you know how to find a word in a dictionary based on your knowledge of the ABC?
$endgroup$
– badjohn
Jan 20 at 14:40
$begingroup$
What I am saying is this : for example if you have two strings $a_1a_2a_3...$ and $b_1b_2b_3...$ of natural numbers, which correspond to two elements of $mathbb N^mathbb N$. You first compare $a_1$ and $b_1$ : if $a_1<b_1$ then the first string is smaller than the second string. If $a_1 > b_1$ then the first string is larger than the second string. If $a_1 = b_1$, then move to comparing $a_2$ and $b_2$ (repeat the above steps with $a_1,b_1$ replaced by $a_2,b_2$). If these are equal then move to $a_3$ and $b_3$. I hope you see how to compare the strings from here.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:43
$begingroup$
What I am saying is this : for example if you have two strings $a_1a_2a_3...$ and $b_1b_2b_3...$ of natural numbers, which correspond to two elements of $mathbb N^mathbb N$. You first compare $a_1$ and $b_1$ : if $a_1<b_1$ then the first string is smaller than the second string. If $a_1 > b_1$ then the first string is larger than the second string. If $a_1 = b_1$, then move to comparing $a_2$ and $b_2$ (repeat the above steps with $a_1,b_1$ replaced by $a_2,b_2$). If these are equal then move to $a_3$ and $b_3$. I hope you see how to compare the strings from here.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:43
$begingroup$
@badjohn, so, this set is a set of natural numbers strings, tell me if I'm wrong. So those would be sets of natural numbers, so how exactly do I use the alphabet there?
$endgroup$
– Karol
Jan 20 at 14:43
$begingroup$
@badjohn, so, this set is a set of natural numbers strings, tell me if I'm wrong. So those would be sets of natural numbers, so how exactly do I use the alphabet there?
$endgroup$
– Karol
Jan 20 at 14:43
|
show 6 more comments
1 Answer
1
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$begingroup$
First, think of words in a dictionary: how do you order them? Which comes first: Apple or Orange? You compare the first letters and since A is before O, you say that Apple is before Orange. If the first letters, are the same then compare the second so apple is before avocado. If they are the same until one ends then you need a rule, e.g. the shorter one is first so grape is before grapefruit.
Your functions can be written as a list of numbers. $f(1), f(2), f(3), ... $ To compare $f$ and $g$, compare $f(1)$ and $g(1)$. If $f(1)$ is less then $f$ is less. If they are the same then compare $f(2)$ and $g(2)$, etc. Continue until you find a difference.
answered Jan 20 at 14:52
badjohnbadjohn
4,3521720
$endgroup$
$begingroup$
Thank you, just quick question. Is a linear order a lexicographical order?
$endgroup$
– Karol
Jan 20 at 15:05
$begingroup$
I'd said the reverse. A lexicographical order will be a linear but maybe not the reverse.
$endgroup$
– badjohn
Jan 20 at 15:16
$begingroup$
why is a lexicographical order a linear order? I cannot find any explanation and I'm beginning to understand it so I want to be 100% sure.
$endgroup$
– Karol
Jan 20 at 15:24
$begingroup$
Well, because the procedure I described will satisfy the axioms of a linear order. en.m.wikipedia.org/wiki/Total_order
$endgroup$
– badjohn
Jan 20 at 15:37
add a comment |
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votes
$begingroup$
First, think of words in a dictionary: how do you order them? Which comes first: Apple or Orange? You compare the first letters and since A is before O, you say that Apple is before Orange. If the first letters, are the same then compare the second so apple is before avocado. If they are the same until one ends then you need a rule, e.g. the shorter one is first so grape is before grapefruit.
Your functions can be written as a list of numbers. $f(1), f(2), f(3), ... $ To compare $f$ and $g$, compare $f(1)$ and $g(1)$. If $f(1)$ is less then $f$ is less. If they are the same then compare $f(2)$ and $g(2)$, etc. Continue until you find a difference.
answered Jan 20 at 14:52
badjohnbadjohn
4,3521720
$endgroup$
$begingroup$
Thank you, just quick question. Is a linear order a lexicographical order?
$endgroup$
– Karol
Jan 20 at 15:05
$begingroup$
I'd said the reverse. A lexicographical order will be a linear but maybe not the reverse.
$endgroup$
– badjohn
Jan 20 at 15:16
$begingroup$
why is a lexicographical order a linear order? I cannot find any explanation and I'm beginning to understand it so I want to be 100% sure.
$endgroup$
– Karol
Jan 20 at 15:24
$begingroup$
Well, because the procedure I described will satisfy the axioms of a linear order. en.m.wikipedia.org/wiki/Total_order
$endgroup$
– badjohn
Jan 20 at 15:37
add a comment |
$begingroup$
First, think of words in a dictionary: how do you order them? Which comes first: Apple or Orange? You compare the first letters and since A is before O, you say that Apple is before Orange. If the first letters, are the same then compare the second so apple is before avocado. If they are the same until one ends then you need a rule, e.g. the shorter one is first so grape is before grapefruit.
Your functions can be written as a list of numbers. $f(1), f(2), f(3), ... $ To compare $f$ and $g$, compare $f(1)$ and $g(1)$. If $f(1)$ is less then $f$ is less. If they are the same then compare $f(2)$ and $g(2)$, etc. Continue until you find a difference.
answered Jan 20 at 14:52
badjohnbadjohn
4,3521720
$endgroup$
$begingroup$
Thank you, just quick question. Is a linear order a lexicographical order?
$endgroup$
– Karol
Jan 20 at 15:05
$begingroup$
I'd said the reverse. A lexicographical order will be a linear but maybe not the reverse.
$endgroup$
– badjohn
Jan 20 at 15:16
$begingroup$
why is a lexicographical order a linear order? I cannot find any explanation and I'm beginning to understand it so I want to be 100% sure.
$endgroup$
– Karol
Jan 20 at 15:24
$begingroup$
Well, because the procedure I described will satisfy the axioms of a linear order. en.m.wikipedia.org/wiki/Total_order
$endgroup$
– badjohn
Jan 20 at 15:37
add a comment |
$begingroup$
First, think of words in a dictionary: how do you order them? Which comes first: Apple or Orange? You compare the first letters and since A is before O, you say that Apple is before Orange. If the first letters, are the same then compare the second so apple is before avocado. If they are the same until one ends then you need a rule, e.g. the shorter one is first so grape is before grapefruit.
Your functions can be written as a list of numbers. $f(1), f(2), f(3), ... $ To compare $f$ and $g$, compare $f(1)$ and $g(1)$. If $f(1)$ is less then $f$ is less. If they are the same then compare $f(2)$ and $g(2)$, etc. Continue until you find a difference.
answered Jan 20 at 14:52
badjohnbadjohn
4,3521720
$endgroup$
First, think of words in a dictionary: how do you order them? Which comes first: Apple or Orange? You compare the first letters and since A is before O, you say that Apple is before Orange. If the first letters, are the same then compare the second so apple is before avocado. If they are the same until one ends then you need a rule, e.g. the shorter one is first so grape is before grapefruit.
Your functions can be written as a list of numbers. $f(1), f(2), f(3), ... $ To compare $f$ and $g$, compare $f(1)$ and $g(1)$. If $f(1)$ is less then $f$ is less. If they are the same then compare $f(2)$ and $g(2)$, etc. Continue until you find a difference.
answered Jan 20 at 14:52
badjohnbadjohn
4,3521720
answered Jan 20 at 14:52
badjohnbadjohn
4,3521720
answered Jan 20 at 14:52
badjohnbadjohn
4,3521720
answered Jan 20 at 14:52
badjohnbadjohn
4,3521720
4,3521720
$begingroup$
Thank you, just quick question. Is a linear order a lexicographical order?
$endgroup$
– Karol
Jan 20 at 15:05
$begingroup$
I'd said the reverse. A lexicographical order will be a linear but maybe not the reverse.
$endgroup$
– badjohn
Jan 20 at 15:16
$begingroup$
why is a lexicographical order a linear order? I cannot find any explanation and I'm beginning to understand it so I want to be 100% sure.
$endgroup$
– Karol
Jan 20 at 15:24
$begingroup$
Well, because the procedure I described will satisfy the axioms of a linear order. en.m.wikipedia.org/wiki/Total_order
$endgroup$
– badjohn
Jan 20 at 15:37
add a comment |
$begingroup$
Thank you, just quick question. Is a linear order a lexicographical order?
$endgroup$
– Karol
Jan 20 at 15:05
$begingroup$
I'd said the reverse. A lexicographical order will be a linear but maybe not the reverse.
$endgroup$
– badjohn
Jan 20 at 15:16
$begingroup$
why is a lexicographical order a linear order? I cannot find any explanation and I'm beginning to understand it so I want to be 100% sure.
$endgroup$
– Karol
Jan 20 at 15:24
$begingroup$
Well, because the procedure I described will satisfy the axioms of a linear order. en.m.wikipedia.org/wiki/Total_order
$endgroup$
– badjohn
Jan 20 at 15:37
$begingroup$
Thank you, just quick question. Is a linear order a lexicographical order?
$endgroup$
– Karol
Jan 20 at 15:05
$begingroup$
Thank you, just quick question. Is a linear order a lexicographical order?
$endgroup$
– Karol
Jan 20 at 15:05
$begingroup$
I'd said the reverse. A lexicographical order will be a linear but maybe not the reverse.
$endgroup$
– badjohn
Jan 20 at 15:16
$begingroup$
I'd said the reverse. A lexicographical order will be a linear but maybe not the reverse.
$endgroup$
– badjohn
Jan 20 at 15:16
$begingroup$
why is a lexicographical order a linear order? I cannot find any explanation and I'm beginning to understand it so I want to be 100% sure.
$endgroup$
– Karol
Jan 20 at 15:24
$begingroup$
why is a lexicographical order a linear order? I cannot find any explanation and I'm beginning to understand it so I want to be 100% sure.
$endgroup$
– Karol
Jan 20 at 15:24
$begingroup$
Well, because the procedure I described will satisfy the axioms of a linear order. en.m.wikipedia.org/wiki/Total_order
$endgroup$
– badjohn
Jan 20 at 15:37
$begingroup$
Well, because the procedure I described will satisfy the axioms of a linear order. en.m.wikipedia.org/wiki/Total_order
$endgroup$
– badjohn
Jan 20 at 15:37
add a comment |
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$begingroup$
Hint : There is an order on the alphabet $a,b,c,d,...,x,y,z$, and hence there is an order on words made out of $a,b,c,d,...,x,y,z$ which is reflected in the English dictionary, for example. Think about using this approach on your set : there is a natural order on $mathbb N$, and it should lead to an ordering on $mathbb N^mathbb N$
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:29
$begingroup$
@астонвіллаолофмэллбэрг, thanks for answering but haven't learned about the alphabet. Any other ideas?
$endgroup$
– Karol
Jan 20 at 14:38
$begingroup$
I think that he just means the letters that we write with, not a mathematical term. Do you know how to find a word in a dictionary based on your knowledge of the ABC?
$endgroup$
– badjohn
Jan 20 at 14:40
$begingroup$
What I am saying is this : for example if you have two strings $a_1a_2a_3...$ and $b_1b_2b_3...$ of natural numbers, which correspond to two elements of $mathbb N^mathbb N$. You first compare $a_1$ and $b_1$ : if $a_1<b_1$ then the first string is smaller than the second string. If $a_1 > b_1$ then the first string is larger than the second string. If $a_1 = b_1$, then move to comparing $a_2$ and $b_2$ (repeat the above steps with $a_1,b_1$ replaced by $a_2,b_2$). If these are equal then move to $a_3$ and $b_3$. I hope you see how to compare the strings from here.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 20 at 14:43
$begingroup$
@badjohn, so, this set is a set of natural numbers strings, tell me if I'm wrong. So those would be sets of natural numbers, so how exactly do I use the alphabet there?
$endgroup$
– Karol
Jan 20 at 14:43