Mixing
How to draw Hasse diagram for divisibilty?
$begingroup$
Please help me out.. Is there some appropriate method to draw Hasse diagram
My question is $L={1,2,3,4,5,6,10,12,15,30,60}$
Please explain me by step by step solution... Thanks for help..
elementary-set-theory graph-theory
edited Dec 4 '12 at 6:05
Asaf Karagila♦
303k32429761
asked Dec 4 '12 at 5:37
TasneemTasneem
1813
$endgroup$
add a comment |
$begingroup$
Please help me out.. Is there some appropriate method to draw Hasse diagram
My question is $L={1,2,3,4,5,6,10,12,15,30,60}$
Please explain me by step by step solution... Thanks for help..
elementary-set-theory graph-theory
edited Dec 4 '12 at 6:05
Asaf Karagila♦
303k32429761
asked Dec 4 '12 at 5:37
TasneemTasneem
1813
$endgroup$
$begingroup$
Hint: writing each number as its prime factorisation (eg writing $12$ as $2^2 cdot 3$) is often helpful when considering divisibility.
$endgroup$
– Peter LeFanu Lumsdaine
Dec 4 '12 at 5:43
1
$begingroup$
$20$ is also a divisor of $60$. Should it also be a member of the set $L$ ?
$endgroup$
– miracle173
Dec 4 '12 at 5:50
$begingroup$
Please draw a hasse diagram for me.
$endgroup$
– Tasneem
Dec 4 '12 at 5:58
$begingroup$
I'm far from sure that [graph-theory] fits this question, but I'm not that sure.
$endgroup$
– Asaf Karagila♦
Dec 4 '12 at 6:06
add a comment |
$begingroup$
Please help me out.. Is there some appropriate method to draw Hasse diagram
My question is $L={1,2,3,4,5,6,10,12,15,30,60}$
Please explain me by step by step solution... Thanks for help..
elementary-set-theory graph-theory
edited Dec 4 '12 at 6:05
Asaf Karagila♦
303k32429761
asked Dec 4 '12 at 5:37
TasneemTasneem
1813
$endgroup$
Please help me out.. Is there some appropriate method to draw Hasse diagram
My question is $L={1,2,3,4,5,6,10,12,15,30,60}$
Please explain me by step by step solution... Thanks for help..
elementary-set-theory graph-theory
elementary-set-theory graph-theory
edited Dec 4 '12 at 6:05
Asaf Karagila♦
303k32429761
asked Dec 4 '12 at 5:37
TasneemTasneem
1813
edited Dec 4 '12 at 6:05
Asaf Karagila♦
303k32429761
asked Dec 4 '12 at 5:37
TasneemTasneem
1813
edited Dec 4 '12 at 6:05
Asaf Karagila♦
303k32429761
edited Dec 4 '12 at 6:05
Asaf Karagila♦
303k32429761
edited Dec 4 '12 at 6:05
Asaf Karagila♦
303k32429761
303k32429761
asked Dec 4 '12 at 5:37
TasneemTasneem
1813
asked Dec 4 '12 at 5:37
TasneemTasneem
1813
asked Dec 4 '12 at 5:37
TasneemTasneem
1813
1813
$begingroup$
Hint: writing each number as its prime factorisation (eg writing $12$ as $2^2 cdot 3$) is often helpful when considering divisibility.
$endgroup$
– Peter LeFanu Lumsdaine
Dec 4 '12 at 5:43
1
$begingroup$
$20$ is also a divisor of $60$. Should it also be a member of the set $L$ ?
$endgroup$
– miracle173
Dec 4 '12 at 5:50
$begingroup$
Please draw a hasse diagram for me.
$endgroup$
– Tasneem
Dec 4 '12 at 5:58
$begingroup$
I'm far from sure that [graph-theory] fits this question, but I'm not that sure.
$endgroup$
– Asaf Karagila♦
Dec 4 '12 at 6:06
add a comment |
$begingroup$
Hint: writing each number as its prime factorisation (eg writing $12$ as $2^2 cdot 3$) is often helpful when considering divisibility.
$endgroup$
– Peter LeFanu Lumsdaine
Dec 4 '12 at 5:43
1
$begingroup$
$20$ is also a divisor of $60$. Should it also be a member of the set $L$ ?
$endgroup$
– miracle173
Dec 4 '12 at 5:50
$begingroup$
Please draw a hasse diagram for me.
$endgroup$
– Tasneem
Dec 4 '12 at 5:58
$begingroup$
I'm far from sure that [graph-theory] fits this question, but I'm not that sure.
$endgroup$
– Asaf Karagila♦
Dec 4 '12 at 6:06
$begingroup$
Hint: writing each number as its prime factorisation (eg writing $12$ as $2^2 cdot 3$) is often helpful when considering divisibility.
$endgroup$
– Peter LeFanu Lumsdaine
Dec 4 '12 at 5:43
$begingroup$
Hint: writing each number as its prime factorisation (eg writing $12$ as $2^2 cdot 3$) is often helpful when considering divisibility.
$endgroup$
– Peter LeFanu Lumsdaine
Dec 4 '12 at 5:43
1
1
$begingroup$
$20$ is also a divisor of $60$. Should it also be a member of the set $L$ ?
$endgroup$
– miracle173
Dec 4 '12 at 5:50
$begingroup$
$20$ is also a divisor of $60$. Should it also be a member of the set $L$ ?
$endgroup$
– miracle173
Dec 4 '12 at 5:50
$begingroup$
Please draw a hasse diagram for me.
$endgroup$
– Tasneem
Dec 4 '12 at 5:58
$begingroup$
Please draw a hasse diagram for me.
$endgroup$
– Tasneem
Dec 4 '12 at 5:58
$begingroup$
I'm far from sure that [graph-theory] fits this question, but I'm not that sure.
$endgroup$
– Asaf Karagila♦
Dec 4 '12 at 6:06
$begingroup$
I'm far from sure that [graph-theory] fits this question, but I'm not that sure.
$endgroup$
– Asaf Karagila♦
Dec 4 '12 at 6:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I would start at the bottom. Clearly $1$ divides every member of $L$, so $1$ is the minimum element in the divisibility order on $L$. Now what is the next layer up of the Hasse diagram? It must contain those members of $L$ that are divisible by $1$ (and of course by themselves) but not by any other element of $L$. Those elements are $2,3$, and $5$: every other element of $L$ is divisible by at least one of these three numbers. Thus, your partial Hasse diagram now looks like this:
2 3 5
| /
|/
1
What numbers will make up the third row? I claim that the third row contains $4,6,10$, and $15$. First, why not $12,30$, or $60$? $12$ is out because $6$ must be below $12$, and $6$ isn’t in the second row; $30$ is out because $15$ must be below $30$, and $15$ isn’t in the second row; and $60$ is out because every other element of $L$ must be below it. A bit of thought should convince you that no member of $L$ must come below $4,6,10$, or $15$ but above the second row, so that $4,6,10$, and $15$ really do belong in the third row. Specifically, $4$ must lie above $2$; $6$ must lie above $2$ and $3$; $10$ must lie above $2$ and $5$; and $15$ must lie above $3$ and $5$.
You can probably finish it up pretty easily from there: $12$ and $30$ go in the fourth row, with $12$ above $4$ and $3$, and $30$ above $6$ and $5$. Finally, $60$ will be the top element, sitting above $12$ and $30$. As a quick cross-check that $12$ and $30$ are the only members of $L$ that should be immediately below $60$, note that every other member of $L$ (besides $12,30$, and $60$) is a divisor of either $12$ or $30$ and therefore goes somewhere below at least one of these two numbers.
answered Dec 4 '12 at 6:05
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
$begingroup$
@Tasneem: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '12 at 6:26
add a comment |
$begingroup$
You have $L={1,2,3,4,5,6,10,12,15,20,30,60}$. $|L|$ = 12. This tells you the first important bit of information: there will be $12$ points in your Hasse diagram. (Please note that 20 was missing from the set in your question).
The next step is to find all the different ratios between the divisors as these will be the lines in your Hasse diagram.
I start with the highest divisor and work my way down.
1) 1:60 (Nothing is parallel to 1->60, so we will not draw this line in our diagram).
2) 1:30, 2:60 (This tells us the the lines 1->30 and 2->60 must be parallel).
3) 1:20, 3:60 (This tells us the the lines 1->20 and 3->60 must be parallel).
4) 1:15, 2:30, 4:60 (This tells us the the lines 1->15, 2->30, and 4->60 must be parallel).
5) 1:12, 5:60
6) 1:10, 2:20, 3:30, 6:60
7) 1:6, 2:12, 5:30, 10:60
8) 1:5, 2:10, 4:20, 6:30, 12:60
9) 1:4, 3:12, 5:20, 15:60
10) 1:3, 2:6, 4:12, 5:15, 10:30, 20:60
11) 1:2, 2:4, 3:6, 5:10, 6:12, 10:20, 15:30, 30:60
Now that we've calculated all of that, a fun trick about Hasse diagrams for divisors is that you can generally guess the resulting structure by simply looking at the set of ratios for the smallest divisor (in this case, the 1:2 ratios).
This tells us that we should have 4 parallel lines of 3 points in our diagram:
1 ---- 2 --- 4
3 ---- 6 --- 12
5 --- 10 --- 20
15 -- 30 --- 60
How can we arrange these lines such that they make sense with the other ratios? Let's look at the 1:3 ratios. We know that 1 must be connected to 3, 2 to 6, 4 to 12, etc. I'll try to display this visually.
1 --- 2 --- 4
| | |
3 --- 6 --- 12
5 -- 10 -- 20
| | |
15 -- 30 -- 60
Notice that we get the 1:4 ratios for free. 1->4 is already parallel to 3->12, 5->20, etc.
Let's apply the 1:5 ratios and see what happens.
3 --- 6 --- 12
| | |
1 --- 2 --- 4
| | |
5 -- 10 -- 20
| | |
15 -- 30 -- 60
| | | (connects to 3->6->12)
Okay, the resulting diagram is starting to look like two stacked cubes. From here, I would draw the cubes and label the vertices with the numbers we deduced above. Then use the list of divisor ratios and make sure that all lines are present and parallel to lines of the same ratio.
For $L={1,2,3,4,5,6,10,12,15,20,30,60}$, you can check your answer on Wolfram.
EDIT: I discovered that the Wolfram solution is missing all of the diagonals. Namely, 1->20, 3->60, 1->10, 2->20, 3->30, 6->60, 1->6, 2->12, 5->30, 10->60. The vertices, however, are all in the correct position.
answered Jan 9 at 22:18
user2205316user2205316
12
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2 Answers
2
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2 Answers
2
active
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$begingroup$
I would start at the bottom. Clearly $1$ divides every member of $L$, so $1$ is the minimum element in the divisibility order on $L$. Now what is the next layer up of the Hasse diagram? It must contain those members of $L$ that are divisible by $1$ (and of course by themselves) but not by any other element of $L$. Those elements are $2,3$, and $5$: every other element of $L$ is divisible by at least one of these three numbers. Thus, your partial Hasse diagram now looks like this:
2 3 5
| /
|/
1
What numbers will make up the third row? I claim that the third row contains $4,6,10$, and $15$. First, why not $12,30$, or $60$? $12$ is out because $6$ must be below $12$, and $6$ isn’t in the second row; $30$ is out because $15$ must be below $30$, and $15$ isn’t in the second row; and $60$ is out because every other element of $L$ must be below it. A bit of thought should convince you that no member of $L$ must come below $4,6,10$, or $15$ but above the second row, so that $4,6,10$, and $15$ really do belong in the third row. Specifically, $4$ must lie above $2$; $6$ must lie above $2$ and $3$; $10$ must lie above $2$ and $5$; and $15$ must lie above $3$ and $5$.
You can probably finish it up pretty easily from there: $12$ and $30$ go in the fourth row, with $12$ above $4$ and $3$, and $30$ above $6$ and $5$. Finally, $60$ will be the top element, sitting above $12$ and $30$. As a quick cross-check that $12$ and $30$ are the only members of $L$ that should be immediately below $60$, note that every other member of $L$ (besides $12,30$, and $60$) is a divisor of either $12$ or $30$ and therefore goes somewhere below at least one of these two numbers.
answered Dec 4 '12 at 6:05
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
$begingroup$
@Tasneem: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '12 at 6:26
add a comment |
$begingroup$
I would start at the bottom. Clearly $1$ divides every member of $L$, so $1$ is the minimum element in the divisibility order on $L$. Now what is the next layer up of the Hasse diagram? It must contain those members of $L$ that are divisible by $1$ (and of course by themselves) but not by any other element of $L$. Those elements are $2,3$, and $5$: every other element of $L$ is divisible by at least one of these three numbers. Thus, your partial Hasse diagram now looks like this:
2 3 5
| /
|/
1
What numbers will make up the third row? I claim that the third row contains $4,6,10$, and $15$. First, why not $12,30$, or $60$? $12$ is out because $6$ must be below $12$, and $6$ isn’t in the second row; $30$ is out because $15$ must be below $30$, and $15$ isn’t in the second row; and $60$ is out because every other element of $L$ must be below it. A bit of thought should convince you that no member of $L$ must come below $4,6,10$, or $15$ but above the second row, so that $4,6,10$, and $15$ really do belong in the third row. Specifically, $4$ must lie above $2$; $6$ must lie above $2$ and $3$; $10$ must lie above $2$ and $5$; and $15$ must lie above $3$ and $5$.
You can probably finish it up pretty easily from there: $12$ and $30$ go in the fourth row, with $12$ above $4$ and $3$, and $30$ above $6$ and $5$. Finally, $60$ will be the top element, sitting above $12$ and $30$. As a quick cross-check that $12$ and $30$ are the only members of $L$ that should be immediately below $60$, note that every other member of $L$ (besides $12,30$, and $60$) is a divisor of either $12$ or $30$ and therefore goes somewhere below at least one of these two numbers.
answered Dec 4 '12 at 6:05
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
$begingroup$
@Tasneem: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '12 at 6:26
add a comment |
$begingroup$
I would start at the bottom. Clearly $1$ divides every member of $L$, so $1$ is the minimum element in the divisibility order on $L$. Now what is the next layer up of the Hasse diagram? It must contain those members of $L$ that are divisible by $1$ (and of course by themselves) but not by any other element of $L$. Those elements are $2,3$, and $5$: every other element of $L$ is divisible by at least one of these three numbers. Thus, your partial Hasse diagram now looks like this:
2 3 5
| /
|/
1
What numbers will make up the third row? I claim that the third row contains $4,6,10$, and $15$. First, why not $12,30$, or $60$? $12$ is out because $6$ must be below $12$, and $6$ isn’t in the second row; $30$ is out because $15$ must be below $30$, and $15$ isn’t in the second row; and $60$ is out because every other element of $L$ must be below it. A bit of thought should convince you that no member of $L$ must come below $4,6,10$, or $15$ but above the second row, so that $4,6,10$, and $15$ really do belong in the third row. Specifically, $4$ must lie above $2$; $6$ must lie above $2$ and $3$; $10$ must lie above $2$ and $5$; and $15$ must lie above $3$ and $5$.
You can probably finish it up pretty easily from there: $12$ and $30$ go in the fourth row, with $12$ above $4$ and $3$, and $30$ above $6$ and $5$. Finally, $60$ will be the top element, sitting above $12$ and $30$. As a quick cross-check that $12$ and $30$ are the only members of $L$ that should be immediately below $60$, note that every other member of $L$ (besides $12,30$, and $60$) is a divisor of either $12$ or $30$ and therefore goes somewhere below at least one of these two numbers.
answered Dec 4 '12 at 6:05
Brian M. ScottBrian M. Scott
457k38509909
$endgroup$
I would start at the bottom. Clearly $1$ divides every member of $L$, so $1$ is the minimum element in the divisibility order on $L$. Now what is the next layer up of the Hasse diagram? It must contain those members of $L$ that are divisible by $1$ (and of course by themselves) but not by any other element of $L$. Those elements are $2,3$, and $5$: every other element of $L$ is divisible by at least one of these three numbers. Thus, your partial Hasse diagram now looks like this:
2 3 5
| /
|/
1
What numbers will make up the third row? I claim that the third row contains $4,6,10$, and $15$. First, why not $12,30$, or $60$? $12$ is out because $6$ must be below $12$, and $6$ isn’t in the second row; $30$ is out because $15$ must be below $30$, and $15$ isn’t in the second row; and $60$ is out because every other element of $L$ must be below it. A bit of thought should convince you that no member of $L$ must come below $4,6,10$, or $15$ but above the second row, so that $4,6,10$, and $15$ really do belong in the third row. Specifically, $4$ must lie above $2$; $6$ must lie above $2$ and $3$; $10$ must lie above $2$ and $5$; and $15$ must lie above $3$ and $5$.
You can probably finish it up pretty easily from there: $12$ and $30$ go in the fourth row, with $12$ above $4$ and $3$, and $30$ above $6$ and $5$. Finally, $60$ will be the top element, sitting above $12$ and $30$. As a quick cross-check that $12$ and $30$ are the only members of $L$ that should be immediately below $60$, note that every other member of $L$ (besides $12,30$, and $60$) is a divisor of either $12$ or $30$ and therefore goes somewhere below at least one of these two numbers.
answered Dec 4 '12 at 6:05
Brian M. ScottBrian M. Scott
457k38509909
answered Dec 4 '12 at 6:05
Brian M. ScottBrian M. Scott
457k38509909
answered Dec 4 '12 at 6:05
Brian M. ScottBrian M. Scott
457k38509909
answered Dec 4 '12 at 6:05
Brian M. ScottBrian M. Scott
457k38509909
457k38509909
$begingroup$
@Tasneem: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '12 at 6:26
add a comment |
$begingroup$
@Tasneem: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '12 at 6:26
$begingroup$
@Tasneem: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '12 at 6:26
$begingroup$
@Tasneem: You’re welcome.
$endgroup$
– Brian M. Scott
Dec 4 '12 at 6:26
add a comment |
$begingroup$
You have $L={1,2,3,4,5,6,10,12,15,20,30,60}$. $|L|$ = 12. This tells you the first important bit of information: there will be $12$ points in your Hasse diagram. (Please note that 20 was missing from the set in your question).
The next step is to find all the different ratios between the divisors as these will be the lines in your Hasse diagram.
I start with the highest divisor and work my way down.
1) 1:60 (Nothing is parallel to 1->60, so we will not draw this line in our diagram).
2) 1:30, 2:60 (This tells us the the lines 1->30 and 2->60 must be parallel).
3) 1:20, 3:60 (This tells us the the lines 1->20 and 3->60 must be parallel).
4) 1:15, 2:30, 4:60 (This tells us the the lines 1->15, 2->30, and 4->60 must be parallel).
5) 1:12, 5:60
6) 1:10, 2:20, 3:30, 6:60
7) 1:6, 2:12, 5:30, 10:60
8) 1:5, 2:10, 4:20, 6:30, 12:60
9) 1:4, 3:12, 5:20, 15:60
10) 1:3, 2:6, 4:12, 5:15, 10:30, 20:60
11) 1:2, 2:4, 3:6, 5:10, 6:12, 10:20, 15:30, 30:60
Now that we've calculated all of that, a fun trick about Hasse diagrams for divisors is that you can generally guess the resulting structure by simply looking at the set of ratios for the smallest divisor (in this case, the 1:2 ratios).
This tells us that we should have 4 parallel lines of 3 points in our diagram:
1 ---- 2 --- 4
3 ---- 6 --- 12
5 --- 10 --- 20
15 -- 30 --- 60
How can we arrange these lines such that they make sense with the other ratios? Let's look at the 1:3 ratios. We know that 1 must be connected to 3, 2 to 6, 4 to 12, etc. I'll try to display this visually.
1 --- 2 --- 4
| | |
3 --- 6 --- 12
5 -- 10 -- 20
| | |
15 -- 30 -- 60
Notice that we get the 1:4 ratios for free. 1->4 is already parallel to 3->12, 5->20, etc.
Let's apply the 1:5 ratios and see what happens.
3 --- 6 --- 12
| | |
1 --- 2 --- 4
| | |
5 -- 10 -- 20
| | |
15 -- 30 -- 60
| | | (connects to 3->6->12)
Okay, the resulting diagram is starting to look like two stacked cubes. From here, I would draw the cubes and label the vertices with the numbers we deduced above. Then use the list of divisor ratios and make sure that all lines are present and parallel to lines of the same ratio.
For $L={1,2,3,4,5,6,10,12,15,20,30,60}$, you can check your answer on Wolfram.
EDIT: I discovered that the Wolfram solution is missing all of the diagonals. Namely, 1->20, 3->60, 1->10, 2->20, 3->30, 6->60, 1->6, 2->12, 5->30, 10->60. The vertices, however, are all in the correct position.
answered Jan 9 at 22:18
user2205316user2205316
12
$endgroup$
add a comment |
$begingroup$
You have $L={1,2,3,4,5,6,10,12,15,20,30,60}$. $|L|$ = 12. This tells you the first important bit of information: there will be $12$ points in your Hasse diagram. (Please note that 20 was missing from the set in your question).
The next step is to find all the different ratios between the divisors as these will be the lines in your Hasse diagram.
I start with the highest divisor and work my way down.
1) 1:60 (Nothing is parallel to 1->60, so we will not draw this line in our diagram).
2) 1:30, 2:60 (This tells us the the lines 1->30 and 2->60 must be parallel).
3) 1:20, 3:60 (This tells us the the lines 1->20 and 3->60 must be parallel).
4) 1:15, 2:30, 4:60 (This tells us the the lines 1->15, 2->30, and 4->60 must be parallel).
5) 1:12, 5:60
6) 1:10, 2:20, 3:30, 6:60
7) 1:6, 2:12, 5:30, 10:60
8) 1:5, 2:10, 4:20, 6:30, 12:60
9) 1:4, 3:12, 5:20, 15:60
10) 1:3, 2:6, 4:12, 5:15, 10:30, 20:60
11) 1:2, 2:4, 3:6, 5:10, 6:12, 10:20, 15:30, 30:60
Now that we've calculated all of that, a fun trick about Hasse diagrams for divisors is that you can generally guess the resulting structure by simply looking at the set of ratios for the smallest divisor (in this case, the 1:2 ratios).
This tells us that we should have 4 parallel lines of 3 points in our diagram:
1 ---- 2 --- 4
3 ---- 6 --- 12
5 --- 10 --- 20
15 -- 30 --- 60
How can we arrange these lines such that they make sense with the other ratios? Let's look at the 1:3 ratios. We know that 1 must be connected to 3, 2 to 6, 4 to 12, etc. I'll try to display this visually.
1 --- 2 --- 4
| | |
3 --- 6 --- 12
5 -- 10 -- 20
| | |
15 -- 30 -- 60
Notice that we get the 1:4 ratios for free. 1->4 is already parallel to 3->12, 5->20, etc.
Let's apply the 1:5 ratios and see what happens.
3 --- 6 --- 12
| | |
1 --- 2 --- 4
| | |
5 -- 10 -- 20
| | |
15 -- 30 -- 60
| | | (connects to 3->6->12)
Okay, the resulting diagram is starting to look like two stacked cubes. From here, I would draw the cubes and label the vertices with the numbers we deduced above. Then use the list of divisor ratios and make sure that all lines are present and parallel to lines of the same ratio.
For $L={1,2,3,4,5,6,10,12,15,20,30,60}$, you can check your answer on Wolfram.
EDIT: I discovered that the Wolfram solution is missing all of the diagonals. Namely, 1->20, 3->60, 1->10, 2->20, 3->30, 6->60, 1->6, 2->12, 5->30, 10->60. The vertices, however, are all in the correct position.
answered Jan 9 at 22:18
user2205316user2205316
12
$endgroup$
add a comment |
$begingroup$
You have $L={1,2,3,4,5,6,10,12,15,20,30,60}$. $|L|$ = 12. This tells you the first important bit of information: there will be $12$ points in your Hasse diagram. (Please note that 20 was missing from the set in your question).
The next step is to find all the different ratios between the divisors as these will be the lines in your Hasse diagram.
I start with the highest divisor and work my way down.
1) 1:60 (Nothing is parallel to 1->60, so we will not draw this line in our diagram).
2) 1:30, 2:60 (This tells us the the lines 1->30 and 2->60 must be parallel).
3) 1:20, 3:60 (This tells us the the lines 1->20 and 3->60 must be parallel).
4) 1:15, 2:30, 4:60 (This tells us the the lines 1->15, 2->30, and 4->60 must be parallel).
5) 1:12, 5:60
6) 1:10, 2:20, 3:30, 6:60
7) 1:6, 2:12, 5:30, 10:60
8) 1:5, 2:10, 4:20, 6:30, 12:60
9) 1:4, 3:12, 5:20, 15:60
10) 1:3, 2:6, 4:12, 5:15, 10:30, 20:60
11) 1:2, 2:4, 3:6, 5:10, 6:12, 10:20, 15:30, 30:60
Now that we've calculated all of that, a fun trick about Hasse diagrams for divisors is that you can generally guess the resulting structure by simply looking at the set of ratios for the smallest divisor (in this case, the 1:2 ratios).
This tells us that we should have 4 parallel lines of 3 points in our diagram:
1 ---- 2 --- 4
3 ---- 6 --- 12
5 --- 10 --- 20
15 -- 30 --- 60
How can we arrange these lines such that they make sense with the other ratios? Let's look at the 1:3 ratios. We know that 1 must be connected to 3, 2 to 6, 4 to 12, etc. I'll try to display this visually.
1 --- 2 --- 4
| | |
3 --- 6 --- 12
5 -- 10 -- 20
| | |
15 -- 30 -- 60
Notice that we get the 1:4 ratios for free. 1->4 is already parallel to 3->12, 5->20, etc.
Let's apply the 1:5 ratios and see what happens.
3 --- 6 --- 12
| | |
1 --- 2 --- 4
| | |
5 -- 10 -- 20
| | |
15 -- 30 -- 60
| | | (connects to 3->6->12)
Okay, the resulting diagram is starting to look like two stacked cubes. From here, I would draw the cubes and label the vertices with the numbers we deduced above. Then use the list of divisor ratios and make sure that all lines are present and parallel to lines of the same ratio.
For $L={1,2,3,4,5,6,10,12,15,20,30,60}$, you can check your answer on Wolfram.
EDIT: I discovered that the Wolfram solution is missing all of the diagonals. Namely, 1->20, 3->60, 1->10, 2->20, 3->30, 6->60, 1->6, 2->12, 5->30, 10->60. The vertices, however, are all in the correct position.
answered Jan 9 at 22:18
user2205316user2205316
12
$endgroup$
You have $L={1,2,3,4,5,6,10,12,15,20,30,60}$. $|L|$ = 12. This tells you the first important bit of information: there will be $12$ points in your Hasse diagram. (Please note that 20 was missing from the set in your question).
The next step is to find all the different ratios between the divisors as these will be the lines in your Hasse diagram.
I start with the highest divisor and work my way down.
1) 1:60 (Nothing is parallel to 1->60, so we will not draw this line in our diagram).
2) 1:30, 2:60 (This tells us the the lines 1->30 and 2->60 must be parallel).
3) 1:20, 3:60 (This tells us the the lines 1->20 and 3->60 must be parallel).
4) 1:15, 2:30, 4:60 (This tells us the the lines 1->15, 2->30, and 4->60 must be parallel).
5) 1:12, 5:60
6) 1:10, 2:20, 3:30, 6:60
7) 1:6, 2:12, 5:30, 10:60
8) 1:5, 2:10, 4:20, 6:30, 12:60
9) 1:4, 3:12, 5:20, 15:60
10) 1:3, 2:6, 4:12, 5:15, 10:30, 20:60
11) 1:2, 2:4, 3:6, 5:10, 6:12, 10:20, 15:30, 30:60
Now that we've calculated all of that, a fun trick about Hasse diagrams for divisors is that you can generally guess the resulting structure by simply looking at the set of ratios for the smallest divisor (in this case, the 1:2 ratios).
This tells us that we should have 4 parallel lines of 3 points in our diagram:
1 ---- 2 --- 4
3 ---- 6 --- 12
5 --- 10 --- 20
15 -- 30 --- 60
How can we arrange these lines such that they make sense with the other ratios? Let's look at the 1:3 ratios. We know that 1 must be connected to 3, 2 to 6, 4 to 12, etc. I'll try to display this visually.
1 --- 2 --- 4
| | |
3 --- 6 --- 12
5 -- 10 -- 20
| | |
15 -- 30 -- 60
Notice that we get the 1:4 ratios for free. 1->4 is already parallel to 3->12, 5->20, etc.
Let's apply the 1:5 ratios and see what happens.
3 --- 6 --- 12
| | |
1 --- 2 --- 4
| | |
5 -- 10 -- 20
| | |
15 -- 30 -- 60
| | | (connects to 3->6->12)
Okay, the resulting diagram is starting to look like two stacked cubes. From here, I would draw the cubes and label the vertices with the numbers we deduced above. Then use the list of divisor ratios and make sure that all lines are present and parallel to lines of the same ratio.
For $L={1,2,3,4,5,6,10,12,15,20,30,60}$, you can check your answer on Wolfram.
EDIT: I discovered that the Wolfram solution is missing all of the diagonals. Namely, 1->20, 3->60, 1->10, 2->20, 3->30, 6->60, 1->6, 2->12, 5->30, 10->60. The vertices, however, are all in the correct position.
answered Jan 9 at 22:18
user2205316user2205316
12
edited Jan 9 at 22:31
answered Jan 9 at 22:18
user2205316user2205316
12
answered Jan 9 at 22:18
user2205316user2205316
12
answered Jan 9 at 22:18
user2205316user2205316
12
12
add a comment |
add a comment |
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$begingroup$
Hint: writing each number as its prime factorisation (eg writing $12$ as $2^2 cdot 3$) is often helpful when considering divisibility.
$endgroup$
– Peter LeFanu Lumsdaine
Dec 4 '12 at 5:43
1
$begingroup$
$20$ is also a divisor of $60$. Should it also be a member of the set $L$ ?
$endgroup$
– miracle173
Dec 4 '12 at 5:50
$begingroup$
Please draw a hasse diagram for me.
$endgroup$
– Tasneem
Dec 4 '12 at 5:58
$begingroup$
I'm far from sure that [graph-theory] fits this question, but I'm not that sure.
$endgroup$
– Asaf Karagila♦
Dec 4 '12 at 6:06