Mixing
How do you find a common multiple (which could be irrational) of a two numbers?
$begingroup$
I wasn't really sure how to phrase the question nor know how I should tag it.
Lets say I have a piece of paper (42x25) that I need to cut into squares (the size and number of the squares do not matter). How many pieces can I cut out with having little to no remainder?
I did this problem by following an art book's instructions with the dimensions of 21x12.5 cm instead of 42x25 cm and doing these steps.
- lw = R ; 21 - 12.5 = 8.5
- R/2 = length of segment ; 8.5/2 = 4.25
- Then I measured out 4.25 cm along the edges of a paper until I reached 5 segments on the length and 3 segments on the width.
- Then I divided the length and width by their respective segments. And found out their lengths were around 4.2 cm
What I am asking:
What is the formula to find the common multiple when multiplied with a whole number it roughly equals to given lengths? The only thing I can this of is that if:
x = the segment length
y = the number of segments of the length of an edge1
z = the number of segments of the length of an edge2
Then
xy ~= Edge1
xz ~= Edge2
But I can't figure out how to actually solve it since there are two variables.
geometry
asked Jan 22 at 20:27
BlueBlue
1114
$endgroup$
add a comment |
$begingroup$
I wasn't really sure how to phrase the question nor know how I should tag it.
Lets say I have a piece of paper (42x25) that I need to cut into squares (the size and number of the squares do not matter). How many pieces can I cut out with having little to no remainder?
I did this problem by following an art book's instructions with the dimensions of 21x12.5 cm instead of 42x25 cm and doing these steps.
- lw = R ; 21 - 12.5 = 8.5
- R/2 = length of segment ; 8.5/2 = 4.25
- Then I measured out 4.25 cm along the edges of a paper until I reached 5 segments on the length and 3 segments on the width.
- Then I divided the length and width by their respective segments. And found out their lengths were around 4.2 cm
What I am asking:
What is the formula to find the common multiple when multiplied with a whole number it roughly equals to given lengths? The only thing I can this of is that if:
x = the segment length
y = the number of segments of the length of an edge1
z = the number of segments of the length of an edge2
Then
xy ~= Edge1
xz ~= Edge2
But I can't figure out how to actually solve it since there are two variables.
geometry
asked Jan 22 at 20:27
BlueBlue
1114
$endgroup$
add a comment |
$begingroup$
I wasn't really sure how to phrase the question nor know how I should tag it.
Lets say I have a piece of paper (42x25) that I need to cut into squares (the size and number of the squares do not matter). How many pieces can I cut out with having little to no remainder?
I did this problem by following an art book's instructions with the dimensions of 21x12.5 cm instead of 42x25 cm and doing these steps.
- lw = R ; 21 - 12.5 = 8.5
- R/2 = length of segment ; 8.5/2 = 4.25
- Then I measured out 4.25 cm along the edges of a paper until I reached 5 segments on the length and 3 segments on the width.
- Then I divided the length and width by their respective segments. And found out their lengths were around 4.2 cm
What I am asking:
What is the formula to find the common multiple when multiplied with a whole number it roughly equals to given lengths? The only thing I can this of is that if:
x = the segment length
y = the number of segments of the length of an edge1
z = the number of segments of the length of an edge2
Then
xy ~= Edge1
xz ~= Edge2
But I can't figure out how to actually solve it since there are two variables.
geometry
asked Jan 22 at 20:27
BlueBlue
1114
$endgroup$
I wasn't really sure how to phrase the question nor know how I should tag it.
Lets say I have a piece of paper (42x25) that I need to cut into squares (the size and number of the squares do not matter). How many pieces can I cut out with having little to no remainder?
I did this problem by following an art book's instructions with the dimensions of 21x12.5 cm instead of 42x25 cm and doing these steps.
- lw = R ; 21 - 12.5 = 8.5
- R/2 = length of segment ; 8.5/2 = 4.25
- Then I measured out 4.25 cm along the edges of a paper until I reached 5 segments on the length and 3 segments on the width.
- Then I divided the length and width by their respective segments. And found out their lengths were around 4.2 cm
What I am asking:
What is the formula to find the common multiple when multiplied with a whole number it roughly equals to given lengths? The only thing I can this of is that if:
x = the segment length
y = the number of segments of the length of an edge1
z = the number of segments of the length of an edge2
Then
xy ~= Edge1
xz ~= Edge2
But I can't figure out how to actually solve it since there are two variables.
geometry
geometry
asked Jan 22 at 20:27
BlueBlue
1114
asked Jan 22 at 20:27
BlueBlue
1114
asked Jan 22 at 20:27
BlueBlue
1114
asked Jan 22 at 20:27
BlueBlue
1114
asked Jan 22 at 20:27
BlueBlue
1114
1114
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You want to find two small integers $y$ and $z$ such that
$${text{Edge}_1overtext{Edge}_2}approx {yover z}.$$
The standard way is that of expressing ${text{Edge}_1overtext{Edge}_2}$ as a continued fraction: its convergents will then give the best rational approximations.
Example: for ${text{Edge}_1overtext{Edge}_2}={42over25}$, the continued fraction representation is
$$
{42over25}=1+{1over1+displaystyle{1over2+displaystyle{1over8}}}
$$
and its convergents are
$$
1,quad 2,quad{5over3},quad{42over25}.
$$
The best one is probably ${5over3}$, which gives your solution: divide $42$ into $5$ parts and $25$ into $3$ parts.
answered Jan 22 at 21:44
AretinoAretino
25k21445
$endgroup$
$begingroup$
I looked into continued fractions wouldn't 42/25 = [1 , 1, 2, 8] <-- assuming this is the convergent ? So how did you get 5/3? Was it approximated from 25/17?
$endgroup$
– Blue
Jan 22 at 22:19
2
$begingroup$
$${5over3}=1+{1over1+displaystyle{1over2}}$$ is the third convergent.
$endgroup$
– Aretino
Jan 22 at 22:45
$begingroup$
So if I'm understanding this right, if I were to change the lengths to ${39over17}$ the convergents would be [2;3,2,5]. with the second convergent being ${7over3}$ and third being ${16over7}$. Which will find the lengths of the segments 39 and 17?
$endgroup$
– Blue
Jan 22 at 22:58
1
$begingroup$
With $7/3$ you find $39/7=5.57$, $17/3=5.67$. With $16/7$ you have $39/16=2.44$, $17/7=2.43$. Higher convergents give better accuracy but imply a larger number of pieces.
$endgroup$
– Aretino
Jan 22 at 23:06
add a comment |
$begingroup$
Leat $a$ and $b$ be the sides in question. Their ratio $frac{a}{b}$ is either rational or irrational number. If it is rational, let say $frac{p}{q}$, the "unit" length will be
$$
u=frac{a}{p}=frac{b}{q}.
$$
If the ratio is irrational you can take any rational approximation for it, e.g.
$$
sqrt2simfrac{141}{100}
$$
and proceed as before.
answered Jan 22 at 21:20
useruser
5,23811030
$endgroup$
add a comment |
$begingroup$
From your examples, it looks like you're dealing with rational lengths for the paper, $l$ and $w$. It even looks like you're using a finite decimal expansion (the decimal doesn't go on forever).
From this, I suggest that you multiply $l$ and $w$ by powers of $10$ until you have two whole numbers, I'll call them $L$ and $W$. Let $G$ be the greatest common divisor of $L$ and $W$.
Let $g$ be $G$ divided by the power of $10$ that you multiplied in the previous step. This is the length of the side of a square. Now, $l/g$ and $w/g$ will tell you the number of squares along an edge of the paper.
If $l$ and $w$ are rational, but aren't decimals with finite expansions, multiply them by the least common multiple of their denominators, and mimic the steps above.
If all your side lengths are rational, then you will never have an irrational side length for your squares. If your side lengths are irrational, then the only way to get squares without anything left over is that the ratio of the side lengths is rational.
answered Jan 22 at 21:28
Michael BurrMichael Burr
27k23262
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You want to find two small integers $y$ and $z$ such that
$${text{Edge}_1overtext{Edge}_2}approx {yover z}.$$
The standard way is that of expressing ${text{Edge}_1overtext{Edge}_2}$ as a continued fraction: its convergents will then give the best rational approximations.
Example: for ${text{Edge}_1overtext{Edge}_2}={42over25}$, the continued fraction representation is
$$
{42over25}=1+{1over1+displaystyle{1over2+displaystyle{1over8}}}
$$
and its convergents are
$$
1,quad 2,quad{5over3},quad{42over25}.
$$
The best one is probably ${5over3}$, which gives your solution: divide $42$ into $5$ parts and $25$ into $3$ parts.
answered Jan 22 at 21:44
AretinoAretino
25k21445
$endgroup$
$begingroup$
I looked into continued fractions wouldn't 42/25 = [1 , 1, 2, 8] <-- assuming this is the convergent ? So how did you get 5/3? Was it approximated from 25/17?
$endgroup$
– Blue
Jan 22 at 22:19
2
$begingroup$
$${5over3}=1+{1over1+displaystyle{1over2}}$$ is the third convergent.
$endgroup$
– Aretino
Jan 22 at 22:45
$begingroup$
So if I'm understanding this right, if I were to change the lengths to ${39over17}$ the convergents would be [2;3,2,5]. with the second convergent being ${7over3}$ and third being ${16over7}$. Which will find the lengths of the segments 39 and 17?
$endgroup$
– Blue
Jan 22 at 22:58
1
$begingroup$
With $7/3$ you find $39/7=5.57$, $17/3=5.67$. With $16/7$ you have $39/16=2.44$, $17/7=2.43$. Higher convergents give better accuracy but imply a larger number of pieces.
$endgroup$
– Aretino
Jan 22 at 23:06
add a comment |
$begingroup$
You want to find two small integers $y$ and $z$ such that
$${text{Edge}_1overtext{Edge}_2}approx {yover z}.$$
The standard way is that of expressing ${text{Edge}_1overtext{Edge}_2}$ as a continued fraction: its convergents will then give the best rational approximations.
Example: for ${text{Edge}_1overtext{Edge}_2}={42over25}$, the continued fraction representation is
$$
{42over25}=1+{1over1+displaystyle{1over2+displaystyle{1over8}}}
$$
and its convergents are
$$
1,quad 2,quad{5over3},quad{42over25}.
$$
The best one is probably ${5over3}$, which gives your solution: divide $42$ into $5$ parts and $25$ into $3$ parts.
answered Jan 22 at 21:44
AretinoAretino
25k21445
$endgroup$
$begingroup$
I looked into continued fractions wouldn't 42/25 = [1 , 1, 2, 8] <-- assuming this is the convergent ? So how did you get 5/3? Was it approximated from 25/17?
$endgroup$
– Blue
Jan 22 at 22:19
2
$begingroup$
$${5over3}=1+{1over1+displaystyle{1over2}}$$ is the third convergent.
$endgroup$
– Aretino
Jan 22 at 22:45
$begingroup$
So if I'm understanding this right, if I were to change the lengths to ${39over17}$ the convergents would be [2;3,2,5]. with the second convergent being ${7over3}$ and third being ${16over7}$. Which will find the lengths of the segments 39 and 17?
$endgroup$
– Blue
Jan 22 at 22:58
1
$begingroup$
With $7/3$ you find $39/7=5.57$, $17/3=5.67$. With $16/7$ you have $39/16=2.44$, $17/7=2.43$. Higher convergents give better accuracy but imply a larger number of pieces.
$endgroup$
– Aretino
Jan 22 at 23:06
add a comment |
$begingroup$
You want to find two small integers $y$ and $z$ such that
$${text{Edge}_1overtext{Edge}_2}approx {yover z}.$$
The standard way is that of expressing ${text{Edge}_1overtext{Edge}_2}$ as a continued fraction: its convergents will then give the best rational approximations.
Example: for ${text{Edge}_1overtext{Edge}_2}={42over25}$, the continued fraction representation is
$$
{42over25}=1+{1over1+displaystyle{1over2+displaystyle{1over8}}}
$$
and its convergents are
$$
1,quad 2,quad{5over3},quad{42over25}.
$$
The best one is probably ${5over3}$, which gives your solution: divide $42$ into $5$ parts and $25$ into $3$ parts.
answered Jan 22 at 21:44
AretinoAretino
25k21445
$endgroup$
You want to find two small integers $y$ and $z$ such that
$${text{Edge}_1overtext{Edge}_2}approx {yover z}.$$
The standard way is that of expressing ${text{Edge}_1overtext{Edge}_2}$ as a continued fraction: its convergents will then give the best rational approximations.
Example: for ${text{Edge}_1overtext{Edge}_2}={42over25}$, the continued fraction representation is
$$
{42over25}=1+{1over1+displaystyle{1over2+displaystyle{1over8}}}
$$
and its convergents are
$$
1,quad 2,quad{5over3},quad{42over25}.
$$
The best one is probably ${5over3}$, which gives your solution: divide $42$ into $5$ parts and $25$ into $3$ parts.
answered Jan 22 at 21:44
AretinoAretino
25k21445
answered Jan 22 at 21:44
AretinoAretino
25k21445
answered Jan 22 at 21:44
AretinoAretino
25k21445
answered Jan 22 at 21:44
AretinoAretino
25k21445
25k21445
$begingroup$
I looked into continued fractions wouldn't 42/25 = [1 , 1, 2, 8] <-- assuming this is the convergent ? So how did you get 5/3? Was it approximated from 25/17?
$endgroup$
– Blue
Jan 22 at 22:19
2
$begingroup$
$${5over3}=1+{1over1+displaystyle{1over2}}$$ is the third convergent.
$endgroup$
– Aretino
Jan 22 at 22:45
$begingroup$
So if I'm understanding this right, if I were to change the lengths to ${39over17}$ the convergents would be [2;3,2,5]. with the second convergent being ${7over3}$ and third being ${16over7}$. Which will find the lengths of the segments 39 and 17?
$endgroup$
– Blue
Jan 22 at 22:58
1
$begingroup$
With $7/3$ you find $39/7=5.57$, $17/3=5.67$. With $16/7$ you have $39/16=2.44$, $17/7=2.43$. Higher convergents give better accuracy but imply a larger number of pieces.
$endgroup$
– Aretino
Jan 22 at 23:06
add a comment |
$begingroup$
I looked into continued fractions wouldn't 42/25 = [1 , 1, 2, 8] <-- assuming this is the convergent ? So how did you get 5/3? Was it approximated from 25/17?
$endgroup$
– Blue
Jan 22 at 22:19
2
$begingroup$
$${5over3}=1+{1over1+displaystyle{1over2}}$$ is the third convergent.
$endgroup$
– Aretino
Jan 22 at 22:45
$begingroup$
So if I'm understanding this right, if I were to change the lengths to ${39over17}$ the convergents would be [2;3,2,5]. with the second convergent being ${7over3}$ and third being ${16over7}$. Which will find the lengths of the segments 39 and 17?
$endgroup$
– Blue
Jan 22 at 22:58
1
$begingroup$
With $7/3$ you find $39/7=5.57$, $17/3=5.67$. With $16/7$ you have $39/16=2.44$, $17/7=2.43$. Higher convergents give better accuracy but imply a larger number of pieces.
$endgroup$
– Aretino
Jan 22 at 23:06
$begingroup$
I looked into continued fractions wouldn't 42/25 = [1 , 1, 2, 8] <-- assuming this is the convergent ? So how did you get 5/3? Was it approximated from 25/17?
$endgroup$
– Blue
Jan 22 at 22:19
$begingroup$
I looked into continued fractions wouldn't 42/25 = [1 , 1, 2, 8] <-- assuming this is the convergent ? So how did you get 5/3? Was it approximated from 25/17?
$endgroup$
– Blue
Jan 22 at 22:19
2
2
$begingroup$
$${5over3}=1+{1over1+displaystyle{1over2}}$$ is the third convergent.
$endgroup$
– Aretino
Jan 22 at 22:45
$begingroup$
$${5over3}=1+{1over1+displaystyle{1over2}}$$ is the third convergent.
$endgroup$
– Aretino
Jan 22 at 22:45
$begingroup$
So if I'm understanding this right, if I were to change the lengths to ${39over17}$ the convergents would be [2;3,2,5]. with the second convergent being ${7over3}$ and third being ${16over7}$. Which will find the lengths of the segments 39 and 17?
$endgroup$
– Blue
Jan 22 at 22:58
$begingroup$
So if I'm understanding this right, if I were to change the lengths to ${39over17}$ the convergents would be [2;3,2,5]. with the second convergent being ${7over3}$ and third being ${16over7}$. Which will find the lengths of the segments 39 and 17?
$endgroup$
– Blue
Jan 22 at 22:58
1
1
$begingroup$
With $7/3$ you find $39/7=5.57$, $17/3=5.67$. With $16/7$ you have $39/16=2.44$, $17/7=2.43$. Higher convergents give better accuracy but imply a larger number of pieces.
$endgroup$
– Aretino
Jan 22 at 23:06
$begingroup$
With $7/3$ you find $39/7=5.57$, $17/3=5.67$. With $16/7$ you have $39/16=2.44$, $17/7=2.43$. Higher convergents give better accuracy but imply a larger number of pieces.
$endgroup$
– Aretino
Jan 22 at 23:06
add a comment |
$begingroup$
Leat $a$ and $b$ be the sides in question. Their ratio $frac{a}{b}$ is either rational or irrational number. If it is rational, let say $frac{p}{q}$, the "unit" length will be
$$
u=frac{a}{p}=frac{b}{q}.
$$
If the ratio is irrational you can take any rational approximation for it, e.g.
$$
sqrt2simfrac{141}{100}
$$
and proceed as before.
answered Jan 22 at 21:20
useruser
5,23811030
$endgroup$
add a comment |
$begingroup$
Leat $a$ and $b$ be the sides in question. Their ratio $frac{a}{b}$ is either rational or irrational number. If it is rational, let say $frac{p}{q}$, the "unit" length will be
$$
u=frac{a}{p}=frac{b}{q}.
$$
If the ratio is irrational you can take any rational approximation for it, e.g.
$$
sqrt2simfrac{141}{100}
$$
and proceed as before.
answered Jan 22 at 21:20
useruser
5,23811030
$endgroup$
add a comment |
$begingroup$
Leat $a$ and $b$ be the sides in question. Their ratio $frac{a}{b}$ is either rational or irrational number. If it is rational, let say $frac{p}{q}$, the "unit" length will be
$$
u=frac{a}{p}=frac{b}{q}.
$$
If the ratio is irrational you can take any rational approximation for it, e.g.
$$
sqrt2simfrac{141}{100}
$$
and proceed as before.
answered Jan 22 at 21:20
useruser
5,23811030
$endgroup$
Leat $a$ and $b$ be the sides in question. Their ratio $frac{a}{b}$ is either rational or irrational number. If it is rational, let say $frac{p}{q}$, the "unit" length will be
$$
u=frac{a}{p}=frac{b}{q}.
$$
If the ratio is irrational you can take any rational approximation for it, e.g.
$$
sqrt2simfrac{141}{100}
$$
and proceed as before.
answered Jan 22 at 21:20
useruser
5,23811030
answered Jan 22 at 21:20
useruser
5,23811030
answered Jan 22 at 21:20
useruser
5,23811030
answered Jan 22 at 21:20
useruser
5,23811030
5,23811030
add a comment |
add a comment |
$begingroup$
From your examples, it looks like you're dealing with rational lengths for the paper, $l$ and $w$. It even looks like you're using a finite decimal expansion (the decimal doesn't go on forever).
From this, I suggest that you multiply $l$ and $w$ by powers of $10$ until you have two whole numbers, I'll call them $L$ and $W$. Let $G$ be the greatest common divisor of $L$ and $W$.
Let $g$ be $G$ divided by the power of $10$ that you multiplied in the previous step. This is the length of the side of a square. Now, $l/g$ and $w/g$ will tell you the number of squares along an edge of the paper.
If $l$ and $w$ are rational, but aren't decimals with finite expansions, multiply them by the least common multiple of their denominators, and mimic the steps above.
If all your side lengths are rational, then you will never have an irrational side length for your squares. If your side lengths are irrational, then the only way to get squares without anything left over is that the ratio of the side lengths is rational.
answered Jan 22 at 21:28
Michael BurrMichael Burr
27k23262
$endgroup$
add a comment |
$begingroup$
From your examples, it looks like you're dealing with rational lengths for the paper, $l$ and $w$. It even looks like you're using a finite decimal expansion (the decimal doesn't go on forever).
From this, I suggest that you multiply $l$ and $w$ by powers of $10$ until you have two whole numbers, I'll call them $L$ and $W$. Let $G$ be the greatest common divisor of $L$ and $W$.
Let $g$ be $G$ divided by the power of $10$ that you multiplied in the previous step. This is the length of the side of a square. Now, $l/g$ and $w/g$ will tell you the number of squares along an edge of the paper.
If $l$ and $w$ are rational, but aren't decimals with finite expansions, multiply them by the least common multiple of their denominators, and mimic the steps above.
If all your side lengths are rational, then you will never have an irrational side length for your squares. If your side lengths are irrational, then the only way to get squares without anything left over is that the ratio of the side lengths is rational.
answered Jan 22 at 21:28
Michael BurrMichael Burr
27k23262
$endgroup$
add a comment |
$begingroup$
From your examples, it looks like you're dealing with rational lengths for the paper, $l$ and $w$. It even looks like you're using a finite decimal expansion (the decimal doesn't go on forever).
From this, I suggest that you multiply $l$ and $w$ by powers of $10$ until you have two whole numbers, I'll call them $L$ and $W$. Let $G$ be the greatest common divisor of $L$ and $W$.
Let $g$ be $G$ divided by the power of $10$ that you multiplied in the previous step. This is the length of the side of a square. Now, $l/g$ and $w/g$ will tell you the number of squares along an edge of the paper.
If $l$ and $w$ are rational, but aren't decimals with finite expansions, multiply them by the least common multiple of their denominators, and mimic the steps above.
If all your side lengths are rational, then you will never have an irrational side length for your squares. If your side lengths are irrational, then the only way to get squares without anything left over is that the ratio of the side lengths is rational.
answered Jan 22 at 21:28
Michael BurrMichael Burr
27k23262
$endgroup$
From your examples, it looks like you're dealing with rational lengths for the paper, $l$ and $w$. It even looks like you're using a finite decimal expansion (the decimal doesn't go on forever).
From this, I suggest that you multiply $l$ and $w$ by powers of $10$ until you have two whole numbers, I'll call them $L$ and $W$. Let $G$ be the greatest common divisor of $L$ and $W$.
Let $g$ be $G$ divided by the power of $10$ that you multiplied in the previous step. This is the length of the side of a square. Now, $l/g$ and $w/g$ will tell you the number of squares along an edge of the paper.
If $l$ and $w$ are rational, but aren't decimals with finite expansions, multiply them by the least common multiple of their denominators, and mimic the steps above.
If all your side lengths are rational, then you will never have an irrational side length for your squares. If your side lengths are irrational, then the only way to get squares without anything left over is that the ratio of the side lengths is rational.
answered Jan 22 at 21:28
Michael BurrMichael Burr
27k23262
answered Jan 22 at 21:28
Michael BurrMichael Burr
27k23262
answered Jan 22 at 21:28
Michael BurrMichael Burr
27k23262
answered Jan 22 at 21:28
Michael BurrMichael Burr
27k23262
27k23262
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