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Ways of geometrical multiplication
Construct circle tangent to two lineshyperbolic geometry (and circle ) construction problemConstruct a circle with straight edge and compass with some given conditions.Finding tangents to a circle with a straightedgeInscribing square in circle in just seven compass-and-straightedge stepsfinding the center of a circle (elementary geometry)Algebra - Construct circle with radius AB around distinct point CConstruct a perpendicular to a given line from a given (external) point, using a compass only onceDoes this mean that one can construct the cube root of two in three dimensions?Approximation of the quadratic formula with straightedge and compass
$begingroup$
There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).
Let $A,B$ be two integer points on the line $O1$:

Method 1
- Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).

Draw a circle with radius $|OB|$ around $B$.
Let $C$ be the (other) intersection point of this circle with the line $O1$.

Draw a circle with radius $|OB|$ around $C$.
Do this $a-1$ times.
The last intersection point $C$ is the product $A times B$.

Method 2
- Construct a rectangle with side lengths $|OA|$, $|OB|$.

- Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).

Draw a circle with radius $|O1|$ around $0$.
Let $C$ be the intersection point of this circle with the line $O1$.

Draw a circle with radius $|O1|$ around $C$.
Do this $c$ times.
The last intersection point $C$ is the product $A times B$.

Method 3
Construct the line perpendicular to $O1$ through $O$.
Construct the points $1'$ and $B'$.

Draw the line $1'A$.
Construct the parallel to $1'A$ through $B'$.
The intersection point of this parallel with the line $O1$ is the product $A times B$.

Method 4
Construct the perpendicular line to $O1$ through $O$.
Construct the point $1'$.
Construct the circle through $1'$, $A$ and $B$.
The intersection point of this circle with the line $O1'$ is the product $A times B$.

Method 5
This method makes use of the parabola, i.e. goes beyond compass-ruler constructions.
Construct the unit parabola $(x,y)$ with $y = x^2$.
Construct $B'$.
Construct the line perpendicular to $O1$ through $A$.
Construct the line perpendicular to $O1$ through $B'$.
Draw the line through the intersection points of these two lines with the parabola.
The intersection point of this line with the line $O1'$ is the product $A times B$.

For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point).
Note that the different methods take different amounts $sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):
Method 1: $sigma sim ab^2$
Method 2: $sigma sim ab$
Method 3: $sigma sim ab^2$
Method 4: $sigma sim a^2b^2$
Method 5: $sigma sim a^3b$
This is space complexity. Compare this to time complexity, i.e. the number $tau$ of essential construction steps that are needed:
Method 1: $tau sim a$
Method 2: $tau sim ab$
Method 3: $tau sim 1$
Method 4: $tau sim 1$
Method 5: $tau sim 1$
From this point of view method 3 would be the most efficient.
Once again:
I'm looking for other geometrical methods to multiply two numbers
given as points on the number line $O1$ (is there one using the
hyperbola?) and trying to understand better the "deeper" reasons why
they all yield the same result (i.e. point).
Those answers I managed to visualize I will add here:
Method 6 (due to Cia Pan)

Method 7 (due to celtschk)

euclidean-geometry arithmetic conic-sections big-list geometric-construction
$endgroup$
add a comment |
$begingroup$
There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).
Let $A,B$ be two integer points on the line $O1$:

Method 1
- Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).

Draw a circle with radius $|OB|$ around $B$.
Let $C$ be the (other) intersection point of this circle with the line $O1$.

Draw a circle with radius $|OB|$ around $C$.
Do this $a-1$ times.
The last intersection point $C$ is the product $A times B$.

Method 2
- Construct a rectangle with side lengths $|OA|$, $|OB|$.

- Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).

Draw a circle with radius $|O1|$ around $0$.
Let $C$ be the intersection point of this circle with the line $O1$.

Draw a circle with radius $|O1|$ around $C$.
Do this $c$ times.
The last intersection point $C$ is the product $A times B$.

Method 3
Construct the line perpendicular to $O1$ through $O$.
Construct the points $1'$ and $B'$.

Draw the line $1'A$.
Construct the parallel to $1'A$ through $B'$.
The intersection point of this parallel with the line $O1$ is the product $A times B$.

Method 4
Construct the perpendicular line to $O1$ through $O$.
Construct the point $1'$.
Construct the circle through $1'$, $A$ and $B$.
The intersection point of this circle with the line $O1'$ is the product $A times B$.

Method 5
This method makes use of the parabola, i.e. goes beyond compass-ruler constructions.
Construct the unit parabola $(x,y)$ with $y = x^2$.
Construct $B'$.
Construct the line perpendicular to $O1$ through $A$.
Construct the line perpendicular to $O1$ through $B'$.
Draw the line through the intersection points of these two lines with the parabola.
The intersection point of this line with the line $O1'$ is the product $A times B$.

For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point).
Note that the different methods take different amounts $sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):
Method 1: $sigma sim ab^2$
Method 2: $sigma sim ab$
Method 3: $sigma sim ab^2$
Method 4: $sigma sim a^2b^2$
Method 5: $sigma sim a^3b$
This is space complexity. Compare this to time complexity, i.e. the number $tau$ of essential construction steps that are needed:
Method 1: $tau sim a$
Method 2: $tau sim ab$
Method 3: $tau sim 1$
Method 4: $tau sim 1$
Method 5: $tau sim 1$
From this point of view method 3 would be the most efficient.
Once again:
I'm looking for other geometrical methods to multiply two numbers
given as points on the number line $O1$ (is there one using the
hyperbola?) and trying to understand better the "deeper" reasons why
they all yield the same result (i.e. point).
Those answers I managed to visualize I will add here:
Method 6 (due to Cia Pan)

Method 7 (due to celtschk)

euclidean-geometry arithmetic conic-sections big-list geometric-construction
$endgroup$
$begingroup$
IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
$endgroup$
– CiaPan
2 hours ago
$begingroup$
@CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
$endgroup$
– Hans-Peter Stricker
2 hours ago
$begingroup$
@CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
$endgroup$
– Hans-Peter Stricker
1 hour ago
add a comment |
$begingroup$
There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).
Let $A,B$ be two integer points on the line $O1$:

Method 1
- Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).

Draw a circle with radius $|OB|$ around $B$.
Let $C$ be the (other) intersection point of this circle with the line $O1$.

Draw a circle with radius $|OB|$ around $C$.
Do this $a-1$ times.
The last intersection point $C$ is the product $A times B$.

Method 2
- Construct a rectangle with side lengths $|OA|$, $|OB|$.

- Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).

Draw a circle with radius $|O1|$ around $0$.
Let $C$ be the intersection point of this circle with the line $O1$.

Draw a circle with radius $|O1|$ around $C$.
Do this $c$ times.
The last intersection point $C$ is the product $A times B$.

Method 3
Construct the line perpendicular to $O1$ through $O$.
Construct the points $1'$ and $B'$.

Draw the line $1'A$.
Construct the parallel to $1'A$ through $B'$.
The intersection point of this parallel with the line $O1$ is the product $A times B$.

Method 4
Construct the perpendicular line to $O1$ through $O$.
Construct the point $1'$.
Construct the circle through $1'$, $A$ and $B$.
The intersection point of this circle with the line $O1'$ is the product $A times B$.

Method 5
This method makes use of the parabola, i.e. goes beyond compass-ruler constructions.
Construct the unit parabola $(x,y)$ with $y = x^2$.
Construct $B'$.
Construct the line perpendicular to $O1$ through $A$.
Construct the line perpendicular to $O1$ through $B'$.
Draw the line through the intersection points of these two lines with the parabola.
The intersection point of this line with the line $O1'$ is the product $A times B$.

For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point).
Note that the different methods take different amounts $sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):
Method 1: $sigma sim ab^2$
Method 2: $sigma sim ab$
Method 3: $sigma sim ab^2$
Method 4: $sigma sim a^2b^2$
Method 5: $sigma sim a^3b$
This is space complexity. Compare this to time complexity, i.e. the number $tau$ of essential construction steps that are needed:
Method 1: $tau sim a$
Method 2: $tau sim ab$
Method 3: $tau sim 1$
Method 4: $tau sim 1$
Method 5: $tau sim 1$
From this point of view method 3 would be the most efficient.
Once again:
I'm looking for other geometrical methods to multiply two numbers
given as points on the number line $O1$ (is there one using the
hyperbola?) and trying to understand better the "deeper" reasons why
they all yield the same result (i.e. point).
Those answers I managed to visualize I will add here:
Method 6 (due to Cia Pan)

Method 7 (due to celtschk)

euclidean-geometry arithmetic conic-sections big-list geometric-construction
$endgroup$
There are at least five ways to multiply two natural numbers $a$ and $b$ given as integer points $A$ and $B$ on the number line by geometrical means. Two of them include counting, the others are purely geometric. I wonder (i) if there are other ways and (ii) how to deeply understand the interrelationship between the different methods (i.e. recipes).
Let $A,B$ be two integer points on the line $O1$:

Method 1
- Count how often the unit length $|O1|$ fits into $|OA|$. Let this number be $a$ (here $a = 3$).

Draw a circle with radius $|OB|$ around $B$.
Let $C$ be the (other) intersection point of this circle with the line $O1$.

Draw a circle with radius $|OB|$ around $C$.
Do this $a-1$ times.
The last intersection point $C$ is the product $A times B$.

Method 2
- Construct a rectangle with side lengths $|OA|$, $|OB|$.

- Count how often the unit square (with side length $|O1|$) fits into the rectangle. Let this number be $c$ (here $c=6$).

Draw a circle with radius $|O1|$ around $0$.
Let $C$ be the intersection point of this circle with the line $O1$.

Draw a circle with radius $|O1|$ around $C$.
Do this $c$ times.
The last intersection point $C$ is the product $A times B$.

Method 3
Construct the line perpendicular to $O1$ through $O$.
Construct the points $1'$ and $B'$.

Draw the line $1'A$.
Construct the parallel to $1'A$ through $B'$.
The intersection point of this parallel with the line $O1$ is the product $A times B$.

Method 4
Construct the perpendicular line to $O1$ through $O$.
Construct the point $1'$.
Construct the circle through $1'$, $A$ and $B$.
The intersection point of this circle with the line $O1'$ is the product $A times B$.

Method 5
This method makes use of the parabola, i.e. goes beyond compass-ruler constructions.
Construct the unit parabola $(x,y)$ with $y = x^2$.
Construct $B'$.
Construct the line perpendicular to $O1$ through $A$.
Construct the line perpendicular to $O1$ through $B'$.
Draw the line through the intersection points of these two lines with the parabola.
The intersection point of this line with the line $O1'$ is the product $A times B$.

For me it's something like a miracle that these five methods – seemingly very different (as recipes) and not obviously equivalent – yield the very same result (i.e. point).
Note that the different methods take different amounts $sigma$ of Euclidean space (to completely show all intermediate points and (semi-)circles involved, assuming that $a >b$):
Method 1: $sigma sim ab^2$
Method 2: $sigma sim ab$
Method 3: $sigma sim ab^2$
Method 4: $sigma sim a^2b^2$
Method 5: $sigma sim a^3b$
This is space complexity. Compare this to time complexity, i.e. the number $tau$ of essential construction steps that are needed:
Method 1: $tau sim a$
Method 2: $tau sim ab$
Method 3: $tau sim 1$
Method 4: $tau sim 1$
Method 5: $tau sim 1$
From this point of view method 3 would be the most efficient.
Once again:
I'm looking for other geometrical methods to multiply two numbers
given as points on the number line $O1$ (is there one using the
hyperbola?) and trying to understand better the "deeper" reasons why
they all yield the same result (i.e. point).
Those answers I managed to visualize I will add here:
Method 6 (due to Cia Pan)

Method 7 (due to celtschk)

euclidean-geometry arithmetic conic-sections big-list geometric-construction
euclidean-geometry arithmetic conic-sections big-list geometric-construction
edited 34 mins ago
Hans-Peter Stricker
asked 2 hours ago
Hans-Peter StrickerHans-Peter Stricker
6,53443994
6,53443994
$begingroup$
IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
$endgroup$
– CiaPan
2 hours ago
$begingroup$
@CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
$endgroup$
– Hans-Peter Stricker
2 hours ago
$begingroup$
@CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
$endgroup$
– Hans-Peter Stricker
1 hour ago
add a comment |
$begingroup$
IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
$endgroup$
– CiaPan
2 hours ago
$begingroup$
@CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
$endgroup$
– Hans-Peter Stricker
2 hours ago
$begingroup$
@CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
$endgroup$
– CiaPan
2 hours ago
$begingroup$
IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
$endgroup$
– CiaPan
2 hours ago
$begingroup$
@CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
$endgroup$
– Hans-Peter Stricker
2 hours ago
$begingroup$
@CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
$endgroup$
– Hans-Peter Stricker
2 hours ago
$begingroup$
@CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
@CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
$endgroup$
– Hans-Peter Stricker
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
- Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.
- Construct the perpendicular at $O$.
- Construct the semicircle on the diameter $A'B$.
- Find $H$ at the intersection of the semicircle and the perpendicular.
$(OH)^2 = OA'cdot OB = OAcdot OB$. - Draw line $1H$ and construct a perpendicular to it through $H$.
- Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$
$endgroup$
$begingroup$
Thanks! I'll try to visualize this (or do you have a diagram at hand)?
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Done (have a look at my edited question).
$endgroup$
– Hans-Peter Stricker
1 hour ago
add a comment |
$begingroup$
The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).
- Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).
- Select on $g$ an arbitrary point $P$ other than the origin.
- Draw a line through $1$ and $P$.
- Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.
- Draw a line through $P$ and $B$.
- Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.
Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).
With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.
$endgroup$
$begingroup$
I have added your construction to my list (see above). Thanks a lot!
$endgroup$
– Hans-Peter Stricker
25 mins ago
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
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active
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votes
$begingroup$
- Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.
- Construct the perpendicular at $O$.
- Construct the semicircle on the diameter $A'B$.
- Find $H$ at the intersection of the semicircle and the perpendicular.
$(OH)^2 = OA'cdot OB = OAcdot OB$. - Draw line $1H$ and construct a perpendicular to it through $H$.
- Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$
$endgroup$
$begingroup$
Thanks! I'll try to visualize this (or do you have a diagram at hand)?
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Done (have a look at my edited question).
$endgroup$
– Hans-Peter Stricker
1 hour ago
add a comment |
$begingroup$
- Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.
- Construct the perpendicular at $O$.
- Construct the semicircle on the diameter $A'B$.
- Find $H$ at the intersection of the semicircle and the perpendicular.
$(OH)^2 = OA'cdot OB = OAcdot OB$. - Draw line $1H$ and construct a perpendicular to it through $H$.
- Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$
$endgroup$
$begingroup$
Thanks! I'll try to visualize this (or do you have a diagram at hand)?
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Done (have a look at my edited question).
$endgroup$
– Hans-Peter Stricker
1 hour ago
add a comment |
$begingroup$
- Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.
- Construct the perpendicular at $O$.
- Construct the semicircle on the diameter $A'B$.
- Find $H$ at the intersection of the semicircle and the perpendicular.
$(OH)^2 = OA'cdot OB = OAcdot OB$. - Draw line $1H$ and construct a perpendicular to it through $H$.
- Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$
$endgroup$
- Construct the point $A'$ on the given line such that $O$ is a midpoint of the line segment $AA'$.
- Construct the perpendicular at $O$.
- Construct the semicircle on the diameter $A'B$.
- Find $H$ at the intersection of the semicircle and the perpendicular.
$(OH)^2 = OA'cdot OB = OAcdot OB$. - Draw line $1H$ and construct a perpendicular to it through $H$.
- Find point $K$ at the intersection of the last constructed line and the first given line. We have $(OH)^2 = 1cdot OK,$ hence $OK = OAcdot OB.$
answered 1 hour ago
CiaPanCiaPan
10.1k11247
10.1k11247
$begingroup$
Thanks! I'll try to visualize this (or do you have a diagram at hand)?
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Done (have a look at my edited question).
$endgroup$
– Hans-Peter Stricker
1 hour ago
add a comment |
$begingroup$
Thanks! I'll try to visualize this (or do you have a diagram at hand)?
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Done (have a look at my edited question).
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Thanks! I'll try to visualize this (or do you have a diagram at hand)?
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Thanks! I'll try to visualize this (or do you have a diagram at hand)?
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Done (have a look at my edited question).
$endgroup$
– Hans-Peter Stricker
1 hour ago
$begingroup$
Done (have a look at my edited question).
$endgroup$
– Hans-Peter Stricker
1 hour ago
add a comment |
$begingroup$
The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).
- Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).
- Select on $g$ an arbitrary point $P$ other than the origin.
- Draw a line through $1$ and $P$.
- Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.
- Draw a line through $P$ and $B$.
- Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.
Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).
With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.
$endgroup$
$begingroup$
I have added your construction to my list (see above). Thanks a lot!
$endgroup$
– Hans-Peter Stricker
25 mins ago
add a comment |
$begingroup$
The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).
- Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).
- Select on $g$ an arbitrary point $P$ other than the origin.
- Draw a line through $1$ and $P$.
- Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.
- Draw a line through $P$ and $B$.
- Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.
Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).
With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.
$endgroup$
$begingroup$
I have added your construction to my list (see above). Thanks a lot!
$endgroup$
– Hans-Peter Stricker
25 mins ago
add a comment |
$begingroup$
The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).
- Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).
- Select on $g$ an arbitrary point $P$ other than the origin.
- Draw a line through $1$ and $P$.
- Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.
- Draw a line through $P$ and $B$.
- Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.
Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).
With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.
$endgroup$
The following is quite similar to your method 3, but only requires you to draw parallels, not circles (see remark below).
- Draw an arbitrary line $g$ other than the number line through $O$. (The “number line” here is the line through $O$ and $1$).
- Select on $g$ an arbitrary point $P$ other than the origin.
- Draw a line through $1$ and $P$.
- Draw a parallel to that line through $A$. Call the intersection with $g$ $Q$.
- Draw a line through $P$ and $B$.
- Draw a parallel to that line through $Q$. The intersection with the number line is then $Atimes B$.
Remark: In standard geometry (that is, construction with compass and ruler), you of course need to draw circles to construct the parallel. But one might instead consider using no compass, but a “parallels-ruler" (I have no idea what it is actually called; it's basically a ruler that has a built-in roll, allowing you to move the ruler without rotating, and thus to construct parallels).
With only a parallels-ruler you cannot construct circles (so it's strictly weaker than compass and ruler), but as the construction above shows, you can multiply.
edited 1 hour ago
answered 1 hour ago
celtschkceltschk
30.3k755101
30.3k755101
$begingroup$
I have added your construction to my list (see above). Thanks a lot!
$endgroup$
– Hans-Peter Stricker
25 mins ago
add a comment |
$begingroup$
I have added your construction to my list (see above). Thanks a lot!
$endgroup$
– Hans-Peter Stricker
25 mins ago
$begingroup$
I have added your construction to my list (see above). Thanks a lot!
$endgroup$
– Hans-Peter Stricker
25 mins ago
$begingroup$
I have added your construction to my list (see above). Thanks a lot!
$endgroup$
– Hans-Peter Stricker
25 mins ago
add a comment |
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$begingroup$
IMVHO methods 1 and 2 do not count: when you use the word 'count', the method becomes arithmetical instead of geometrical.
$endgroup$
– CiaPan
2 hours ago
$begingroup$
@CiaPan: a) I didn't claim that methods 1 and 2 are purely geometrical. But they are at least partially. b) What else is done in methods 1 and 2 in the "count" steps?
$endgroup$
– Hans-Peter Stricker
2 hours ago
$begingroup$
@CiaPan: This is why I believe that methods 1 and 2 are more geometrical than arithmetical: It's really only counting that is needed, but no "true" arithmetic, i.e. addition or multiplication. You may ask: But how does one really count the number of unit squares (by which geometrical means), doesn't one essentially count $a$ and $b$ and then multiply them? If this must be so, you have won.
$endgroup$
– Hans-Peter Stricker
1 hour ago