1/5 is one fifth of the length of a line segment of one unit – but how?

This comes from the Common Core

Develop understanding of fractions as numbers.
1. ……
2. Understand a fraction as a number on the number line; represent
fractions on a number line diagram.
a. Represent a fraction 1/b on a number line diagram by defining the
interval from 0 to 1 as the whole and partitioning it into b equal
parts. Recognize that each part has size 1/b and that the endpoint
of the part based at 0 locates the number 1/b on the number line.
b. Represent a fraction a/b on a number line diagram by marking off
a lengths 1/b from 0. …….

….but how do you do it ?????

The mystery is solved…….

Here is the line, with 0 and 1 marked. /You chose it already !

Here is a numbered line, any size, equally spaced, at intervals of one unit.

It only has to start from zero.

Now construct the line from point 5 to the “fraction” line at point 1, and a parallel line from point 1 on the numbered line.

The point of intersection of the parallel line and the “fraction” line is then 1/5 of the distance from 0 to 1 on the “fraction” line.

1/5, 2/5, 3/5, 4/5 and 1 are equally spaced on the fraction line.

L cannot be moved in the static picture.

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A minus times a minus is a plus -Are you sure you know why?

What exactly are negative numbers?
A reference , from Wikipedia:
In A.D. 1759, Francis Maseres, an English mathematician, wrote that negative numbers “darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple”.
He came to the conclusion that negative numbers were nonsensical.[25]

A minus times a minus is a plus
Two minuses make a plus
Dividing by a negative, especially a negative fraction !!!!
(10 – 2) x (7 – 3) = 10 x 7 – 2 x 7 + 10 x -3 + 2 x 3, really? How do we know?
Or we use “the area model”, or some hand waving with the number line.

It’s time for some clear thinking about this stuff.

Mathematically speaking, the only place that requires troublesome calculations with negative numbers is in algebra, either in evaluation or in rearrangement, but what about the real world ?
Where in the real world does one encounter negative x negative ?
I found two situations, in electricity and in mechanics:

1: “volts x amps = watts”, as it it popularly remembered really means “voltage drop x current flowing = power”
It is sensible to choose a measurement system (scale) for each of these so that a current flowing from a higher to a lower potential point is treated as positive, as is the voltage drop.

Part of simple circuit A———–[resistors etc in here]————–B
Choosing point A, at potential a, as the reference, and point B, at potential b, as the “other” point, then the potential drop from A to B is a – b
If b<a then a current flows from A to B, and its value is positive, just as a – b is positive
If b>a then a current flows from B to A, and its value is negative, just as a – b is negative

In each case the formula for power, voltage drop x current flowing = power, must yield an unsigned number, as negative power is a nonsense. Power is an “amount”.
So when dealing with reality minus times minus is plus (in this case nosign at all).

The mechanics example is about the formula “force times distance = work done”
You can fill in the details.

Now let’s do multiplication on the number line, or to be more precise, two number lines:
Draw two number lines, different directions, starting together at the zero. The scales do not have to be the same.
To multiply 2 by three (3 times 2):
1: Draw a line from the 1 on line A to the 2 on line B
2: Draw a line from the 3 on line A parallel to the first line.
3: It meets line B at the point 6
4: Done: 3 times 2 is 6

Number line A holds the multipliers, number line B holds the numbers being multiplied.

To multiply a negative number by a positive number we need a pair of signed number lines, crossing at their zero points.

So to multiply -2 by 3 (3 times -2) we do the same as above, but the number being multiplied is now -2, so 1 on line A is joined to -2 on line B

The diagram below is for -2 times 3. Wow, it ends in the same place.

Finally, and you can see where this is going, we do -2 times -3.

Join the 1 on line A to the -3 on line B, and then the parallel to this line passing through the -2 on line A:

and as hoped for, this line passes through the point 6 on the number line B.

Does this “prove” the general case? Only in the proverbial sense. The reason is that we do not have a proper definition of signed numbers. (There is one).

Incidentally, the numbering on the scales above is very poor. The positive numbers are NOT NOT NOT the same things as the unsigned numbers 1, 1.986, 234.5 etc

Each of them should have a + in front, but mathematicians are Lazy. More on this another day.

Problem for you: Show that (a-b)(c-d) = ac – bc – ad + bd without using anything to do with “negative numbers”

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References.
Wikipedia:
Reference direction for current
Since the current in a wire or component can flow in either direction, when a variable I is defined to represent
that current, the direction representing positive current must be specified, usually by an arrow on the circuit
schematic diagram. This is called the reference direction of current I. If the current flows in the opposite
direction, the variable I has a negative value.

Yahoo Answers: Reference direction for potential difference
Best Answer: Potential difference can be negative. It depends on which direction you measure the voltage – e.g.
which way round you connect a voltmeter. (if this is the best answer, I hate to think of what the worst answer is)
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Linear transformations, geometrically

 

Following a recent blog post relating a transformation of points on a line to points on another line to the graph of the equation relating the input and output I thought it would be interesting to explore the linear and affine mappings of a plane to itself from a geometrical construction perspective.

It was ! (To me anyway)

These linear mappings  (rigid and not so rigid motions) are usually  approached in descriptive and manipulative  ways, but always very specifically. I wanted to go directly from the transformation as equations directly to the transformation as geometry.

Taking an example, (x,y) maps to (X,Y) with the linear equations

X = x + y + 1 and Y = -0.5x +y

it was necessary to construct a point on the x axis with the value of X, and likewise a point on the y axis with the value of Y. The transformed (x,y) is then the point (X,Y) on the plane.

The construction below shows the points and lines needed to establish the point(X,0), which is G in the picture, starting with the point D as the (x,y)

 

The corresponding construction was done for Y, and the resulting point (X,Y) is point J. Point D was then forced to lie on a line, the sloping blue line, and as it is moved along the line the transformed point J moves on another line

Now the (x,y) point (B in the picture below, don’t ask why!) is forced to move on the blue circle. What does the transformed point do? It moves on an ellipse, whose size and orientation are determined by the actual transformation. At this point matrix methods become very handy.(though the 2D matrix methods cannot deal with translations)

All this was constructed with my geometrical construction program (APP if you like) called GEOSTRUCT and available as a free web based application from

http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

The program produces a listing of all the actions requested, and these are listed below for this application:

Line bb moved to pass through Point A
New line cc created, through points B and C
New Point D
New line dd created, through Point D, at right angles to Line aa
New line ee created, through Point D, at right angles to Line bb
New line ff created, through Point D, parallel to Line cc
New point E created as the intersection of Line ff and Line aa
New line gg created, through Point E, at right angles to Line aa
New line hh created, through Point B, at right angles to Line bb
New point F created as the intersection of Line hh and Line gg
New line ii created, through Point F, parallel to Line cc
New point G created as the intersection of Line ii and Line aa

G is the X coordinate, from X = x + y + 1 (added by me)

New line jj created, through Point G, at right angles to Line aa
New line kk created, through Point D, at right angles to Line cc
New point H created as the intersection of Line kk and Line bb
New point I created, as midpoint of points H and B
New line ll created, through Point I, at right angles to Line bb
New point J created as the intersection of Line ll and Line jj

J is the Y coordinate, from Y = -x/2 + y  (added by me)
and K is the transformed point (X,Y) Point J chosen as the tracking point (added by me)

New Line mm
Point D moved and placed on Line mm

 

Real problems with conic sections (ellipse, parabola) part two

So suppose we have a parabolic curve and we want to find out stuff about it.

Its equation … Oh, we have no axes.

Its focus … That would be nice, but it is a bit out of reach.

Its axis, in fact its axis of symmetry … Fold it in half? But how?

Try the method of part one, with the ellipse. (previous post)

This looks promising. I even get another axis, for my coordinate system, if I really want the equation.

Now, analysis of the standard equation for a parabola (see later) says that a line at 45 deg to the axis, as shown, cuts the parabola at a point four focal lengths from the axis. In the picture, marked on the “vertical”axis, this is the length DH

So I need a point one quarter of the way from D to H. Easy !

and then the circle center D, with radius DH/4 cuts the axis of the parabola at the focal point (the focus).

Even better, we get the directrix as well …

and now for the mathy bit (well, you do the algebra, I did the picture)

Yes, I know that this one points up and the previous one pointed to the right !

All diagrams were created with my geometrical construction program, GEOSTRUCT

You will find it here:

www.mathcomesalive.com/geostruct/geostructforbrowser1.html

GEOSTRUCT, a program for investigative geometry

I have been developing this computer software / program / application for some years now, and it is now accessible as a web page, to run in your browser.

It provides basic geometric construction facilities, with lines, points and circles, from which endless possibilities follow.

Just try it out, it’s free.

Click on this or copy and paste for later : www.mathcomesalive.com/geostruct/geostructforbrowser1.html

.Here are some of the basic features, and examples of more advanced constructions, almost all based on straightedge and compass, from “make line pass through a point” to “intersection of two circles”, and dynamic constructions with rolling and rotating circles.

Two lines, with points placed on them

Three random lines with two points of intersection generated

Five free points, three generated circles and a center point

Three free points, connected as point pairs, medians generated

Two free circles and three free points, point pairs and centers generated

GIF showing points of intersection of a line with a circle

Construction for locus of hypocycloid


GIF showing a dilation (stretch) in the horizontal direction

Piston and flywheel

Construction for circle touching two circles

Construction for the locus of a parabola, focus-directrix definition.

 

 

 

 

 

Maths(UK) is full of surprises

The radical axis theorem

How come that after 55 or so years of involment with math this came as a big surprise !!!!!

Duality, fundamental and profound, but here’s a starter for you.

Duality, how things are connected in unexpected ways. The simplest case is that of the five regular Platonic solids, the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron. They all look rather different, BUT…..

take any one of them and find the mid point of each of the faces, join these points up, and you get one of the five regular Platonic solids. Do it to this new one and you get back to the original one. Calling the operation “Doit” we get

tetrahedron –Doit–> tetrahedron –Doit–> tetrahedron
cube –Doit–> octahedron –Doit–> cube
dodecahedron –Doit–> icosahedron –Doit–> dodecahedron

The sizes may change, but we are only interested in the shapes.

This is called a Duality relationship, in which the tetrahedron is the dual of itself, the cube and octahedron are duals of each other, and the dodecahedron and icosahedron are also duals of each other.

Now we will look at lines and points in the x-y plane.

3x – 2y = 4 and y = (3/2)x + 2 and 3x – 2y – 4 = 0 are different ways of describing the same line, but there are many more. We can multiply every coefficient, including the constant, by any number not 0 and the result describes the same line, for example 6x – 4y = 8, or 0.75x – 0.5y = 1, or -0.75x + 0.5y + 1 = 0

This means that a line can be described entirely by two numbers, the x and the y coefficients found when the line equation is written in the last of the forms given above. Generally this is ax + by + 1 = 0

Now any point in the plane needs two numbers to specify it, the x and the y coordinates, for example (2,3)

So if a line needs two numbers and a point needs two numbers then given two numbers p and q I can choose to use then to describe a point or a line. So the numbers p and q can be the point (p,q) or the line px + qy + 1 = 0

The word “dual” is used in this situation. The point (p,q) is the dual of the line px + qy + 1 = 0, and vice versa.

The line joining the points C and D is dual to the point K, in red.  The line equation is 2x + y = 3, and we rewrite it in the “standard” form as  -0.67x – 0.33y +1 = 0  so we get  (-0.67, -0.33) for the coordinates of the dual point K.

A quick calculation (using the well known formula) shows that the distance of the line from the origin multiplied by the distance of the point from the origin is a constant (in this case 1).

The second picture shows the construction of the dual point.

What happens as we move the line about ? Parallel to itself, the dual point moves out and in.

More interesting is what happens when we rotate the line around a fixed point on the line:

The line passes through the fixed point C.  The dual point traces out a straight line, shown in green.

This can be interpreted as “A point can be seen as a set of concurrent lines”, just as a line can be seen as a set of collinear points (we have fewer problems with the latter).

It gets more interesting when we consider a curve. There are two ways of looking at a curve, one as a (fairly nicely) organized set of points ( a locus), and the other as a set of (fairly nicely) arranged lines (an envelope).

A circle is a set of points equidistant from a central point, but it is also the envelope of a set of lines equidistant from a central point (the tangent lines).

So what happens when we look for the dual of a circle? We can either find the line dual to each point on the circle, or find the point dual to each tangent line to the circle. Here’s both:

In this case the circle being dualled is the one with center C, and the result is a hyperbola, shown in green.  The result can be deduced analytically, but it is a pain to do so.

The hyperbola again.  It doesn’t look quite perfect, probably due to rounding errors.

The question remains – If I do the dualling operation on the hyperbola, will I get back to the circle ?

Also, why a hyperbola and not an ellipse ? Looking at what is going on suggests that if the circle to be dualled has the origin inside then we will get an ellipse. This argument can be made more believable with a little care !

If you get this far and want more, try this very heavy article:

http://en.wikipedia.org/wiki/Duality_(mathematics)

What is Algebra really for ?

An example tells a good tale.

Translation of a line in an x-y coordinate system:

Take a line  y = 2x -3, and translate it by 4 up and 5 to the right.

Simple approach : The point P = (2, 1) is on the line (so are some others!). Let us translate the point to get Q = (2+5, 1+4), which is Q = (7, 5), and find the line through Q parallel to the original line.  The only thing that changes is the c value, so the new equation is  y = 2x + c, and it must pass through Q.  So we require  5 = 14 + c, giving the value of c as -9.

Not much algebra there, but a horrible question remains – “What happened to all the other points on the line ?”

We try a more algebraic approach – with any old line  ax + by + c = 0, and any old translation, q up and p to the right.

First thing is to find a point on the line – “What ? We don’t know ANYTHING about the line.”

This is where algebra comes to the rescue. Let us suppose (state) that a point P = (d, e) IS on the line.

Then ad + be + c = 0

Now we can move the point P to Q = (d + p, e + q)  (as with the numbers earlier), and make the new line pass through this point:  This requires a new constant c (call it newc) and we then have  a(d + p) + b(e + q) + ‘newc’ = 0

Expand the parentheses (UK brackets, and it’s shorter) to get  ad + ap + be + bq + ‘newc’ = 0

Some inspired rearranging gives  ‘newc’ = -ap – bq – ad – be, which is equal to -(ap + bq) – (ad + be)

“Why did you do that last step ?” – “Because I looked back a few lines and figured that  (ad + be) = -c, which not only simplifies the expression, it also disposes of the unspecified point  P.

End result is:  Translated line equation is  ax + by + ‘newc’ =0,  that is,  ax + by + c – (ap + bq) = 0

and the job is done for ALL lines, even the vertical ones, and ALL translations. Also we can be sure that we know what has happened to ALL the points on the line.

I am not going to check this with the numerical example, you are !

Angle between two lines in the plane……Vector product in 3D…….connections???????

So I was in the middle of converting my geometry application Geostruct (we used to call them programs) into javascript

 get it here with the introduction .doc file here

when I decided that the “angle between two lines” routine needed a rewrite. Some surprises ensued !

Two lines,  ax + by + c  = 0  and  px + qy + r = 0

Their slopes (gradients) are  -a/b = tan(θ)  and  -p/q = tan(φ)

The angle between the lines is  φ – θ,

so it would be nice to know something about  tan(φ – θ)

Back to basics, where  tan(φ – θ) = sin(φ– θ)/cos(φ– θ),

and we have the two expansions

sin(φ– θ) = sin(φ)cos(θ) – cos(φ)sin(θ)   and

cos(φ– θ) = cos(φ)cos(θ) + sin(φ)sin(θ)

So we have  tan(φ – θ) = (sin(φ)cos(θ) – cos(φ)sin(θ))/( cos(φ)cos(θ) + sin(φ)sin(θ))

Dividing top and bottom by  cos(φ)sin(θ)  and skipping some tedious algebra we get

tan(φ – θ)   =  (tan(φ) – tan(θ))/(1 +  tan(φ)tan(θ))

This is where the books stop, which turns out to be a real shame !

Going back to the two lines and their equations, the two lines

ax + by = 0  and  px + qy = 0

have the same angle between them (some things are toooo obvious)

Things are simpler if we look at these two lines through the origin when they both have positive slope.

Take b and q as positive and write the equations as   ax – by = 0  and  px – qy = 0

Then the point whose coordinates are (b,a) lies on the first and (q,p) lies on the second.

Also, the slopes of the two lines are now  a/b , tan(θ)   and  p/q , tan(φ)

Let us put these into the  tan(φ – θ)   equation above, and once more after tedious algebra

tan(φ – θ)  = (bp – aq)/(ap + bq)

which is a very nice formula for the tan of the angle between two lines.

This is ok if we are interested just in “the angle between the lines”,  but if we are considering rotations, and one of the lines is the “first” one, then the tangent is inadequate. We need both the sine and the cosine of the angle to establish size AND direction (clockwise or anticlockwise).

The formula above can be seen as showing  cos(φ– θ)  as  (ap + bq) divided by something

and  sin(φ– θ)  as  (bp – aq) divided by the same something.

Calling the something  M  it is fairly clear that    (ap + bq)² + (bp – aq) ² = M²

and more tedious algebra and some “observation and making use of structure” gives

M²  = (a² + b²)(p² + q²)

and we now have

sin(φ– θ)  = (bp – aq)/M  and  cos(φ– θ) = (bq + ap)/M

and M is the product of the lengths of the two line segments, from the origin to (b,a) and from the origin to (q,p)

It was at this point that I saw M times the sine of the “angle between” as twice the well known formula for the area of a triangle. “half a b sin(C)”, or, if you prefer, the area of the parallelogram defined by the two line segments.

Suddenly I saw all this in 2D vector terms, with bq + ap being the dot product of (b,a) and (q,p) , and bp – aq as being part of the definition of the 3D vector or cross product, in fact the only non zero component (and in the z direction), since in 3D terms our two vectors lie in the xy plane.

Why is the “vector product” not considered in the 2D case ??? It is simpler, and looking at the formula for sine , above, we have a 2D interpretation of the “vector”or cross product as twice the area of the triangle formed by (b,a) and (q,p). (just as in the standard 3D definition, but treated as a scalar).

So “bang goes” the common terms, scalar product for c . d  and vector product for  c X d

Dot product and cross product are much better anyway, and a bit of ingenuity will lead you to the reason for the word “cross”.

This is one of the things implemented using this approach:

Anyway, the end result of all this, for rotating points on a circle, was a calculation process which did not require the actual calculation of any angle. No arctan( ) !

Rigid motions are actually useful !

I am currently reconstructing my geometrical construction application, Geostruct, to run in a web page using javascript.

One of the actions is to find the points of intersection of a straight line with a circle. Here is a gif showing the result:

The algebra needed to solve the two simultaneous equations is straightforward, but a pain in the butt to get right and code up, so I thought “Why not solve the equations for the very simple case of the circle centered at the origin and the line vertical, at the same distance (a) from the centre of the circle

Then it is a simple matter of  rotating the two points (a,b) and (a,-b) about the origin, through the angle made by the original line to the vertical, and then translating the circle back to its original position, the translated points are then the desired points of intersection.

The same routine can be used for the intersection of two circles, with a little bit of prior calculation.