Definition:
A linear equation with associated field [math] \mathbb{F} [/math] is an equation of the following form:
[math] a_1 x_1+a_2x_2+... +a_nx_n = b [/math]
The x terms are the variables of the equation, and the other terms are constants. The a terms are called coefficients.
All the constants are elements of the field [math] \mathbb{F} [/math].
The two most common fields are the real numbers [math] \mathbb{R} [/math], and the complex numbers [math] \mathbb{C} [/math].
Consider the case where n=2, and [math] \mathbb{F=R} [/math]. For this case we have:
[math] a_1 x +a_2y = b [/math]
Where [math] a_1,a_2,b \in \mathbb{R} [/math].
This is the equation of a straight line in the XY plane.
Once the variables are instantiated with real numbers, the resulting expression is an example of a mathematical statement, and as with any statement, it is either true or false. Before the variables are instantiated, it's not a statement. After the variables have been instantiated, either the LHS will equal the RHS or not. In the case that the LHS=RHS the statement is true, otherwise the statement is false. The point is, we can use binary logic to analyze linear equations over a field.
Now, a straight line in the XY plane is actually a subset of the entire plane. We can define the XY plane, using the set theoretic notion of the Cartesian product.
Definition: The cartesian product A X B, is the set of all two tuples (x,y) where x is an element of set A, and y is an element of set B. And we can write this symbolically, as follows:
[math] \mathcal{f} (x,y)| x \in A \ \& \ y \in B \mathcal{g} [/math]
So in the case of the XY plane, set A as well as set B are the real number system [math] \mathbb{R} [/math]. So, for the XY plane we have:
[math] XY \ \ plane = \mathcal{f} (x,y)| x \in \mathbb{R} \ \& \ y \in \mathbb{R} \mathcal{g} [/math]
So, the XY plane is the set of all ordered pairs (x,y), where x is an element of the real number system, and y is an element of the real number system.
Now, the equation [math] a_1x + a_2 y = b [/math] is going to be a true statement for some ordered pairs, and a false statement for others. But the point is, that if you plot all the points (x,y) which make the equation a true statement, rather than a false statement, a straight line emerges.
Now, it may occur to you that two points determine a unique straight line. So if you want to check this, you can use graph paper. First plot two points, then draw the straight line through them. In principle, the line you drew is the set of all points for which the equation is true. All other points in the XY plane are locations where the equation is false.
However real space is three dimensional, and we may wonder how to express a formula for an arbitrary straight line in three dimensional Euclidean space.
Before tackling that question, let's first consider what is called a system of simultaneous linear equations.
It may occur to you, that given two arbitrary staight lines in the XY plane, that either those straight lines are parallel, or not. If two straight lines in the XY plane are not parallel, then they must intersect, and conversely if two lines in the XY plane intersect then they aren't parallel.
So now, suppose that you are given two straight lines in the XY plane chosen at random by your professor. Suppose he gave you the following:
[math] L1 \equiv \mathcal {f} (x,y)| y=3x+1 \mathcal {g} [/math]
Thus, straight line L1 is the set of all points (x,y) such that the equation y=3x+1 is true.
Some of you will recognize the equation as the point/slope form of a straight line. In an equation of the form y=mx+b, m is the slope of the straight line, and b is the y-intercept of the line.
Now, here is the other line he chose:
[math] L2 \equiv \mathcal {f} (x,y)| y=2x-4 \mathcal {g} [/math]
Now, here is the test question:
Do the two given straight lines intersect, and if so, where?
At this point, you want to solve a system of simultaneous linear equations. Now, there is already a standard procedure for doing this, but it involves matrix algebra. But you are bright student, and you already know it. Nonetheless, lets pretend you don't, and solve a problem or two, perhaps you may need a refresher.
First write the two equations he gave you as follows:
[math] y-3x=1 [/math]
[math] y-2x=4 [/math]
Now, utilize matrix notation, and write the linear system as follows:
[math]\left[ \begin{array}{ccc}
1 & -3\\
1 & -2\\
\end{array} \right]
\left[ \begin{array}{c}
x\\
y\\
\end{array} \right] =
\left[ \begin{array}{c}
1\\
4\\
\end{array} \right]
[/math]
The first matrix is usually called the coefficient matrix, for obvious reasons. In this case here, we have a two-by-two matrix.
Now, in order to answer your test question, write the system in the following form:
[math]\left[ \begin{array}{ccc}
1 & -3\\
1 & -2\\
\end{array} \right|
\left \begin{array}{c}
1\\
4\\
\end{array} \right]
[/math]
Now, you have to perform a finite sequence of elementary row operations, to get the system into the following form:
[math]\left[ \begin{array}{ccc}
1 & 0\\
0 & 1\\
\end{array} \right|
\left \begin{array}{c}
a\\
b\\
\end{array} \right]
[/math]
There are three standard elementary row operations they are:
1. Multiplication of any row by a scalar, this is called scalar multiplication.
2. Row addition (In this process, you add the elements of one row, to the elements of another row of your choice.)
3. Row interchange (Here, all you do is switch two rows)
So let us solve your professor's system of simultaneous linear equations, to determine whether or not these two straight lines are parallel, and if not, then where in the XY plane they intersect. We shall do this by performing a finite sequence of elementary row operations.
[math]\left[ \begin{array}{ccc}
1 & -3\\
1 & -2\\
\end{array} \right|
\left \begin{array}{c}
1\\
4\\
\end{array} \right]=
\left[ \begin{array}{ccc}
1 & -3\\
-1 & 2\\
\end{array} \right|
\left \begin{array}{c}
1\\
-4\\
\end{array} \right]
[/math]
The RHS of the equation above, was obtained using the elementary row operation known as scalar multiplication. The second row of the matrix equation was multiplied by the scalar -1.
[math]\left[ \begin{array}{ccc}
1 & -3\\
-1 & 2\\
\end{array} \right|
\left \begin{array}{c}
1\\
-4\\
\end{array} \right]=
\left[ \begin{array}{ccc}
1 & -3\\
0 & -1\\
\end{array} \right|
\left \begin{array}{c}
1\\
-3\\
\end{array} \right]
[/math]
The RHS of the equation above, was obtained using the elementary row operation known as row addition. The first row was added to the second row.
[math]\left[ \begin{array}{ccc}
1 & -3\\
0 & -1\\
\end{array} \right|
\left \begin{array}{c}
1\\
-3\\
\end{array} \right]=
\left[ \begin{array}{ccc}
1 & -3\\
0 & 1\\
\end{array} \right|
\left \begin{array}{c}
1\\
3\\
\end{array} \right]
[/math]
The RHS of the equation above, was obtained using the elementary row operation known as scalar multiplication. The second row was multiplied by negative one.
[math]\left[ \begin{array}{ccc}
1 & -3\\
0 & 1\\
\end{array} \right|
\left \begin{array}{c}
1\\
3\\
\end{array} \right]=
\left[ \begin{array}{ccc}
1 & -2\\
0 & 1\\
\end{array} \right|
\left \begin{array}{c}
4\\
3\\
\end{array} \right]
[/math]
The RHS of the equation above, was obtained using row addition, row two was added to row one.
Now, we can streamline the process, if we are willing to perform two or more elementary row operations at a time. In the next step, scalar multiplication, as well as row addition will be used in the same step. First row two will be multiplied by two, and then the result will be added to row one.
[math]\left[ \begin{array}{ccc}
1 & -2\\
0 & 1\\
\end{array} \right|
\left \begin{array}{c}
4\\
3\\
\end{array} \right]=
\left[ \begin{array}{ccc}
1 & 0\\
0 & 1\\
\end{array} \right|
\left \begin{array}{c}
10\\
3\\
\end{array} \right]
[/math]
Notice that the coefficent matrix is now the 2X2 identity matrix I.
Now, through the repeated use of the transitive property of equality, it follows that:
[math]\left[ \begin{array}{ccc}
1 & -3\\
1 & -2\\
\end{array} \right|
\left \begin{array}{c}
1\\
4\\
\end{array} \right]=
\left[ \begin{array}{ccc}
1 & 0\\
0 & 1\\
\end{array} \right|
\left \begin{array}{c}
10\\
3\\
\end{array} \right]
[/math]
Now, write the RHS of the equation above, as a matrix equation:
[math]\left[ \begin{array}{ccc}
1 & 0\\
0 & 1\\
\end{array} \right ]
\left [ \begin{array}{c}
x\\
y\\
\end{array} \right] = \left [ \begin{array}{c}
10\\
3\\
\end{array} \right]
[/math]
Now, convert back to linear equations:
1x+0y=10
0x+1y=3
From which we can rapidly infer that:
x=10 and y=3, is the solution of the original system of simultaneous linear equations.
What this means geometrically, is that the two given straight lines are not parallel, instead they intersect at one and only one point in the XY plane, and that point is the ordered pair (10,3).
And we are done.
Now, in the example above, we converted from matrix equations to linear equations, at the end of the problem.
The name of the procedure for doing this is called, matrix multiplication.
The basis of matrix multiplication, is the dot product of two vectors.
Here is the definition of the dot product of two vectors:
Definition:
[math] \vec A \bullet \vec B \equiv |\vec A| |\vec B| cos(\vec A,\vec B) [/math]
Translation: The dot product of vector A with vector B, is defined to be the magnitude of vector A, times the magnitude of vector B, times the cosine of the angle between the two vectors.
Matrix Multiplication
Computing the product of two matrices is actually quite simple, if you know the dot product.
I will quickly review the dot product of two arbitrary vectors in three dimensional Euclidean space.
Suppose you are asked to compute the dot product of the following two vectors:
[math] 3 \hat i + 2 \hat j - 7 \hat k [/math]
[math] 2 \hat i + 5 \hat j + 3 \hat k [/math]
What is needed now, is a formula which explains how to compute the dot product, in terms of the x,y,z components of the vectors in your frame.
Now, the magnitude of a vector comes from the three dimensional version of the Pythagorean theorem.
The three dimensional version actually comes from the two dimensional version, so if you know the two dimensional form of the Pythagorean theorem, then you can easily understand the three dimensional version.
I will briefly digress, to prove this.
An arbitrary position vector R in a given reference frame S, can be looked at, as the sum of three vectors, one purely in the i^ direction, another purely in the j^ direction, and the last in the purely k^ direction. That is,
[math] \vec R = x\hat i + y \hat j + z \hat k [/math]
So temporarily ignore the k^ direction, and focus on the projection of R onto the XY plane of the frame.
[math] \vec R = (x\hat i + y \hat j) + z \hat k [/math]
[math] proj(\vec R)_{XY} = (x\hat i + y \hat j) [/math]
The projection of R onto the XY plane, is itself is a vector entirely in the XY plane.
For the sake of simplicity, suppose that the head of the projection vector lies in the first quadrant of the XY plane. Thus, this vector is the hypotenuse of a right triangle, all the points of which lie in the XY plane.
Using the definition of vector addition, the following vector equation is a true statement:
[math] \vec h = x \hat i + y \hat j [/math]
Where the magnitude of h vector can be clearly seen to come from the Pythagorean theorem. And of course h vector, is the vector projection of R onto the XY plane, that is:
[math] \vec h \equiv proj(\vec R)_{XY} [/math]
So we now have a vector triangle, with the following sides:
[math] 1. \ x \hat i [/math]
[math] 2. \ y \hat j [/math]
[math] 3. \ \vec h [/math]
The magnitude of the first vector is x.
The magnitude of the second vector is y.
And we treat the magnitude of vector h as unknown, for now.
The proof that the triangle contains a right angle is as follows:
One side of the triangle has length x, and points in the i^ direction. As a vector, it's tail is located at the origin of the frame, and its head is a distance x away from the origin, at location (x,0,0) in the frame.
Then there is another vector, whose tail is located at (x,0,0) in the frame, but whose head is located somewhere in the first quadrant of the frame.
And, by stipulation, it's direction is equivalent to j^.
So in order to prove that the angle between these two vectors is a right angle, it suffices to prove that the angle between i^ and j^ is a right angle.
Now, in the design of the reference frame, the Y axis is stipulated to be perpendicular to the X axis. And what that means comes directly from Euclid.
It means this... it means that a straight line has fallen upon another straight line, and made the adjacent angles equal. And when this happens, each of the angles is called a right angle, and all right angles are equal
Euclid Postulate 4.
The point is this, the leg of the triangle with length y, lies upon a straight line which is parallel to the y axis. It cannot be the y axis, because it contains a point which isn't on the y axis, namely the point located at (x,0,0) in the frame.
In the limit as x approaches zero, this line becomes the Y axis. Now, if we let x reach zero, then this line will totally coincide with the y axis of the frame, and therefore, its angle inside the triangle will be equal to the angle the Y axis makes with the X axis, which is stipulated to be right. Therefore, there is an angle inside the triangle, which is a right angle.
And now, by the Pythagorean theorem, the sum of areas of squares constructed on the legs of the triangle, will be equivalent to the area of a square constructed on the hypotenuse.
Now, here is the definition of the area of a square:
Definition: Let S denote the length of a side of a square.
The area of the square is defined to be equal to the product of S with itself. As a formula...
[math] A_{square} = S \cdot S = S^2 [/math]
So one leg of the right triangle has length x, therefore, the square on that leg has area x2.
Another leg of the right triangle has length y, therefore, a square constructed on that leg has area y2.
The sum of these two areas is:
[math] x^2 + y^2 [/math]
The area of the square on the hypotenuse is, by definition, equal to:
[math] h^2 [/math]
And by the Pythagorean theorem , they are equal. That is...
[math] x^2 + y^2 = h^2 [/math]
And now, we can solve for h, by taking the square root of each side of the equation above. By the theory of equations, we must get two answers which make the statement above true, one positive, and one negative. But distance is purely a positive quantity, so using the interpretation that h is the length of the side of something, h must be positive.
[math] h = \pm \sqrt{x^2 + y^2} [/math]
Disregarding the negative root, we have:
[math] h = \sqrt{x^2+y^2} [/math]
So, the formula above for h, is equal to the length of the projection of R, onto the XY plane.
Now, the point of this digression, was to demonstrate the three dimensional version of the Pythagorean theorem.
Now, the hypotenuse of the right triangle, can also be viewed as a vector, whose tail is located at the origin, and whose head is located somewhere in the first quadrant of the XY plane. And we have expressed this using vector notation, as follows:
[math] \vec h [/math]
Now, R vector also has its tail located at the origin of the frame, however, the head of R vector lies outside of the XY plane.
In fact, the straight line which contains R vector intersects the XY plane at one and only one point, namely the origin of the frame.
To summarize:
Both R, and h have their tails located at the origin, but their heads are located at two different points in space, or perhaps I should say in the frame.
So there must be a vector triangle that can be constructed as follows:
[math] \vec h + z \hat k = \vec R [/math]
Because, now, by simple substitution, we get back the original definition of R...
[math] \vec h + z \hat k = \vec R [/math]
[math] \vec h = x \hat i + y \hat j [/math]
[math] \therefore [/math]
[math] \vec R = x \hat i + y \hat j + z \hat k [/math]
So now, there is another right triangle formed.
I know this is getting a bit long, but it can be simplified, that can occur later, once we have a formula for the dot product.
In this second right triangle, the legs are:
[math] \vec h [/math]
[math] z \hat k [/math]
And the hypotenuse is
[math] \vec R [/math]
It would be nice to first show that the triangle contains a right angle. Then we know that the Pythagorean theorem holds for it.
The direction of any vector which lies on the Z axis of the frame, is stipulated to be j^.
Now, the length of [math] z \hat k [/math] is just z, and its direction is also identical to that of any vector which lies entirely upon the Z axis of the frame.
Now, what needs to get done, is to show that the angle between vector h, and vector zk^, is a right angle.
The way I will accomplish this, is to resort to using Euclid.
First, notice that three non-collinear points lie in exactly one plane.
Now, focus on the following three points in the frame:
1. The origin.
2. The head of [math] \vec R [/math]
3. The head of [math] \vec h [/math]
Now, focus on the unique plane, which contains these three points.
Now, here is Euclid's second theorem:
Euclid, Book I, proposition 2
Now Euclid's second theorem is actually an ingenious construction.
In proposition two, he makes immediate use of proposition one, which was just proven rigorously by him.
Now here is Heath's translation of the ancient Greek, if I'm remembering it right...
"To place a straight line equal to a given finite straight line at a given point as an extremity."
But, the meaning of the sentence comes from the construction.
You are given an arbitrary point, and you are given an arbitrary straight line.
And the given point is not an extremity of the given straight line.
And you are asked to construct a straight line, using the given point as an extremity, such that the line you construct has a length which is equal to the length of the line you were given.
And Euclid's second proposition shows you how to do this.
Now, in Euclid, there is a distinction made between straight line segments, and infinite straight lines.
In Euclid Prop2, you are given a finite straight line, and a point not on that line.
Now, a finite straight line, and a point not on it, lie in one and only one plane.
And this construction takes place in the unique plane which contains the given point, and the given finite straight line.
So now back to our problem taking place in a frame.
There were three points to currently focus upon, the origin of the frame, the head of h vector, and the head of R vector. Those three points form a triangle, once you join each point to the other.
And they are three non-collinear points as well, which means they lie in one and only one plane.
Using Euclid's second proposition, where the given finite straight line is [math] z \hat k [/math] and the given point is the origin of the frame, construct a straight line equal to the given straight line, with the origin of the frame as an extremity. You can disregard the direction of the given vector.
After you have done that, it is unlikely that the straight line you just constructed coincides with the z axis of the frame.
You can now utilize Euclid's third postulate, which is this:
Euclid, Postulate 3
Translation: You can construct a circle with any point as center, and any radius you desire.
So now, using the straight line which was just constructed with the origin of the frame as an extremity, describe a circle, in the same plane the earlier construction was carried out in, such that it's center is the origin of the frame, and use the line just constructed as the radius.
Now, the circle and the z axis will intersect at exactly two points.
One of them will be closer to the head of [math] \vec R [/math] than the other. Call the one which is closer point B, and call the tip of R vector umm call it point C.
Now, join point B to point C, this step actually makes use of Euclid's first axiom, which is this:
Euclid, Postulate 1
Now, call the origin of the frame point A.
The goal is to now show that triangle A,B,C is equivalent to the vector triangle defined by:
[math] \vec h + z\hat k = \vec R [/math]
Because then if you do that, the corresponding parts of congruent triangles are congruent, and there will be more than enough information available to prove that the vector triangle contains a right angle.
Let the frame be a Right handed frame.
Now denote the angle between vector AB, and vector AC by [math] \phi[/math], that you are consistent with this .
The next goal, is to show that angle ABC is a right angle.
Call the head of [math] \vec h [/math] point D.
Now look at the following two propositions of Euclid:
Euclid, Book 1, proposition 29
Euclid, Book 1, Proposition 27
Proposition 27 of Euclid, says this:
If a transversal cuts two straight lines, and makes alternate interior angles equal to one another, then the two straight lines are parallel.
Proposition 29 of Euclid says this:
If a transversal cuts two parallel lines, then the alternate interior angles are equal to one another.
So proposition 29 is essentially the converse of proposition 27.
Thus, the two propositions can be combined, using the "if and only if" of binary logic. In other words:
Let L1,L2, and L3 be three given straight lines, and let it be the case that L3 intersects L1, and L3 also intersects L2.
The alternate interior angles which were formed are equal if an only if L1 and L2 are parallel.
Suppose that we can show that vector BC is parallel to vector AD. Then it will follow that angle BAC is equal to angle ACD. And angle BAC has also been called [math] \phi [/math], or if you prefer, the measure of angle BAC is equal to [math] \phi [/math].
Therefore, it would be the case that the measure of angle ACD is equal to phi.
Now, Euclid showed that the sum of the interior angles of a triangle is 180 degrees in the following proposition:
Euclid, Book 1, Proposition 32
There is a theorem needed, that would make what I am trying to do really really easy. Actually, I'm thinking of a sequence of proofs.
There seems to be a concept missing. Something to do with the notion of different vectors which point at the same location in space.
Suppose you have infinite straight line L1.
Pick two points on it at random, point A, and point B. The distance between them can be measured by a ruler.
Now, suppose that we map the real number system onto line L1.
Line L1 is now a coordinate axis, call it the X axis of a frame.
Let A denote the point 0, of the real number system, and let the coordinate of point B be some arbitrary positive real number.
Distance is a strictly nonnegative quantity.
The distance between point A, and point B, is the real number which corresponds to the position of point B, minus the real number which corresponds to the position of point A.
So let X2 denote the x coordinate of point B, and let X1 denote the x coordinate of point A. The distance between them, is
[math] d(A,B) = X2-X1 [/math]
I haven't yet made use of a fact I am very well aware of, which is:
The vector [math] z \hat k [/math] is parallel to any vector on the Z axis.
And now, if i use Euclid's postulates 27, and 29, then any transversal which cuts the infinite straight line containing vector zk^ and the Z axis, will make the alternate interior angles equal. So that if the line through AD meets the z axis at a right angle, then angle ADC must also be a right angle.
And then I think anything else I want to prove will just follow.
I think the problem is arising, because I have not paid attention to the exact position of points in the frame.
And ultimately, the whole point of this excercise is to discover the formula for the dot product.
Ok what should have been done, is to express the distance between two arbitrary points in a plane, in terms of their coordinates.
So that is just what I am going to do.
Construction of a three dimensional reference frame
I went away and thought about the problem for a bit, and decided the proper way to solve the problem, is to simply first discuss a three dimensional reference frame.
Only introduce as many terms as are necessary to fully describe a frame.
Suppose you are given two points in space A,B.
By Euclid's first postulate, you can construct the straight line from A to B.
Now, a straight line is not a vector.
A vector is a more powerful concept, than that of a straight line, because a vector has both magnitude and direction.
Now, you were given two points, but you were allowed to go from any point to any other point, by Euclid Postulate 1. So there are two possible ways that you could have constructed the straight line. Either you could have drawn from A towards B, or you could have drawn from B towards A.
I do not believe that Euclid distinguished between these two.
We can afford to be more general than Euclid.
If you drew from A to B, then you constructed the vector from A to B, which can be written as:
[math] \vec{AB} [/math]
If you drew from B to A, then you constructed the vector from B to A, which can be written as:
[math] \vec{BA} [/math]
Now, the distance between any two points in a frame, is a temporal constant, a quantity that never changes for any reason. In other words, in any frame, the distance between two randomly chosen points, is constant in time.
So there is no need to ever write that the distance between two points in one frame is a function of the temporal variable t.
And this can be made a postulate.
Now, I should have said, let the given points be in the same frame. I wasn't initially clear enough.
You are given two points A,B in one frame of reference S. Thus, it is impossible for two points in frame S to be in relative motion. Two points in different frames can be in relative motion, but not two points in the same frame.
Here is Euclid's first postulate... yes again.
Euclid's Elements Postulate One
Let's try another source...
Here it is in the actual ancient Greek it was written in...
Euclid's First Postulate, J.L. Heiberg
Pronunciation (i happen to know a tiny bit of ancient Greek)
(ee-tee-sthaw) (ah-paw) (pahn-dos) (sih-may-you) (epi) (pan) (see-may-awn) (ef-thay-ee-ahn) (gra-meen) (ah-yga-ygeen)
yg is a gutteral sound, not an english 'g' sound, more gutteral, kind of like the 'g' sound, but not exactly.
According to the source at Perseus project, the first word of Euclid's sentence could mean: ask for, demand, assume
I don't think it means any of these.
Here is the modern greek word for assume
[math] \upsilon \pi o \theta \epsilon \tau \omega [/math]
Pronunciation: (ee-paw-theh-taw)
This has clearly become the English word 'hypothesize'
Here is a link to a Greek English Dictionary
So I don't think Euclid meant "assume" that... he meant something else.
Let me skip the first word, and move onto the second...
Using an online dictionary, the modern greek word [math] \alpha \pi o[/math] can mean 'of', by' or 'from'.
Moving on to the third word...
Actually, I just found this...
The greek word [math] \eta \tau \iota \sigma [/math] means 'any'.
With that in mind, have a look at the first word again, only this time using Greek letters:
[math] H \tau \iota \sigma \theta o [/math]
The first letter of this word, is the capital Greek letter 'Eta', which in lowercase would read...
[math] \eta \tau \iota \sigma \theta o [/math]
A common practice in ancient civilizations, was to formulate a new word using two old words. In that case, we can analyze this first word of Euclid further...
[math] \eta \tau \iota \sigma -\theta o [/math]
Now sometimes, the endings of words just varied for grammatical reasons, so I still don't know the meaning of the first word. Quite possibly, the meaning of the first word of this sentence is identical to the meaning of the universal quantifier of first order logic, that is...
[math] \forall \equiv \text{for any} [/math]
But i don't know yet.
Let me move on to the third word of Euclid's sentence...
[math] \pi \alpha \nu \tau o \sigma [/math]
pronounced (pahn-dos)
Now, this word is a still common word in modern greek, and it means 'always'.
Moving onto the fourth word, its spelling in Greek is:
[math] \sigma \upsilon \mu \epsilon \iota o \upsilon [/math]
And here is the modern Greek word for 'point'
[math] \sigma \upsilon \mu \epsilon \iota o [/math]
I'm not sure if the first one is plural or not, because it's ancient Greek, not modern Greek, but it means either 'point' or 'points' or maybe set of points, not sure yet.
Let me move on to the fifth word, its spelling in ancient Greek would be...
[math] \epsilon \pi \iota [/math]
Some possible meanings(source online dictionary): on, upon, over, onto
Moving on to the sixth word, its spelling in ancient Greek was:
[math] \pi \alpha \nu [/math]
Possible meanings(source online dictionary): All, entire, whole
Moving on to the seventh word, its spelling in Greek is:
[math] \sigma \upsilon \mu \epsilon \iota o \nu [/math]
This word probably means 'point' in the singular. Then ending of the word.. 'on' is a masculine ending.
Moving on to the eigth word, its spelling in Greek is:
[math] \epsilon \upsilon \theta \epsilon \iota \alpha [/math]
Now, the modern Greek word for 'straight', from the online dictionary, is this:
[math] \epsilon \upsilon \theta \epsilon \upsilon \sigma [/math]
They sound practically the same, and are probably synonymous.
Moving on to the ninth word, its spelling in Greek is:
[math] \gamma \rho \alpha \mu \mu \eta \nu [/math]
Using an online dictionary, the modern greek word for 'line' is:
[math] \gamma \rho \alpha \mu \mu \eta [/math]
The only difference, is that there is no 'n' at the end of this word, i'd call it irrelevent, so they mean the same thing.
And now for the tenth and last word; its spelling in Greek is:
[math] \alpha \gamma \alpha \gamma \eta \nu [/math]
This word isn't in the online dictionary, and the LSJ at Perseus doesn't have it either.
Here is the original sentence as close to ancient Greek as I can get it:
[math] H \tau \iota \sigma \theta o \ \alpha \pi o \ \pi \alpha \nu \tau o \sigma \ \sigma \upsilon \mu \epsilon \iota o \upsilon \ \epsilon \pi \iota \ \pi \alpha \nu \ \sigma \upsilon \mu \epsilon \iota o \nu \ \epsilon \upsilon \theta \epsilon \iota \alpha \ \gamma \rho \alpha \mu \mu \eta [/math]
[math] \ \alpha \gamma \alpha \gamma \eta \nu \text{.} [/math]
Now, I will just try a one-to-one translation, and see what happens:
Any-tho for always points onto whole point straight line agagein(?).
That last word appears to be very important, yet I have no idea what it means or meant...
I have no idea how Heath arrived at his translation, and he left out a temporal logical word 'always.' I doubt that matters in the meaning of the sentence though.
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I put the line here, because everything before it was written yesterday.
Last night, I thought about the first word of Euclid's sentence, I am now attempting to translate his very first postulate, from Ancient Greek into modern English.
The modern Greek word for 'give' is:
[math] \delta \omega \sigma [/math]
Pronunciation: {thawz}
Here is a link to the LSJ dictionary at Perseus project, which gives the ancient Greek word for the English word 'give'.
Look down the list, and you will find the ancient Greek word:
[math] \delta \omega \sigma \omega \nu [/math]
Pronounced: {thaw-sawn}
(It's not exactly the English 'th' sound, its more like the sound you make when you pronounce the English word 'the')
I realized that I used the wrong last letter, on the first word of the sentence. It should have been an omega, not an omicron. In other words, the first word of the postulate is:
[math] H \tau \iota \sigma \delta \omega [/math]
And I know that the word:
[math] \eta \tau \iota \sigma [/math] means 'any' in English.
And the greek letter 'eta' looks like the roman letter H
So, I thought about the possibility that the first word of the postulate is actually composed of two greek words, in other words, I want to carry out a word etymology, but only in Greek. So we have this:
[math] \eta \tau \iota \sigma+\theta \omega [/math]
Now, since the greek word for 'give' is:
[math] \delta \omega \sigma [/math]
It appears to me, that the very first word, of Euclid's first of five postulates, tranlates into English as follows:
[math] H \tau \iota \sigma \theta \omega \equiv \text{Given any} [/math]
And even if it should be an omicron and not an omega, it doesn't matter, because I know I have the pronunciation right in both cases. The sounds are the same for sure. And also, the meaning of the word now makes sense in the sentence. So i was right. The first word of Euclid's first sentence translates into English as...
The universal quantifier of first order logic.
[math] H \tau \iota \sigma \theta \omega \equiv \text{Given any} \equiv \forall [/math]
I continued the translation further.
The next few words are:
[math] \alpha \pi o \ \pi \alpha \nu \tau o \sigma \ \sigma \upsilon \mu \epsilon \iota o \upsilon
[/math]
the possible translations of 'apo' are: of,by, from
Now Heath translated it as, "from", but I think it means 'of'.
Heath translated [math] \pi \alpha \nu \tau o \sigma [/math] as 'any', but its the modern Greek word for 'always.' But the word 'always' just doesn't make any sense in the sentence, no matter how you look at it. But, the word [math] \pi \alpha \nu [/math] translates as 'all or 'whole' or entire, and it happens to be the prefix of the word [math] \pi \alpha \nu \tau o \sigma [/math]. So I just decided to translate [math] \pi \alpha \nu \tau o \sigma [/math] as 'all' instead of 'always,' and I got this...
[Given any of all points]
As a partial translation. It is as if Euclid was using set theory to reason, over two thousand years ago. He postulated a set of points. And then began his first postulate as...
Given any (element of the set of) all points...
Here is the rest of his sentence...
[math] \epsilon \pi \iota \ \pi \alpha \nu \ \sigma \upsilon \mu \epsilon \iota o \nu \ \epsilon \upsilon \theta \epsilon \iota \alpha \ \gamma \rho \alpha \mu \mu \eta \ \alpha \gamma \alpha \gamma \eta \nu [/math]
Given any of all points, to any point a straight line can be made.
(if [math] \sigma \upsilon \mu \epsilon \iota o \nu [/math] is singular)
Or possibly
Given any of all points, to all points a straight line can be made.
(if [math] \sigma \upsilon \mu \epsilon \iota o \nu [/math] is plural)
Now, I had to totally guess on the last word, but I figured it means something like 'join' 'connect' 'draw' 'attach' 'construct' 'produce' something like that, because Heath translated the basic meaning correctly.
Any way enough of that, back to review of linear algebra.
For a reason that will appear clear later, I am now studying triangle similarity.
Here are some links I am looking at.
Similar triangles 1
Similar triangles 2
Here is a nice one...
Similiar triangles 3
Question: Given two triangles, how can you logically analyze them, to inevitably know later on, that they are similiar triangles, or not?
First you need a definition of similiar triangles.
From link three above, we see that to determine that two triangles are similiar, all we need to know, is that they have two angles which are identical in measure. Actually, two triangles will be similiar if and only if all three of the angles of one are congruent to the angles of the other.
The reason for this, is because the sum of the interior angles of any triangle is equal to two right angles (180 degrees), and that was proven by Euclid here:
Euclid, Book 1, proposition 32
So in other words, if at least two angles of one triangle correspond to two angles of another, for example
ABC = DEF
BCA=EFD
then
CAB=FDE
because
ABC+BCA+CAB=DEF+EFD+FDE (from Euclid)
so using algebra, together with logic,
Assume ABC = DEF and that BCA=EFD. This is the scope of the first and only assumption.
From Euclid: ABC+BCA+CAB=DEF+EFD+FDE
Subtract equals from equals to obtain:
ABC+BCA+CAB - ABC = DEF+EFD+FDE-DEF
BCA+CAB = EFD+FDE
Subtract equals from equals to obtain:
BCA+CAB -BCA = EFD+FDE-EFD
CAB = FDE
Now, close the scope of the first assumption, as follows:
If ABC = DEF & BCA=EFD then CAB = FDE
Which was to be proven. QED
So, in other words, if at least two angles of one triangle are congruent to two angles of another triangle, then all three angles of one triangle, are congruent to all three angles of the other. Where we have used a result obtained by Euclid, over two thousand years ago.
Ok, here is what wolfram has to say about similarity:
Similarity from mathworld
Ok, three non-colliner points in space suffice to construct a triangle.
Suppose we are given two triangles:
[math] \triangle ABC [/math]
[math] \triangle EFG [/math]
If the location in space, of their centers of inertia are identical, and the triangles are coplanar, it may be possible to make the sides of the triangles overlap by a rotation. If it is not possible to do this after 2pi degrees have passed, in time, then first what needs to be done, is for one of the triangles to be rotated in a third dimension, relative to the plane they are both initially in, and that rotation needs to be, umm, 180 degrees. Then after one of the triangles has been rotated out of the plane it was initally, but its center of inertia remaining at the same place, when it is again in the original plane, it will be possible to rotate it, so that its sides will coincide with the other triangle.
Since the triangles are not really made out of material, it's incorrect to refer to them as having a center of inertia, instead it should just be called the center of the triangle.
But the point is, that there is a condition for the two triangles to become one triangle, after a finite number of translations and rotations have been performed, and that condition is this:
Again the given triangles are:
[math] \triangle ABC [/math]
[math] \triangle EFG [/math]
Now, if the two triangles are to even have a chance of perfectly coinciding, then at least one of the angles of one of the triangles must be identical in measure, to at least one of the angles of the other triangle.
Suppose we know for certain that angle ABC is equal to angle EFG.
Let me do this from scratch.
Fact 1: Given any three points in space, exactly one triangle can be constructed.
Now, in order to be consistent with modern mathematical language, a triangle is just the straight lines which make up the three sides, and does not include the interior of the triangle.
So
[math] \triangle ABC \neq \triangle EFG [/math]
if and only if there is a point on one, which is not on the other.
We can now use naive set theory to say this too.
Each side of a triangle, has a certain amount of length, which could be measured by a ruler. You have to bring the ruler up to the side of the triangle to actually make the measurement.
We need a way to refer to the sides of a given triangle. There is already a method for doing this, using Latex. We have:
[math] \overline {AB} \equiv \text{side AB} [/math]
[math] \overline {AC} \equiv \text{side AC} [/math]
[math] \overline {BC} \equiv \text{side BC} [/math]
So now, any one of the three sides of a triangle is a set of points in space. And now we can use naive set theory to discuss the sides of our triangle.
So consider side AB. The straight line segment from point A to point B, consists of the point A, the point B, and all the points in between.
Now, to the definition of 'between'. The mathematical concept of 'between' was first seriously focused upon by David Hilbert, and later by Pasch.
Here is a good link about it Hilbert, Pasch, and Betweeness of points
The truth of the matter is, I have David Hilbert's book, and I started to read it, and I didn't like his style of proof. Right now, i don't want to worry about the history of the study of 'betweeness', later yes, now no.
Let a straight line have been constructed from point A, to point B.
Without looking, I am going to see if i can axiomatize 'betweeness' on my own...
A randomly chosen point C in space, will either be between the two given points A,B... or not.
Suppose not, what must be true?
The answer is rather simple.
Let d(A,C) denote the distance from A to C.
Let d(B,C) denote the distance from B to C.
Let d(A,B) denote the distance from A to B.
Now consider the sum of distances, d(A,C)+d(B,C).
Let us assume that the sum is an element of the real number system, i.e.
[math] d(A,C)+d(B,C) \in \mathbb{R} [/math]
And also:
[math] d(A,B) \in \mathbb{R} [/math]
Now, by the trichotomy property of the real number system, one and only one of the following three statements is true:
[math] \text{1.} \ d(A,C)+d(B,C) = d(A,B) [/math]
[math] \text{2.} \ d(A,C)+d(B,C) > d(A,B) [/math]
[math] \text{3.} \ d(A,C)+d(B,C) < d(A,B) [/math]
The shortest path from one point A in space, to another point B in space, is the straight line path of Euclidean Geometry.
Now, I have supposed that a randomly chosen point C in space is not between A and B.
What that means, is that the point C is not on the shortest path from A to B.
But, the union of points comprising the line segment AC, and the line segment CB, is a path from A to B. Since the point C is not on the shortest path from A to B, it therefore is an element of a longer path from A to B. This is what happens when you suppose that the randomly chosen point C, is not an element of the straight line from A to B.
Therefore:
[math] \text{not (C is between A,B)} \Rightarrow d(A,C)+d(B,C) > d(A,B)[/math]
And the previous statement is logically equivalent to the contrapositive, and so has the same truth value as its contrapositive. Its contrapositive is:
[math] \neg [d(A,C)+d(B,C) > d(A,B)] \Rightarrow \text{C is between A,B} [/math]
So now, we need a statement of the form
"If C is between A, B then such and such"
At which point, betweeness of points can be axiomatized using the distance axioms.
Now, if C is between A,B, then it is an element of the straight line segment from A to B, and therefore C would be on the shortest of all possible paths from A to B.
Suppose that
[math] \neg [d(A,C)+d(B,C) > d(A,B)] [/math]
Therefore, by the trichotomy property of the real number system, one of the following statements must be true:
[math] \text{1.} \ d(A,C)+d(B,C) = d(A,B) [/math]
[math] \text{2.} \ d(A,C)+d(B,C) < d(A,B) [/math]
Now, it cannot be statement two which is true, for then C would lie on a path from A to B which is shorter than the shortest of all possible paths. Therefore, statement 1 would be true, under the given supposition.
Therefore, the following statement is true:
[math] \neg [d(A,C)+d(B,C) > d(A,B)] \Rightarrow \ d(A,C)+d(B,C) = d(A,B) [/math]
Where we have closed the scope of the assumption.
So now, let us connect the verbal form of the statement that C is between A,B, to a precise mathematical statement of the same fact. We can accomplish this as follows.
Stipulate that the following statement is true:
[math] \text{C is between A,B} \Rightarrow \neg [d(A,C)+d(B,C) > d(A,B)] [/math]
Using hypothetical syllogism, it now must follow, that the following statement is true:
[math] \text{C is between A,B} \Rightarrow d(A,C)+d(B,C) = d(A,B) [/math]
Now, in this example, we picked one out of infinitely many points in space which could have been chosen at random.
So, we can use universal generalization and write:
[math] \forall C \in \mathbb{R}^3 [\text{C is between A,B} \Rightarrow d(A,C)+d(B,C) = d(A,B)][/math]
Translation: For any point C in three dimensional Euclidean space, if C is in between points A and B, then the distance from point A to point C, plus the distance from point B to point C is equal to the distance from point A to point B.
So again, let us continue with our randomly chosen point C.
In order to have a logical definition of "C is between A,B" we need the converse of [math] \text{C is between A,B} \Rightarrow d(A,C)+d(B,C) = d(A,B)[/math]
Now, if A=B, then it is impossible that A>B, for any quantities A,B.
Therefore:
[math] d(A,C)+d(B,C) = d(A,B) \Rightarrow \neg (d(A,C)+d(B,C) > d(A,B) ) [/math]
We can now use hypothetical syllogism, to infer that the following statement is true:
[math] d(A,C)+d(B,C) = d(A,B) \Rightarrow \text{C is between A,B} [/math]
And this statement is true, for any randomly chosen point C. Therefore:
[math] \forall C \in \mathbb{R}^3[d(A,C)+d(B,C) = d(A,B) \Rightarrow \text{C is between A,B}] [/math]
So we have now proven that the following two statements are true:
[math] \forall C \in \mathbb{R}^3 [\text{C is between A,B} \Rightarrow d(A,C)+d(B,C) = d(A,B)][/math]
[math] \forall C \in \mathbb{R}^3[d(A,C)+d(B,C) = d(A,B) \Rightarrow \text{C is between A,B}] [/math]
And by saying they are true, we mean that they must be true simultaneously, therefore the following statement must be true:
[math] \forall C \in \mathbb{R}^3 [\text{C is between A,B} \Rightarrow d(A,C)+d(B,C) = d(A,B)] [/math]
AND
[math] \forall C \in \mathbb{R}^3[d(A,C)+d(B,C) = d(A,B) \Rightarrow \text{C is between A,B}] [/math].
So let D be a randomly chosen point in a frame where the straight line defined by the given two points is at rest. Therefore, the following statement is true:
[math] \text{D is between A,B} \Rightarrow d(A,D)+d(B,D) = d(A,B)[/math]
AND
[math] d(A,D)+d(B,D) = d(A,B) \Rightarrow \text{D is between A,B}[/math].
Therefore, using the definition of the logical connective "if and only if" the following statement is true:
[math] \text{D is between A,B} \Leftrightarrow d(A,D)+d(B,D) = d(A,B)[/math]
And it is true for any randomly chosen point, not just D. Therefore:
[math] \forall C \in \mathbb{R}^3[\text{C is between A,B} \Leftrightarrow d(A,C)+d(B,C) = d(A,B)][/math]
Now, the proof was carried out under the assumption that point A, and point B were different locations in space.
If one of the points can move relative to the other, that will complicate the logic. But, if we are to permit ourselves the use of frames which can move relative to one another, then we need to allow for the possibility, that one of the given points can move relative to the other.
So we can summarize what we know thus far:
[math] \forall A,B \in \mathbb{R}^3: \text{if not(A=B) then} [/math]
[math] \forall C \in \mathbb{R}^3[\text{C is between A,B} \Leftrightarrow d(A,C)+d(B,C) = d(A,B)][/math]
Now in the general theory of relativity, points in a single frame can move relative to one another, whence some wonder whether or not space can expand.
And also, we want to be able to utilize what we learn in the best of all possible manners, so that we might want to allow point A to represent a point permanently at rest in one frame, say frame S, and let point B denote a point permanently at rest in another frame, say S`, and have the two frames be in relative motion, with relative speed v.
If we permit this, then the coordinates of point B in frame S are functions of time, and also, the coordinates of point A in frame S` are functions of time.
All that was told to us originally, was that the points A,B were given, and we were to axiomatize 'betweeness' without reading the works of David Hilbert and Moritz Pasch.
For right now, only consider the case where A and B are permanently at rest relative to one another.
Now, either A=B or not (A=B).
Suppose that A=B. Therefore:
[math] \forall C \in \mathbb{R}^3[\text{C is between A,A} \Leftrightarrow d(A,C)+d(A,C) = d(A,A)][/math]
Now, we have to decide whether or not d(x,x)=0 for any x should be adopted as a distance axiom.
Here are the standard distance axioms, of a Euclidean metric:
Distance Axioms
The only problem I have with permitting the distance from a point to itself to be equal to zero, is that it conflicts with the operational definition of a ruler. A ruler is something which must have a length greater than zero. So if you hold that distance is what rulers measure, then there is a conflict.
This can be made more analytical, using the special theory of relativity, in combination with first order logic, as follows:
Suppose we wish to verify whether or not the Lorentz contraction formula is true.
First, let us postulate a set of rulers.
We have already postulated a set of points.
Suppose we have two rulers, which when at rest relative to one another have the same length, see below:
A--------------------B
A`-------------------B`
Let the frame which ruler AB is permanently at rest in, be called frame S.
Let the frame which ruler A`B` is permanently at rest in, be called frame S`. Let the origin of S be the center of inertia of ruler AB, and let the origin of frame S` be the center of inertia of ruler A`B`.
Now, according to the theory of special relativity, these two rulers should always have the same length when at rest with respect to each other, disregarding tiny perturbations in length, due to the nonzero temperature of the rulers.
Let us assume the following formula is a true statement in frame S.
[math] L = L_0 \sqrt{1-v^2/c^2} [/math]
Stipulate that frame S is an inertial reference frame. Thus, ruler AB will remain at rest in other inertial reference frames, or in straight line motion at a constant speed, until such time as an external force acts upon ruler AB.
Because the ruler is permanently at rest in frame S, if an external force acts to accelerate ruler AB, frame S will accelerate along with the ruler, however, at that moment in time, the truth value of the statement "S is an inertial reference frame" will switch from true to false.
For the duration of the 'event' to be analyzed here, let it be stipulated that frame S is an inertial reference frame. Thus, all of Newton's laws of motion are true statements in frame S.
Disregarding how S` got into motion relative to S, let it be the case that S` is moving relative to S at relative speed v.
For the sake of definiteness, let the X axis of frame S be parallel to ruler AB, and let the X` axis of frame S` be parallel to ruler A`B`, and let the positive x axis of both rulers point from left to right.
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Let me make a long story short...
I'm not sure that we should define distance from a point to itself as zero, i think it might cause contradiction.
