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||Définition d'une variable || Algèbre || Arithmetique || Logique ||La raison basée sur les règles et modelés||
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YOU are better than YOU think. Show yourself  how: 

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Read  logic chapters 1 to 5  in online volume Three Skills for Algebra  for greater skills & confidence in  work 
and study.

Read notes and textbooks like a lawyer, so that no nuance, no subtlety and no clause escapes your attention. 

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 Logic chapters 1 to 5  re- appear not in sequence, as is or longer,  in  Volume 1A,  Pattern Based Reason, Bon Appetite.

Logic Mastery
 Amazing, Amusing, Amorous,  Delicious, Delightful, Edifying, Strengthening Elixir. 
It eases work & learning difficulties Makes the hard easier. Opens eyes. Leads to greater precision.
in reading and
writing

Logic mastery makes the hard, easier. Logic mastery  leads to better, stronger and richer comprehension.  Logic mastery  improves reading and writing.  Logic mastery ease learning difficulties.  Logic mastery gives a headstart.  In sum, logic mastery  will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck.


After logic  (a) continue reading Three Skills for Algebra, chapters 8 to 14  and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes  & More Math, chapters 2 to 6;

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Caution: Site advice is approximately correct, for some circumstances, not all. That leaves room for thought

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What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts.


Try the Twiddla Whiteboard. In principle, it  allows to people to draw and chat together online on a copy of this webpage or a clean sheet. The chat may be via text or audio.  Visit www.twiddla.com to set up whiteboards to work with the webpage of your choice.

For online automated help in senior high school maths & calculus, visit  quickmath.com  For Automatic Calculus and Algebra Help with derivatives, integrals, graphs, linear equations, matrix algebra, visit calc101.com  With  overlap, each site quickmath & calc101offers a different range of services, some free, some not, all based on webmathematica. Good luck.

Coordinate Formulas for Distance

(I) Points on a Line

The following gives the distance between a pair of points (the length of the line segment between them) on a horizontal line: 

The case of a vertical line is similar. 

(II) Planar Case: Points in a Plane

Suppose  [x1,y1] and [x2,y2]. are points in the plane. Our aim is to compute the distance c between these points.

The line segment between   [x1,y1] and [x2,y2]., if it is not horizontal, nor vertical, provides the hypotenuse of right triangle (or two) with horizontal and vertical sides. The case where the line segment has a negative slope is drawn below.

The lengths of the sides are a =  |x2- x1| and  b = |y2- y1|

Therefore the Pythagorean Theorem (see proof below) says

c2 = a2 + b= |x2- x||2 + |y2- y1|2

So the distance c between the points [x1,y1] and [x2,y2] satisfies

c2   = (x2- x1|2 + (y2- y1)2

           _______________
c    /|x2- x1|2 + |y2- y1|2

      _______________
 /(x2- x1)2 + {y2- y1)2

Verification of this formula in the two cases where the line segment between  [x1,y1] and [x2,y2]. is horizontal or vertical is left for you to explore. Here  |c|2  =  c permits the replacement of |x2- x1|2  and  |y2- y1|by  (x2- x1)2 and  {y2- y1)2 , respectively.

Exercise: Redo the above proof for the case where the slope of the  line segment between  [x1,y1] and [x2,y2]. is positive.

Points in Space

We wish to express the length d in terms of the coordinates of the end points (X,Y, Z) and (x,y,z). which form the vertices of a parallelepiped (box) in space. By Pythogoras theorem twice, once to a vertical right triangle and once to a horizontal right triangle, we have 

d2 = c2 + h2 =  a2 + b2 + h2  = |X- x|2 + |Y- y|2 + |Z- z|2

Therefore 

           _____________________
d   /|X- x|2 + |Y- y|2 + |Z- z|2

 

Another Example. In the following diagram due to the use of orthogonal coordinates, the square of distance of the point at (a,b,c) to the origin is d2 = c2 + h2 =  a2 + b2 + c2

                            _________
Therefore d   /a2 + b2 + c2

 


The Pythagorean Theorem
from chapter 17, Three Skills for Algebra

The Pythagorean theorem is one of the oldest statements in mathematics. This theorem is used to recognize right triangles in the plane and also to compute the distance between points in the plane. There are hundreds of proofs of this theorem. One that seems easiest to follow is the so-called Chinese square (dissection) proof. 

The form below employs a mixes algebra and geometry before coordinates to imply the theorem and also build algebraic-geometric reasoning skills. 


Theorem 17.1 [Pythagorean Theorem] If a right triangle has a hypotenuse of length c and other two sides of lengths a and b,

then

a2+b2 = c2

The Chinese square proof of the Pythagorean theorem is given next. The proof is based on the construction and then dissection or division of a large square into a smaller square plus four corner triangles. This division gives a second way of computing the area of the large square. The first way is to compute the square L2 of the length L of a side. The proof of the Pythagorean theorem follows by equating the formulas for the two different ways and then simplifying the resulting equation. Details follow.

Chinese Square Proof: To see why a2+b2 = c2, first draw a square with four sides of length b+a as follows:


The area of this square is
A = (a+b)2 = a2+2ab+b2
Second, put four copies of the original right triangle into the square in the following manner:


Next
observe   
a+b+90°
= 180°
   
a+b+g
= 180°
and therefore:   
g = 90°
Similarly the interior angles of the region, a quadrilateral, bounded by the four hypotenuses are all equal to g = 90°. The hypotenuses of the four copies of the original right triangle in the corner of the square of sides a+b also have a common length c. Thus they form sides of a square. Finally observe, the area A of the larger square of sides a+b is also given by the area of the interior square plus the areas of the four corner copies.

Now from the previous steps, we have A = c2+4·[(ab)/2] = c2+2ab. From this a2+2ab+b2 = A = c2+2ab. Cancellation now implies the Pythagorean equality a2+b2 = c2 holds.

Question: Suppose the three sides of a triangle have lengths a, b and c satisfying a2+b2 = c2. Is the triangle a right triangle? (The answer is yes. One way to see why requires the cosine law in trigonometry - a subject for further inquiry perhaps.)

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