I was looking at the in Internet now and found in this site a different proof of this fact:
"Another elegant proof:
In 1952, Robert Gauntt, then a freshman at Purdue University, gave the following proof[2].
a 2 = 2b2 cannot have a non-zero solution in integers because the last non-zero digit of a square, written in base 3, must be 1, whereas the last non-zero digit of twice a square is 2."
Interesting! note the "last non-zero digit" part.
This proof would be obvious if we were used to base 3.
Here at cut the knot there is a list of different proofs. Proofs number 4 and 3 (in that order) are the ones I like most. The proofs at cut the knot are also reproduced below:
Proof 4
Following is yet another illuminating proof. As in the standard proof, assume p and q are mutually prime (numbers with no common factors). Their squares are still mutually prime for they are built from the same factors. Therefore, the fraction p 2 /q2 cannot cancel out. In particular, p2 /q2 cannot cancel down to equal 2. Therefore, p2 /q 2 .
Proof 3
Following is another intuitive one [Lasckovich, p. 4]. This one is based on the assertion that every number is uniquely (up to the order of factors) representable as a product of primes. (A prime is a number divisible only by itself and 1.) The fact which is known as The Fundamental Theorem of Arithmetic. Assuming it as an axiom (or a given fact), let (p/q)2 = 2 for some integers p and q. Then p2 = 2q2 . Factor both p and q into a product of primes. p2 is factored into a product of the very same primes as p each taken twice. Therefore, p2 has an even number of prime factors. So does q2 . Therefore, 2q2 has an odd number of prime factors. Contradiction.
Exponents
In this post we will try to understand some properties of exponents.
How much is 2 3 ? = 2*2*2 = 8
2 4 = 2*2*2*2 =16
2 5 = 2*2*2*2*2 =32
2 6 = 2*2*2*2*2*2 =64
2 2 = 2*2 =4
Lets put all that in sequence...
2 2 2 3 2 4 2 5 2 6
4, 8, 16, 32, 64
Note that when we increase in the exponent we multiply by 2!
So the next terms would be:
128, 256, 512,...
But what the previous terms should be?
When we decrease the exponent... we divide by two!!
2 -1 2 0 2 1 2 2 2 3 2 4 2 5 2 6
z, y, x, 4, 8, 16, 32, 64
x = 4/2 =2
y = x/2 =2/2=1
z = y/2 =1/2 =1/2
2 -1 2 0 2 1 2 2 2 3 2 4 2 5 2 6
1/2,1, 2, 4, 8, 16, 32,64
In here notice 2 1 =2 -that does make some sense
and 2 0 =1!
some more terms:
2 -3 2 -2 2 -1 2 0 2 1 2 2 2 3
c, b, 1/2, 1, 2, 4, 8
b = (1/2)/2 = 1/4
c = b/2 = 1/8
1/8,1/4,1/2,1,2,4,8
Notice that the numbers (1,2,4,8) 'repeat' 1/8=2 -3 = 1/(2 3 )
And that is what happens in general with negative exponents.
2 -n = 1/(2 n )
Now some more properties
2 3 *2 5 = (2*2*2)*(2*2*2*2*2)= 2 "eight times" =2 8
So when we multiply two powers in the same base we add the exponents
For division:
2 7/2 4 = (2*2*2*2*2*2*2)/(2*2*2*2)= 2 "three times" =2 3
There is some cancellation, so when we divide two powers in the same base we subtract the exponents.
(2 3 ) 4 = 2 3 2 3 2 3 2 3 = 2 "twelve times=3*4" = 2 12
Kind of strange to put in words (try it!!) but easier to understand.
How much is 2 3 ? = 2*2*2 = 8
2 4 = 2*2*2*2 =16
2 5 = 2*2*2*2*2 =32
2 6 = 2*2*2*2*2*2 =64
2 2 = 2*2 =4
Lets put all that in sequence...
2 2 2 3 2 4 2 5 2 6
4, 8, 16, 32, 64
Note that when we increase in the exponent we multiply by 2!
So the next terms would be:
128, 256, 512,...
But what the previous terms should be?
When we decrease the exponent... we divide by two!!
2 -1 2 0 2 1 2 2 2 3 2 4 2 5 2 6
z, y, x, 4, 8, 16, 32, 64
x = 4/2 =2
y = x/2 =2/2=1
z = y/2 =1/2 =1/2
2 -1 2 0 2 1 2 2 2 3 2 4 2 5 2 6
1/2,1, 2, 4, 8, 16, 32,64
In here notice 2 1 =2 -that does make some sense
and 2 0 =1!
some more terms:
2 -3 2 -2 2 -1 2 0 2 1 2 2 2 3
c, b, 1/2, 1, 2, 4, 8
b = (1/2)/2 = 1/4
c = b/2 = 1/8
1/8,1/4,1/2,1,2,4,8
Notice that the numbers (1,2,4,8) 'repeat' 1/8=2 -3 = 1/(2 3 )
And that is what happens in general with negative exponents.
2 -n = 1/(2 n )
Now some more properties
2 3 *2 5 = (2*2*2)*(2*2*2*2*2)= 2 "eight times" =2 8
So when we multiply two powers in the same base we add the exponents
For division:
2 7/2 4 = (2*2*2*2*2*2*2)/(2*2*2*2)= 2 "three times" =2 3
There is some cancellation, so when we divide two powers in the same base we subtract the exponents.
(2 3 ) 4 = 2 3 2 3 2 3 2 3 = 2 "twelve times=3*4" = 2 12
Kind of strange to put in words (try it!!) but easier to understand.
Sum of the first n squares
In our last post Volume of the Pyramid we showed that the volume of a truncated solid is the Base*(average of the height at the vertices) for the triangular and square prisms.
Lets now use that idea to calculate the sum of the first n squares.
12 + 22 + 32 + 42 + .... + n2
We can split that in:
1 + 2 + 3 + 4 + 5 + ..... + n
2 + 3 + 4 + 5 + ..... + n
3 + 4 + 5 + ..... + n
4 + 5 + ..... + n
5 + ..... + n
..... + n
. + n
n
If you picture the numbers as the heights of a solid it turns to be a truncated "linear" solid - since the variation is linear, there is also a more formal demonstration of the formula below.
The volume V is:
(B)*(average height at the vertexes)
(n*(n +1)/2)*(1+ n + n)/3
So the sum of the first n squares is
n*(n +1)*(2n +1)/6
You can also put this numbers forming a equilateral triangle rotate by 120o and 240 o and then sum the three versions, getting a nice prism,(it is you can check that). All that is a "3D version" of the classic argument.
1 + 2 +...+(n-1)+ n = S
n +(n-1)+....+2 + 1 = S
n*(n +1) = 2S
S = n*(n +1)/2
We also saw that using averages in the post sums.
Lets now use that idea to calculate the sum of the first n squares.
12 + 22 + 32 + 42 + .... + n2
We can split that in:
1 + 2 + 3 + 4 + 5 + ..... + n
2 + 3 + 4 + 5 + ..... + n
3 + 4 + 5 + ..... + n
4 + 5 + ..... + n
5 + ..... + n
..... + n
. + n
n
If you picture the numbers as the heights of a solid it turns to be a truncated "linear" solid - since the variation is linear, there is also a more formal demonstration of the formula below.
The volume V is:
(B)*(average height at the vertexes)
(n*(n +1)/2)*(1+ n + n)/3
So the sum of the first n squares is
n*(n +1)*(2n +1)/6
You can also put this numbers forming a equilateral triangle rotate by 120o and 240 o and then sum the three versions, getting a nice prism,(it is you can check that). All that is a "3D version" of the classic argument.
1 + 2 +...+(n-1)+ n = S
n +(n-1)+....+2 + 1 = S
n*(n +1) = 2S
S = n*(n +1)/2
We also saw that using averages in the post sums.
Volume of the Pyramid
For some time I asked myself why the volume of the pyramid is 1/3*B*h.
In this post I will give an unusual demonstration of this fact.
After some reflection that the volume is directly proportional to the Area "B" and the height "h" seems reasonable but the 1/3 puzzled me. I know some different demonstrations for it but I wasn't completely satisfied with them.
If you already saw this idea/demonstration somewhere else please tell me where so I can check it.
For some (important) motivation let's first take a rectangular prism with square base ABCD and cut it with a plane not parallel to the base (such that the cut is between the two bases at the "bottom" and the "top"), considering below such a plane a truncated solid that at each vertex ABCD the solid has heights h a , h b , h c , and h d .
What is the height of the solid at the center of the square H?
You can show using some geometry (and it should appeal to your intuition) that
H = (h a + h c )/2 = (h b + h d )/2
That is also equal to our idea of the "average height of the solid" and it is also equal to the average at the points A,B,C ,and D.
H =(h a + h b + h c + h d )/4 .
The volume of the truncated solid V should be
V = (Area of base B)*(average height) "*"
And what happens if the base is an equilateral triangle ABC? The same idea should apply so its volume would be:
V= B*(h a + h b + h c )/3.
Lets analyze the case when h a = h b = 0 and h c = h... in that case we have a pyramid!!! So it's volume should be:
V= B*(0 + 0 + h)/3 = B*h/3
That is it!
Unfortunately this demonstration is not rigorous. I think it is intuitive thought, and it can be made rigorous using integration, or more simply the idea that sheering doesn't change the volume.
Read on for the integration point of view.
There are three equivalent pyramids we can make:
h a =0, h b = 0, and h c = h
h a =0, h b = h, and h c = 0
h a =h, h b = 0, and h c = 0
These pyramids are all equal so they have the same volume, and each pyramid is determined by a different plane. We can think the volume as the integral of the height(function) over the base ABC of the pyramid. If we add the functions determined by each plane we will also add their volumes so we arrive at a linear function where
h a = h b = h c = h
Such is the constant function, so that the three pyramids "added" together equal the volume of the prism and since each has the same volume and so the volume of the pyramid must be 1/3 of the prism with the same base**.
Why go to all that trouble? There is a very well know way to fit three pyramids in a triangular prism. The problem with that is these pyramid are not easy to "see", and we still have to argue why they should have the same volume***.
The main advantage here is the connection of our idea of average and the 1/3 factor of the pyramid. We will also see in "the sums of squares" that this point of view has nice implications.
For those who want more detail see Observation below:
* Obs. One can also demonstrate this by dividing the solid in prisms, or using "reflection maintain the volume" reasoning.
** Obs. I made the base of the pyramid an equilateral triangle to ease things up, we can use Cavalieri's Principle to get to the other cases, and we have too anyway to get the height off a vertex at the base. We can also use to the same effect the principle that pyramids of equal base area and height have the equal volumes.
*** Obs. The three equal pyramids from the corner of a cube or the six equal pyramids from the center, may be easier to see, but still we have to argue why this is true for more general pyramids.
In this post I will give an unusual demonstration of this fact.
After some reflection that the volume is directly proportional to the Area "B" and the height "h" seems reasonable but the 1/3 puzzled me. I know some different demonstrations for it but I wasn't completely satisfied with them.
If you already saw this idea/demonstration somewhere else please tell me where so I can check it.
For some (important) motivation let's first take a rectangular prism with square base ABCD and cut it with a plane not parallel to the base (such that the cut is between the two bases at the "bottom" and the "top"), considering below such a plane a truncated solid that at each vertex ABCD the solid has heights h a , h b , h c , and h d .
What is the height of the solid at the center of the square H?
You can show using some geometry (and it should appeal to your intuition) that
H = (h a + h c )/2 = (h b + h d )/2
That is also equal to our idea of the "average height of the solid" and it is also equal to the average at the points A,B,C ,and D.
H =(h a + h b + h c + h d )/4 .
The volume of the truncated solid V should be
V = (Area of base B)*(average height) "*"
And what happens if the base is an equilateral triangle ABC? The same idea should apply so its volume would be:
V= B*(h a + h b + h c )/3.
Lets analyze the case when h a = h b = 0 and h c = h... in that case we have a pyramid!!! So it's volume should be:
V= B*(0 + 0 + h)/3 = B*h/3
That is it!
Unfortunately this demonstration is not rigorous. I think it is intuitive thought, and it can be made rigorous using integration, or more simply the idea that sheering doesn't change the volume.
Read on for the integration point of view.
There are three equivalent pyramids we can make:
h a =0, h b = 0, and h c = h
h a =0, h b = h, and h c = 0
h a =h, h b = 0, and h c = 0
These pyramids are all equal so they have the same volume, and each pyramid is determined by a different plane. We can think the volume as the integral of the height(function) over the base ABC of the pyramid. If we add the functions determined by each plane we will also add their volumes so we arrive at a linear function where
h a = h b = h c = h
Such is the constant function, so that the three pyramids "added" together equal the volume of the prism and since each has the same volume and so the volume of the pyramid must be 1/3 of the prism with the same base**.
Why go to all that trouble? There is a very well know way to fit three pyramids in a triangular prism. The problem with that is these pyramid are not easy to "see", and we still have to argue why they should have the same volume***.
The main advantage here is the connection of our idea of average and the 1/3 factor of the pyramid. We will also see in "the sums of squares" that this point of view has nice implications.
For those who want more detail see Observation below:
* Obs. One can also demonstrate this by dividing the solid in prisms, or using "reflection maintain the volume" reasoning.
** Obs. I made the base of the pyramid an equilateral triangle to ease things up, we can use Cavalieri's Principle to get to the other cases, and we have too anyway to get the height off a vertex at the base. We can also use to the same effect the principle that pyramids of equal base area and height have the equal volumes.
*** Obs. The three equal pyramids from the corner of a cube or the six equal pyramids from the center, may be easier to see, but still we have to argue why this is true for more general pyramids.
tan tan tan
That was a question I saw at Polymathematics raised:
Prove that if A, B, and C are the 3 angles of a triangle,
tan A + tan B + tan C = (tan A)(tan B)(tan C).
There is demonstration for this fact using complex numbers say A + B + C = 180
(cosA + i sinA)*(cosB + i sinB)*(cosC + i sinC)=-1
So this product is real number dividing each factor by the cos we have
(1 + i tanA)*(1 + i tanB)*(1 + i tanC) is real
By foiling and since the imaginary part is 0 we get that identity.
That was the nicest way I could show that identity.
My interest in it was that it can be used to show Hieron's formula, as I was trying to find nicer ways to show it I got the way above.
Prove that if A, B, and C are the 3 angles of a triangle,
tan A + tan B + tan C = (tan A)(tan B)(tan C).
There is demonstration for this fact using complex numbers say A + B + C = 180
(cosA + i sinA)*(cosB + i sinB)*(cosC + i sinC)=-1
So this product is real number dividing each factor by the cos we have
(1 + i tanA)*(1 + i tanB)*(1 + i tanC) is real
By foiling and since the imaginary part is 0 we get that identity.
That was the nicest way I could show that identity.
My interest in it was that it can be used to show Hieron's formula, as I was trying to find nicer ways to show it I got the way above.
Sums
For answers on this post highlight between the xxx's to see
Quickly tell me how much is
95+96+97=
Instead of doing the sum you could just approximate it
Like for example: 3*95
Or even better it could be...
xxx 3*96 xxx
But wait that is exactly 95+96+97
Since xxx 96+96+96 xxx = 95+96+97
As the terms even out.
Now how much is
94+95+96+97+98=
xxx 5*96! xxx
You can guess 93+94+95+96+97+98+99
But wait a second and if you have for example
94+95+96+97=?
There is no central one to average but still
The average should be xxx (95+96)/2 = 95.5 = (94+97)/2 xxx
So the sum is xxx 4*95.5 xxx
It is always the (number of terms)*(average or "middle term")
And now the last sum
37+39+...+79+81
In this case we don't see the middle term but you know what it should be?
xxx (37+81)/2 xxx
That is the (average) is (first+last)/2
So the sum is:
(number of terms)*(average)=(number of terms)*(first+last)/2
That is the sum of an arithmetic progression.
Quickly tell me how much is
95+96+97=
Instead of doing the sum you could just approximate it
Like for example: 3*95
Or even better it could be...
xxx 3*96 xxx
But wait that is exactly 95+96+97
Since xxx 96+96+96 xxx = 95+96+97
As the terms even out.
Now how much is
94+95+96+97+98=
xxx 5*96! xxx
You can guess 93+94+95+96+97+98+99
But wait a second and if you have for example
94+95+96+97=?
There is no central one to average but still
The average should be xxx (95+96)/2 = 95.5 = (94+97)/2 xxx
So the sum is xxx 4*95.5 xxx
It is always the (number of terms)*(average or "middle term")
And now the last sum
37+39+...+79+81
In this case we don't see the middle term but you know what it should be?
xxx (37+81)/2 xxx
That is the (average) is (first+last)/2
So the sum is:
(number of terms)*(average)=(number of terms)*(first+last)/2
That is the sum of an arithmetic progression.
Complex Numbers 3
So we saw that with the rotation-stretching interpretation multiplying and taking powers of complex numbers makes more sense.
Obs. While this interpretation makes multiplication easier it does not help when making addition. Happily addition is easy in the algebraic form "a + bi".
Now we are going to see three bonus of looking at complex numbers in this way.
First bonus: Our theory is somewhat arbitrary why we defined i to rotate 90 o counterclockwise ? It could be clockwise couldn't it ? The answer is yes! Lets say you defined a different clockwise imaginary number j, because you didn't like mine, then you would have a complex number theory just like mine with j's in the place of i's! Your drawings would be the reflection of my drawings. But the arithmetic would be the same.
But your j does the same job as my -i! So in a way the -i does the same job as i, but clockwise. Changing i's to -i's changes the orientation of my rotations. So if I take a number like 3+4i, then 3+4(-i) = 3 - 4i has opposite rotation but the same stretching factor. The operation changing i's to -i's is called conjugation and this reasoning explain most of it's properties directly.
Second bonus: Let's say our stretching factor is 1 (no stretching at all) in this case our complex number only rotate vectors by angles a and b. So they are:
cos(a) + i sin(b)
cos(b) + i sin(b)
An their product by our interpretation:
cos(a+b) + i sin(a+b)
But we can also multiply
cos(a)+i sin(b))*(cos(b)+i sin(b)) by distribution and that is:
(cos(a)cos(b) - sin(a)sin(b)) + i(cos(a)sin(b) + sin(b)cos(a))
And we have the formulas for cos(a+b) and sin(a+b)!,
Although the geometrical demonstration for those formulas is really nice too (there is at least one nice geom. dem. that I know of).
As an exercise you can find what happens when b = -a.
Third bonus: if you remember the First bonus a+bi and a-bi have same stretching but opposite rotation angles u and -u say. Remembering a+bi forms a right triangle with legs a and b lets call c its hypotenuse, that is also its stretching factor.
Then the product of a+bi and a-bi has u + (-u) = 0 rotation and stretching c*c = c2 , so it is really c2 , but algebraically
(a+bi)*(a-bi) = a2 - abi + abi - (bi)2 = a2 -(-b2 )=a2 +b2 .
So a2 + b2 = c2
That is Pythagoras theorem.
Obs. While this interpretation makes multiplication easier it does not help when making addition. Happily addition is easy in the algebraic form "a + bi".
Now we are going to see three bonus of looking at complex numbers in this way.
First bonus: Our theory is somewhat arbitrary why we defined i to rotate 90 o counterclockwise ? It could be clockwise couldn't it ? The answer is yes! Lets say you defined a different clockwise imaginary number j, because you didn't like mine, then you would have a complex number theory just like mine with j's in the place of i's! Your drawings would be the reflection of my drawings. But the arithmetic would be the same.
But your j does the same job as my -i! So in a way the -i does the same job as i, but clockwise. Changing i's to -i's changes the orientation of my rotations. So if I take a number like 3+4i, then 3+4(-i) = 3 - 4i has opposite rotation but the same stretching factor. The operation changing i's to -i's is called conjugation and this reasoning explain most of it's properties directly.
Second bonus: Let's say our stretching factor is 1 (no stretching at all) in this case our complex number only rotate vectors by angles a and b. So they are:
cos(a) + i sin(b)
cos(b) + i sin(b)
An their product by our interpretation:
cos(a+b) + i sin(a+b)
But we can also multiply
cos(a)+i sin(b))*(cos(b)+i sin(b)) by distribution and that is:
(cos(a)cos(b) - sin(a)sin(b)) + i(cos(a)sin(b) + sin(b)cos(a))
And we have the formulas for cos(a+b) and sin(a+b)!,
Although the geometrical demonstration for those formulas is really nice too (there is at least one nice geom. dem. that I know of).
As an exercise you can find what happens when b = -a.
Third bonus: if you remember the First bonus a+bi and a-bi have same stretching but opposite rotation angles u and -u say. Remembering a+bi forms a right triangle with legs a and b lets call c its hypotenuse, that is also its stretching factor.
Then the product of a+bi and a-bi has u + (-u) = 0 rotation and stretching c*c = c2 , so it is really c2 , but algebraically
(a+bi)*(a-bi) = a2 - abi + abi - (bi)2 = a2 -(-b2 )=a2 +b2 .
So a2 + b2 = c2
That is Pythagoras theorem.
Complex Numbers 2
Last post Complex Numbers 1, we saw that:
Positive numbers like 3 stretches vectors.
Pure positive i numbers like 2i rotate by 90o and stretches.
Negative numbers like -3 rotate by 180o and stretches.
Pure negative i numbers like -2i rotate a vector by 270o and stretches.
What you guess composite numbers like 1+i or 2+3i do?
Returning with our vector v we can phrase the question as what is (1+i)*v ? It is (1+i)*v = v + iv, by distribution.
The vector iv forms 90o with v, so v + iv forms a square! and the sum is the diagonal of that square. That is v + iv has 45o, with v and is a little longer by a stretching factor of sqrt(2)=1,41 approximately.
And (1+i)2 should rotate by 45o + 45o = 90o and stretch by sqrt(2) * sqrt(2) = 2.
So it should be 2i!! Checking algebraically
(1+i)2 = (1+i)*(1+i) = 1 + 2i + i2 = 1 + 2i -1 = 2i.
Now what is (3+4i)*v ? It is the diagonal of the rectangle with sides 3v and 4iv, so it is a rotation by the angle of the diagonal with v and a stretching by 5 of v. So what is (3+4i)2 ? Before doing the multiplication we now that the angle will be the double of 3+4i and the stretching will be of 25!
Take a complex number z1 that rotates by 30o and stretches by 7 and another z2 that rotates by 120o and stretches by 4 then z1 * z2 rotates by 30o +120o = 150o and stretches by 4*7= 28.
In the last paragraph I didn't say what those numbers were, with a little trigonometry you find that they are:
7(cos(30o) + i sin(30o))
4(cos(120o) + i sin(120o))
28(cos(150o) + i sin(150o))
And that is the rule of sum the angles and multiply the stretching factors.
This way to write complex numbers is called the polar form it is more useful when we want to multiply or divide complex numbers, and it also makes powering much more easier.
Like if we want to calculate
(2(cos(30o) + i sin(30o)))15
We get 215 of stretching and an angle of 15*30o = 450o = 90o since 360o is a complete turn.
More on the next post.
Positive numbers like 3 stretches vectors.
Pure positive i numbers like 2i rotate by 90o and stretches.
Negative numbers like -3 rotate by 180o and stretches.
Pure negative i numbers like -2i rotate a vector by 270o and stretches.
What you guess composite numbers like 1+i or 2+3i do?
Returning with our vector v we can phrase the question as what is (1+i)*v ? It is (1+i)*v = v + iv, by distribution.
The vector iv forms 90o with v, so v + iv forms a square! and the sum is the diagonal of that square. That is v + iv has 45o, with v and is a little longer by a stretching factor of sqrt(2)=1,41 approximately.
And (1+i)2 should rotate by 45o + 45o = 90o and stretch by sqrt(2) * sqrt(2) = 2.
So it should be 2i!! Checking algebraically
(1+i)2 = (1+i)*(1+i) = 1 + 2i + i2 = 1 + 2i -1 = 2i.
Now what is (3+4i)*v ? It is the diagonal of the rectangle with sides 3v and 4iv, so it is a rotation by the angle of the diagonal with v and a stretching by 5 of v. So what is (3+4i)2 ? Before doing the multiplication we now that the angle will be the double of 3+4i and the stretching will be of 25!
Take a complex number z1 that rotates by 30o and stretches by 7 and another z2 that rotates by 120o and stretches by 4 then z1 * z2 rotates by 30o +120o = 150o and stretches by 4*7= 28.
In the last paragraph I didn't say what those numbers were, with a little trigonometry you find that they are:
7(cos(30o) + i sin(30o))
4(cos(120o) + i sin(120o))
28(cos(150o) + i sin(150o))
And that is the rule of sum the angles and multiply the stretching factors.
This way to write complex numbers is called the polar form it is more useful when we want to multiply or divide complex numbers, and it also makes powering much more easier.
Like if we want to calculate
(2(cos(30o) + i sin(30o)))15
We get 215 of stretching and an angle of 15*30o = 450o = 90o since 360o is a complete turn.
More on the next post.
Complex Numbers 1
In this post we will try to find some light in those strange complex numbers. There is a nice way to present them as rotations and stretchings of vectors, where complex numbers concepts and properties are all very natural, look for that later in this post.
Let's begin by telling my first experiences with complex numbers.
I think my first impressions were based on those mysterious names, after all we would learn about "complex numbers", about the number " i " that was the "square root of -1", that the teacher said before it didn't existed, and also about the "imaginary" numbers in opposition to the more "real" numbers learned before.
The names "complex numbers" and "imaginary numbers" were not a good choice of words from the pedagogical point of view. Well, "i" is the number that squared is -1, what is the logic in that? Lets make some experiments.
For example: 32 = 9 and 52 = 25 so positive numbers squared are positive. (-3)2 = (-3)x(-3) = 9, because negative times negative is positive, so negative numbers squared are also positive!
Concluding that all numbers squared are positive! So no number squared is negative! And in particular no number squared is -1! at worst it is 0.
So they actually invented a number " i " not a "normal number" that squared would be -1, that is what my teacher said to me.
After some algebra facts they let you know that you can plot a complex number like " 3+ 4i " in an x-y axis were the point (3,4) is, unfortunately that was not so helpful. Later on that class my teacher showed me lots of even more bizarre properties of complex numbers.
All that together was a dazzling sequence of "out of the blue" concepts and relations. Actually, complex numbers have a pretty decent history, real flesh and bone people invented it you know, and that history is not "out of the blue" I may tell something of that story latter.
Now I will present complex numbers in a different and I hope better way, first lets explore vectors.
A vector can represent something as velocity or force on a body and lots of other concepts. Lets think of it as a force on an object (which I am not drawing) to fix ideas.
When we apply two forces on an object what happens?
We have a resulting force! That depends on the strength of the original ones and on their direction. There is a way to add them, say u = AB v = AC move the vector u (maintaining direction) to the endpoint of the other C making a vector CD the sum u + v is AD that is the parallelogram law.
If we add v + v we have a vector twice the size in the same direction or v + v = 2v, also, v + v + v = 3v is the vector in the same direction three times longer. 1.5v is the vector with 1.5 the size, -v = -1v is the vector with opposite direction and same size (v + -v = 0), and -5v has opposite direction and is 5 times the size of v.
So when we multiply by a positive number we "stretch" the vector, or maybe shrink it if we multiply by a number smaller than 1, and when we multiply by a negative number the vector inverts its direction and is stretched appropriately.
If we multiply by 3 and after that multiply by 2 we get a vector six times larger, or just 2(3v) = 6v.
Imagine what we get with -3(4v), 5(-1v) and -3(-2v)? Think geometrically!!
-12v, -5v and 6v respectively
Obs. A negative number times a negative number should be positive since we invert the vector two times! - a good justification for that rule/theorem in my opinion.
If we put another vector w somewhere else multiplying by real numbers would also work exactly in the same way.
We can also think that multiplying by -1 rotates vectors by 180o degrees, in that spirit we could rotate vectors by other angles like 90o or 120o . Let's say the number "j" rotates vectors by 90o degrees counterclockwise.
If we multiply our vector by j two times we rotate it by:
90o + 90o = 180o that is -v, or
j*jv = -v = -1v
So j*j = j2 should be -1, and we have a number that squared is -1! j is really our imaginary number i, time for more experiments.
2i rotates 90o and multiplies or stretches by 2
2i*2i rotates by 90o + 90 o =180 o and multiplies by 2x2=4 or 2ix2i = -4.
-i rotates by 90o and inverts the direction so it rotates by 270o .
i3 = i*i*i rotates by 90o + 90o + 90o = 270o also, so i3 = -i.
i4 = i*i*i*i rotates 4*90o = 360o so i4 = 1 = i2 * i2 .
You can guess i5 , i6 , and i37 , note that the algebra coincides with our geometrical ideas.
All properties of complex numbers can be understood in this geometric "rotation-stretching" way, more on that on next post.
Let's begin by telling my first experiences with complex numbers.
I think my first impressions were based on those mysterious names, after all we would learn about "complex numbers", about the number " i " that was the "square root of -1", that the teacher said before it didn't existed, and also about the "imaginary" numbers in opposition to the more "real" numbers learned before.
The names "complex numbers" and "imaginary numbers" were not a good choice of words from the pedagogical point of view. Well, "i" is the number that squared is -1, what is the logic in that? Lets make some experiments.
For example: 32 = 9 and 52 = 25 so positive numbers squared are positive. (-3)2 = (-3)x(-3) = 9, because negative times negative is positive, so negative numbers squared are also positive!
Concluding that all numbers squared are positive! So no number squared is negative! And in particular no number squared is -1! at worst it is 0.
So they actually invented a number " i " not a "normal number" that squared would be -1, that is what my teacher said to me.
After some algebra facts they let you know that you can plot a complex number like " 3+ 4i " in an x-y axis were the point (3,4) is, unfortunately that was not so helpful. Later on that class my teacher showed me lots of even more bizarre properties of complex numbers.
All that together was a dazzling sequence of "out of the blue" concepts and relations. Actually, complex numbers have a pretty decent history, real flesh and bone people invented it you know, and that history is not "out of the blue" I may tell something of that story latter.
Now I will present complex numbers in a different and I hope better way, first lets explore vectors.
A vector can represent something as velocity or force on a body and lots of other concepts. Lets think of it as a force on an object (which I am not drawing) to fix ideas.
When we apply two forces on an object what happens?
We have a resulting force! That depends on the strength of the original ones and on their direction. There is a way to add them, say u = AB v = AC move the vector u (maintaining direction) to the endpoint of the other C making a vector CD the sum u + v is AD that is the parallelogram law.
If we add v + v we have a vector twice the size in the same direction or v + v = 2v, also, v + v + v = 3v is the vector in the same direction three times longer. 1.5v is the vector with 1.5 the size, -v = -1v is the vector with opposite direction and same size (v + -v = 0), and -5v has opposite direction and is 5 times the size of v.
So when we multiply by a positive number we "stretch" the vector, or maybe shrink it if we multiply by a number smaller than 1, and when we multiply by a negative number the vector inverts its direction and is stretched appropriately.
If we multiply by 3 and after that multiply by 2 we get a vector six times larger, or just 2(3v) = 6v.
Imagine what we get with -3(4v), 5(-1v) and -3(-2v)? Think geometrically!!
-12v, -5v and 6v respectively
Obs. A negative number times a negative number should be positive since we invert the vector two times! - a good justification for that rule/theorem in my opinion.
If we put another vector w somewhere else multiplying by real numbers would also work exactly in the same way.
We can also think that multiplying by -1 rotates vectors by 180o degrees, in that spirit we could rotate vectors by other angles like 90o or 120o . Let's say the number "j" rotates vectors by 90o degrees counterclockwise.
If we multiply our vector by j two times we rotate it by:
90o + 90o = 180o that is -v, or
j*jv = -v = -1v
So j*j = j2 should be -1, and we have a number that squared is -1! j is really our imaginary number i, time for more experiments.
2i rotates 90o and multiplies or stretches by 2
2i*2i rotates by 90o + 90 o =180 o and multiplies by 2x2=4 or 2ix2i = -4.
-i rotates by 90o and inverts the direction so it rotates by 270o .
i3 = i*i*i rotates by 90o + 90o + 90o = 270o also, so i3 = -i.
i4 = i*i*i*i rotates 4*90o = 360o so i4 = 1 = i2 * i2 .
You can guess i5 , i6 , and i37 , note that the algebra coincides with our geometrical ideas.
All properties of complex numbers can be understood in this geometric "rotation-stretching" way, more on that on next post.
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