One of the interesting things about Islam is that it is a religion that has "protected" itself from criticism by explicitly stating that the mere act of challenging the Koran (i.e. word of God) is a great sin in of itself. You are not supposed to "try to find loopholes" or "errors" in the Koran, since those kinds of things cannot exist, and that if you think for any reason that you have found a discrepancy, then it is simply Satan convincing you as such.
Having been raised as a devout Muslim for the first 17 years of my life, I of course, never questioned any Islamic teachings that I received. Every thing that was taught to me was the Truth, and if I didn't believe it, I would go to hell. Nice and simple.
Well, I guess it's time to sin. I have recently been analyzing the Koran more and more and finding some interesting things. Let's jump right into a simple question:
Assuming Islam is correct, will Christians go to Heaven or Hell when they die?
A simple word search on the text of the Koran for the word "Christians" brings up occurrences in 12 different aya(s) or verses:
(Search Results)
The weird part is that those 12 ayas are quite contradictory semantically. Take a look at aya 30 from chapter 9:
[9.30] And the Jews say: Uzair is the son of Allah; and the Christians say: The Messiah is the son of Allah; these are the words of their mouths; they imitate the saying of those who disbelieved before; may Allah destroy them; how they are turned away!
That is clearly a negative statement about Christians. It doesn't get much shittier than "may Allah destroy them".
Here are a few more Christian-damning ayas:
[2.135] And they say: Be Jews or Christians, you will be on the right course. Say: Nay! (we follow) the religion of Ibrahim, the Hanif, and he was not one of the polytheists.
[5.51] O you who believe! do not take the Jews and the Christians for friends; they are friends of each other; and whoever amongst you takes them for a friend, then surely he is one of them; surely Allah does not guide the unjust people.
Now on the other hand, consider the following two ayas:
[2.62] Surely those who believe, and those who are Jews, and the Christians, and the Sabians, whoever believes in Allah and the Last day and does good, they shall have their reward from their Lord, and there is no fear for them, nor shall they grieve.
[5.69] Surely those who believe and those who are Jews and the Sabians and the Christians whoever believes in Allah and the last day and does good-- they shall have no fear nor shall they grieve.
Both ayas are very similar to each other and, more importantly, they both claim that it's ok to be Christian.
Then you have some ayas that are just confusing like this one:
[5.82] Certainly you will find the most violent of people in enmity for those who believe (to be) the Jews and those who are polytheists, and you will certainly find the nearest in friendship to those who believe (to be) those who say: We are Christians; this is because there are priests and monks among them and because they do not behave proudly.
What in Allah's name is going on here? Is this just some kind of Arabic-English translation error? A close look at the Arabic script for those ayas reveals that no, there isn't a translation error, and that the original Arabic Koranic text for those ayas does indeed have the same exact semantics as the English translations shown above.
What is even more scientifically frustrating is the fact that discussing such an obvious contradiction with Muslim friends or family results in an immediate panic over how angry Allah must be right now since we are challenging his words.
Hasn't this been seen by many people before me? Why hasn't anyone mentioned anything about this to me when I was growing up? Well, I am sure that people did come across this contradiction (amongst others), but decided that they only think it's a contradiction because Satan is playing with their heads, or because they are not worthy of understanding what Allah truly means.
Give me a break.
Until next time,
--Shafik
Tuesday, October 14, 2008
Wednesday, October 8, 2008
What *is* the square-root of -1?
Have you ever wondered why complex numbers exist? Have you ever wondered why the hell j == square-root of -1? Were you ever dissatisfied by the explanation your math or engineering profs gave you in college? If you are mathematically inclined (you need to have studied a significant amount of college-level mathematics) and are curious about the way the world works, keep reading.
In math we have the concept of the real number line ... a bunch of numbers that lie in a straight line that stretches from negative infinity to positive infinity. These are 1-D numbers:
. . . . . - 3, -2, -1, 0, 1, 2, 3, . . . . . .
We can easily wrap our heads around this concept. Now, what if we consider 2-D numbers? Numbers that are simply a pair of "x-y" coordinates? Well, this was a little harder to grasp in highschool, but we still understood it relatively easily:
Perhaps we can call 2-D numbers the special word 2D vectors. Where a vector is simply two numbers: (x, y), where x denotes the horizontal portion of the vector, and y denotes the vertical portion.
Ok, so far so good. Hold on to your hats, we are about to get way more technical.
Consider what a "2D rotation" operation in computer graphics is: it is a linear transformation and hence can be represented as a matrix multiplication operation. It is well known that rotation by an angle theta in the 2D field is left-multiplication by this 2x2 matrix:
[ cos(theta) -sin(theta) ]
[ sin(theta) cos(theta) ]
What if we set theta == pi/2 ? (90 degrees counter-clockwise) then that matrix, call it A, becomes:
Viola. Now we can rotate any point, (x, y), by left-multiplying it with this matrix A:
where (w, z) is the resultant vector after rotation.
Now, let's do something very common for square matrices: let's compute the eigen values for this matrix A:
Remember, the eigen values of a matrix, are scalar values such that multiplying a vector by those scalar values is the same as multiplying that the matrix with that same vector.
In other words, the eigen values, are all values such that:
Ax = ex
Where A is the matrix whose eigenvalues we are trying to find, and e is the scalar eigen-value we are trying to solve for. x is any non-null (non-zero) vector.
So, from that equation we have:
(A-eI)x = 0
Where "I" is the 2x2 identity matrix.
Ok, since x is not a zero vector, we have:
det(A - eI) = 0, or the determinant of
Now we have to simply find the roots of the characteristic polynomial of this matrix, to find the value of e:
e^2 + 1 == 0, or e = sqrt(-1)
We have arrived at the most beautiful conclusion ever: without ever mentioning anything about complex numbers, abstract algebra, Abelian groups, Euler's formulae, etc, we find ourselves in need of defining the quantity sqrt(-1).
Again, remember what an eigen value really means in this situation: this is saying, to rotate a vector by 90 degrees in 2D-space we can either multiply by this well known matrix, A, or we can multiply by a mysterious quantity, sqrt(-1).
Of course, it makes sense that the eigen value for a rotation matrix wouldn't be a simple real scalar, since a real number multiplied with a vector would simply scale that vector's magnitude, not rotate it in any way. Cool eh?
We can see from this that the fact that j (or i) == sqrt(-1) is not merely a convention that some mathematician came up with ... it is a fundamental part of dealing with numbers. Much like how negative numbers were "discovered" at a point in time where they might have been seen as "useless".
So. How we find a mystery number, m, that when multiplied by itself twice gives us the sqrt(-1)? Easy. We think in two dimensions.
We are basically looking for 1 x m x m = -1
The number 1 represented in 2 dimensions is the vector (0, 1). We know that the vector (0, -1) is the same as the vector (0, 1) but rotated 180 degrees. So, we rotate (0, 1) by 90 degrees twice to get to (0, -1). How do we "rotate by 90 degrees" again? Easy ... we just showed, it's by multiplying by "sqrt(-1)". Hence we have m = j. Or more clearly, the number that represents a 90 degree counter-clockwise rotation is simply this new construct "j".
One last question to take care of: Why can we use j in normal algebraic equations and treat it just like any other ordinary number? Magic? No. The reason is subtle: We arrived at the need for a "real number that is equal to the sqrt(-1)" when we were figuring out the eigen value of that rotation Matrix A. Of course, no such number exists, but if it did exist, then it just be yet another number! We can multiply/divide/add/subtract it just like we can multiply/divide/add/subtract the number 7. The only caveat is that is a rotation and hence lives on a different (and orthogonal) axis to the real-number axis. No problem, we just have to start thinking in 2D when we need the concept of sqrt(-1). There is a very real connection though between j and the real numbers, and that connection is precisely that, when we square j (i.e. j^2), we get -1, a negative number that we are very familiar with.
That's it!
If you understood this post then congratualtions: you have just linked the two seemingly unrelated fields in mathematics: Linear Algebra, and Complex Algebra.
Until next time,
--Shafik
In math we have the concept of the real number line ... a bunch of numbers that lie in a straight line that stretches from negative infinity to positive infinity. These are 1-D numbers:
. . . . . - 3, -2, -1, 0, 1, 2, 3, . . . . . .
We can easily wrap our heads around this concept. Now, what if we consider 2-D numbers? Numbers that are simply a pair of "x-y" coordinates? Well, this was a little harder to grasp in highschool, but we still understood it relatively easily:
.
.
3
2
1
... -3, -2, -1, 0, 1, 2, 3 ...
-1
-2
-3
.
.
Perhaps we can call 2-D numbers the special word 2D vectors. Where a vector is simply two numbers: (x, y), where x denotes the horizontal portion of the vector, and y denotes the vertical portion.
Ok, so far so good. Hold on to your hats, we are about to get way more technical.
Consider what a "2D rotation" operation in computer graphics is: it is a linear transformation and hence can be represented as a matrix multiplication operation. It is well known that rotation by an angle theta in the 2D field is left-multiplication by this 2x2 matrix:
[ cos(theta) -sin(theta) ]
[ sin(theta) cos(theta) ]
What if we set theta == pi/2 ? (90 degrees counter-clockwise) then that matrix, call it A, becomes:
[ 0 -1]
[ 1 0]
Viola. Now we can rotate any point, (x, y), by left-multiplying it with this matrix A:
[ 0 -1] [x] == [w]
[ 1 0] [y] [z]
where (w, z) is the resultant vector after rotation.
Now, let's do something very common for square matrices: let's compute the eigen values for this matrix A:
Remember, the eigen values of a matrix, are scalar values such that multiplying a vector by those scalar values is the same as multiplying that the matrix with that same vector.
In other words, the eigen values, are all values such that:
Ax = ex
Where A is the matrix whose eigenvalues we are trying to find, and e is the scalar eigen-value we are trying to solve for. x is any non-null (non-zero) vector.
So, from that equation we have:
(A-eI)x = 0
Where "I" is the 2x2 identity matrix.
Ok, since x is not a zero vector, we have:
det(A - eI) = 0, or the determinant of
[ -e -1]
[1 -e] == 0
Now we have to simply find the roots of the characteristic polynomial of this matrix, to find the value of e:
e^2 + 1 == 0, or e = sqrt(-1)
We have arrived at the most beautiful conclusion ever: without ever mentioning anything about complex numbers, abstract algebra, Abelian groups, Euler's formulae, etc, we find ourselves in need of defining the quantity sqrt(-1).
Again, remember what an eigen value really means in this situation: this is saying, to rotate a vector by 90 degrees in 2D-space we can either multiply by this well known matrix, A, or we can multiply by a mysterious quantity, sqrt(-1).
Of course, it makes sense that the eigen value for a rotation matrix wouldn't be a simple real scalar, since a real number multiplied with a vector would simply scale that vector's magnitude, not rotate it in any way. Cool eh?
We can see from this that the fact that j (or i) == sqrt(-1) is not merely a convention that some mathematician came up with ... it is a fundamental part of dealing with numbers. Much like how negative numbers were "discovered" at a point in time where they might have been seen as "useless".
So. How we find a mystery number, m, that when multiplied by itself twice gives us the sqrt(-1)? Easy. We think in two dimensions.
We are basically looking for 1 x m x m = -1
The number 1 represented in 2 dimensions is the vector (0, 1). We know that the vector (0, -1) is the same as the vector (0, 1) but rotated 180 degrees. So, we rotate (0, 1) by 90 degrees twice to get to (0, -1). How do we "rotate by 90 degrees" again? Easy ... we just showed, it's by multiplying by "sqrt(-1)". Hence we have m = j. Or more clearly, the number that represents a 90 degree counter-clockwise rotation is simply this new construct "j".
One last question to take care of: Why can we use j in normal algebraic equations and treat it just like any other ordinary number? Magic? No. The reason is subtle: We arrived at the need for a "real number that is equal to the sqrt(-1)" when we were figuring out the eigen value of that rotation Matrix A. Of course, no such number exists, but if it did exist, then it just be yet another number! We can multiply/divide/add/subtract it just like we can multiply/divide/add/subtract the number 7. The only caveat is that is a rotation and hence lives on a different (and orthogonal) axis to the real-number axis. No problem, we just have to start thinking in 2D when we need the concept of sqrt(-1). There is a very real connection though between j and the real numbers, and that connection is precisely that, when we square j (i.e. j^2), we get -1, a negative number that we are very familiar with.
That's it!
If you understood this post then congratualtions: you have just linked the two seemingly unrelated fields in mathematics: Linear Algebra, and Complex Algebra.
Until next time,
--Shafik
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