Saturday, November 29, 2014

Letter to Hom on Christian Salvation

Alright, now we do a bit of theology! Aren't you excited? From math to quantum to cosmology to the heavenly Father of Jesus! Quite a healthy wide range of things for the spirit to grow, wouldn't you say? ;-)

This specific post/thread is for dear brother Hom on twitter who wrote me to discuss and share my thoughts on the issue of whether a Christian who is saved can lose or keep his/her salvation after numerous sins post their commitment to faith in Christ. In his tweets to me, Hom said:
"I was taught that in my early years and I "kinda" maintain that belief. Yet I know NO passages in the Biblical text that make the point clear. Can you help?"
The belief that Hom alluded to here is the view I expressed in our twitter discussion that there can be cases of alleged Christians who could 'lose' their salvation, or grace, by the continued sinful lives that they can lead subsequently -- even as replete that life may be with sins of egregious proportions (such as committing genocide, indulging in the major sins stated in the Bible, etc, without naming them all).

Some believers are of the view that once you have given your life to Christ as your Savior, you are also saved of your subsequent sins too, no matter how big or small they may be after your pronounced belief. Consequently, even if you live a life of crime, rape, murder, stealing, lying, and any sin you can imagine (so much for the Ten Commandments!), you are still saved because you cannot lose it. So consequently, Hitler, being a Christian would still be saved by grace even after all the inhumanity and genocide that he has caused to tens of millions of human beings. So, you are free to do as you please, sin to whatever degree that you wish (to whatever extreme), and you are assured that you are saved and will go to heaven.

Of course, I cannot accept such a view. (And I never have.) That is not at all how I read the Bible, especially the teachings of the New Testament as to how Christians should conduct their lives. (The Hebrew Scriptures already prescribe divine punishment to various of these sins even for alleged believers of the Mosaic community!) Indeed, St Paul has dedicated a fair amount of his time, travels, and writings to some churches in Asia Minor where he heard that numerous egregious sins continue to be committed by (alleged) Christian members who believed that they had been saved so that now they can do whatever they wanted. That is why St Paul explicitly emphasized:
"What shall we say, then? Shall we go on sinning so that grace may increase? By no means! We are those who have died to sin; how can we live in it any longer? Or don’t you know that all of us who were baptized into Christ Jesus were baptized into his death? We were therefore buried with him through baptism into death in order that, just as Christ was raised from the dead through the glory of the Father, we too may live a new life." (Romans 6:1-4, NIV.)
The Letter of James also teaches that a Christian is responsible to showing how his/her life conduct is to be exemplified through their behavior, actions, or works -- backed, of course, through their faith in Christ.

The Lord Jesus taught us to judge a tree by the fruit that it bears - very eloquently, simply, and without any complicated theology. When the tree produces bad fruit, what is to be done unto it? The Master said
"Every tree that does not bear good fruit is cut down and thrown into the fire. 20 Thus, by their fruit you will recognize them." (Matthew 7:19-20, NIV.)
Clearly, such a tree cannot then have been saved or ascribed salvation to begin with, even if that tree lead others to believe that it was a good tree.

Therefore, in extreme cases like Hitler, or anyone allegedly claiming to be a Christian, such continued bearing of bad fruit would at the very least cast serious doubt on their claims to being believers. Would we believe someone who claims allegiance to the US Constitution only to see that individual violate its articles and laws time and again (even in extreme ways)? I would certainly at least question them. So we're not saying that they had grace and then lost it, but that maybe they didn't have grace in the first place (as we assumed by taking their claim at face value).

There are many examples like the above in the Bible that tell me that the view I expressed is far more reasonable (or at least less problematic) than the view that the Christian community could be allowed to harbor such horrific individuals who do such harm to the faith. If Christians are serious about Jesus' teaching, they are responsible to act it out in their hearts and minds as well as with their fellow man, their neighbor. I hope that I have shared my thoughts with you, Hom, in a gentle spirit, even as I am no Bible teacher nor do I have a degree in theology! But I speak as just one believer, sharing my thoughts and experiences. Ultimately, Jesus knows the full precise answers. As St Paul said, we know in part and we prophecy in part, and in another place he says "For now we see through a glass, darkly."

Yours in Christ,

Saturday, November 8, 2014

A game with $\pi$

Here's an image of something I wrote down, took a photo of, and posted here for you.

It's a little game you can play with any irrational number. I took $\pi$ as an example.

You just learned about an important math concept/process called continued fraction expansions.

With it, you can get very precise rational number approximations for any irrational number to whatever degree of error tolerance you wish.

As an example, if you truncate the above last expansion where the 292 appears (so you omit the "1 over 292" part) you get the rational number $\frac{355}{113}$ which approximates $\pi$ to 6 decimal places. (Better than $\frac{22}{7}$.)

You can do the same thing for other irrational numbers like the square root of 2 or 3. You get their own sequences of whole numbers.

Exercise: for the square root of 2, show that the sequence you get is
1, 2, 2, 2, 2, ...
(all 2's after the 1). For the square root of 3 the continued fraction sequence is
1, 1, 2, 1, 2, 1, 2, 1, 2, ... 
(so it starts with 1 and then the pair "1, 2" repeat periodically forever).

Monday, August 18, 2014

3-4-5 complex number has Infinite order

This is a good exercise/challenge with complex numbers.

Consider the complex number  $Z = \large \frac35 + \frac45 i$. (Where $\large i = \sqrt{-1}$.)

Prove that $Z^n$ is never equal to 1 for any positive whole number $n = 1, 2, 3, 4, \dots $.

This complex number $Z$ comes from the familiar 3-4-5 right triangle that you all know: $3^2 + 4^2 = 5^2$.

In math we sometimes say that an object $X$ has "infinite order" when no positive power of it can be the identity (1, in this multiplicative case). For example, $i$ itself has finite order 4 since $i^4 = 1$, while 2 has infinite order since no positive power of 2 can be equal to 1. The distinct feature of $Z$ above is that it has modulus 1, so is on the unit circle $\mathbb T$ in the complex plane.

Wednesday, July 30, 2014

Multiplying Spaces!

Believe it or not, in Math we can not only multiply numbers but we can multiply spaces! We can multiply two spaces to get bigger spaces - usually of bigger dimensions.

The 'multiplication' that I'm referring to here is known as Tensor products. The things/objects in these spaces are called tensors. (Tensors are like vectors in a way.)

Albert Einstein used tensors in his Special and his General Theory of Relativity (his theory of gravity). Tensors are also used in several branches of Physics, like the theory of elasticity where various stresses and forces act in various ways. And definitely in quantum field theory.

It may sound crazy to say you can "multiply spaces," as we would multiply numbers, but it can be done in a precise and logical way. But here I will spare you the technical details and try to manage to show you the idea that makes it possible to do.

Q. What do you mean by 'spaces'?

I mean a set of things that behave like 'vectors' so that you can add two vectors and get a third vector, and where you can scale a vector by any real number. The latter is called scalar multiplication, so if $v$ is a vector, you can multiply it by $0.23$ or by $-300.87$ etc and get another vector: $0.23v$, $-300.87v$, etc.) The technical name is vector space.

A straight line that extends in both directions indefinitely would be a good example (an Euclidean line).

Another example is you take the $xy$-plane, 2D-space or simply 2-space, or you can take $xyz$-space, or if you like you can take $xyzt$-spacetime known also as Minkowski space which has 4 dimensions.

Q. How do you 'multiply' such spaces?

First, the notation. If $U$ and $V$ are spaces, their tensor product space is written as $U \otimes V$. (It's the multiplication symbol with a circle around it.)

If this is to be an actual multiplication of spaces there is one natural requirement we would want. That the dimensions of this tensor product space $U \otimes V$ should turn out to be the multiplication of the dimensions of U and of V.

So if $U$ has dimension 2 and $V$ has dimension 3, then $U \otimes V$ ought to have dimension $2 \times 3 = 6$.  And if $U$ and $V$ are straight lines, so each of dimension 1, then $U \otimes V$ will also be of dimension 1.

Q. Hey, wait a second, that doesn't quite answer my question. Are you dodging the issue!?

Ha! Yeah, just wanted to see if you're awake! ;-) And you are! Ok, here's the deal without going into too much detail. We pointed out above how you can scale vectors by real numbers. So if you have a vector $v$ from the space $V$ you can scale it by $0.23$ and get the vector $0.23v$. Now just imagine if we can scale the vector $v$ by the vectors in the other space $U$! So if $u$ is a vector from $U$ and $v$ a vector from $V$, then you can scale $v$ by $u$ to get what we call their tensor product which we usually write like

$u \otimes v$.

So with numbers used to scale vectors, e.g. $0.23v$, we could also write it as $0.23 \otimes v$. But we don't normally write it that way when numbers are involved, only when non-number vectors are.

Q. So can you also turn this around and refer to $u \otimes v$ as the vector $u$ scaled by the vector $v$?

Absolutely! So we have two approaches to this and you can show (by a proof) that the two approaches are in fact equivalent. In fact, that's what gives rise to a theorem that says

Theorem. $U \otimes V$ is isomorphic to $V \otimes U$.

(In Math, the word 'isomorphism' gives a precise meaning to what I mean by 'equivalent'.)

Anyway, the point has been made to describe multiplying spaces: you take their vectors and you 'scale' those of one space by the vectors of the other space.

There's a neat way to actually see and appreciate this if we use matrices as our vectors. (Yes, matrices can be viewed as vectors!) Matrices are called arrays in computer science.

One example / experiment should drive the point home:

Let's take these two $2 \times 2$ matrices $A$ and $B$:

$A = \begin{bmatrix} 2 & 3 \\ -1 & 5 \end{bmatrix}, \ \ \ \ \ \ \  B = \begin{bmatrix} -5 & 4 \\ 6 & 7 \end{bmatrix}$

To calculate their tensor product $A \otimes B$, you can take $B$ and scale it by each of the numbers contained in $A$! Like this:

$A\otimes B = \begin{bmatrix} 2B & 3B \\ -1B & 5B \end{bmatrix}$

If you write this out you will get a 4 x 4 matrix when you plug B into it:

$A\otimes B = \begin{bmatrix} -10 & 8 & -15 & 12 \\ 12 & 14 & 18 & 21 \\ 5 & -4 & -25 & 20 \\ -6 & -7 & 30 & 35 \end{bmatrix}$

Oh, and 4 times 4 is 16, yes so the matrix $A\otimes B$ does in fact have 16 entries in it! Check!

Q. You could also do this the other way, by scaling $A$ using each of the numbers in $B$, right?

Right! That would then give $B\otimes A$.

When you do this you will get different matrices/arrays but if you look closely you'll see that they have the very same set of numbers except that they're permuted around in a rather simple way.  How? Well, if you switch the two inner columns and the two inner rows of $B\otimes A$ you will get exactly $A\otimes B$!

Try this experiment with the above $A$ and $B$ examples by working out $B\otimes A$ as we've done. This illustrates what we mean in Math by 'isomorphism': that even though the results may look different, they are actually related to one another in a sort of 'linear' or 'algebraic' fashion.

Ok, that's enough. We get the idea. You can multiply spaces by scaling their vectors by each other. Amazing how such an abstract idea turns out to be a powerful tool in understanding the geometry of spaces, in Relativity Theory, and also in quantum mechanics (quantum field theory).

Warm Regards,

Saturday, July 26, 2014

Bertrand's "postulate" and Legendre's Conjecture

Bertrand's "postulate" states that for any positive integer $n > 1$, you can always find a prime number $p$ in the interval

$n < p < 2n$.

It use to be called "postulate" until it became a theorem when Chebyshev proved it in 1850.

(I saw this while browsing thru a group theory book and got interested to read up a little more.)

What if instead of looking at $n$ and $2n$ you looked at consecutive squares? So for example you take a positive integer $n$ and you ask whether we can always find at least one prime number between $n^2$ and $(n+1)^2$.

Turns out this is a much harder problem and it's still an open question called:

Legendre's Conjecture. For each positive integer $n$ there is at least one prime $p$ such that

$n^2 < p < (n+1)^2$.

People have used programming to check this for large numbers and have always found such primes, but no proof (or counterexample) is known.

If you compare Legendre's with Bertrand's you will notice that $(n+1)^2$ is a lot less than $2n^2$. (At least for $n > 2$.) In fact, the asymptotic ratio of the latter divided by the former is 2 (not 1) for large $n$'s. This shows that the range of numbers in the Legendre case is much narrower than in Bertrand's.

The late great mathematician Erdos proved similar results by obtaining k primes in certain ranges similar to Bertand's.

A deep theorem related to this is the Prime Number Theorem which gives an asymptotic approximation for the number of primes up to $x$. That approximating function is the well-known $x/\ln(x)$.

Great sources:
[1] Bertrand's "postulate"
[2] Legendre's Conjecture
(See also wiki's entries under these topics.)

Friday, July 25, 2014

Direct sum of finite cyclic groups

The purpose of this post is to show how a finite direct sum of finite cyclic groups

$\Large \Bbb Z_{m_1} \oplus \Bbb Z_{m_2} \oplus \dots \oplus \Bbb Z_{m_n}$

can be rearranged so that their orders are in increasing divisional form: $m_1|m_2|\dots | m_n$.

We use the fact that if $p, q$ are coprime, then $\large \Bbb Z_p \oplus \Bbb Z_q = \Bbb Z_{pq}$.

(We'll use equality $=$ for isomorphism $\cong$ of groups.)

Let $p_1, p_2, \dots p_k$ be the list of prime numbers in the prime factorizations of all the integers $m_1, \dots, m_n$.

Write each $m_j$ in its prime power factorization $\large m_j = p_1^{a_{j1}}p_2^{a_{j2}} \dots p_k^{a_{jk}}$. Therefore

$\Large \Bbb Z_{m_j} = \Bbb Z_{p_1^{a_{j1}}} \oplus \Bbb Z_{p_2^{a_{j2}}} \oplus \dots \oplus \Bbb Z_{p_k^{a_{jk}}}$

and so the above direct sum  $\large \Bbb Z_{m_1} \oplus \Bbb Z_{m_2} \oplus \dots \oplus \Bbb Z_{m_n}$ can be written out in matrix/row form as the direct sum of the following rows:

$\Large\Bbb Z_{p_1^{a_{11}}} \oplus \Bbb Z_{p_2^{a_{12}}} \oplus \dots \oplus \Bbb Z_{p_k^{a_{1k}}}$

$\Large\Bbb Z_{p_1^{a_{21}}} \oplus \Bbb Z_{p_2^{a_{22}}} \oplus \dots \oplus \Bbb Z_{p_k^{a_{2k}}}$
$\Large \vdots$
$\Large\Bbb Z_{p_1^{a_{n1}}} \oplus \Bbb Z_{p_2^{a_{n2}}} \oplus \dots \oplus \Bbb Z_{p_k^{a_{nk}}}$

Here, look at the powers of $p_1$ in the first column. They can be permuted / arranged so that their powers are in increasing order. The same with the powers of $p_2$ and the other $p_j$, arrange their groups so that the powers are increasing order. So we get the above direct sum isomorphic to

$\Large\Bbb Z_{p_1^{b_{11}}} \oplus \Bbb Z_{p_2^{b_{12}}} \oplus \dots \oplus \Bbb Z_{p_k^{b_{1k}}}$

$\Large\Bbb Z_{p_1^{b_{21}}} \oplus \Bbb Z_{p_2^{b_{22}}} \oplus \dots \oplus \Bbb Z_{p_k^{b_{2k}}}$
$\Large \vdots$
$\Large\Bbb Z_{p_1^{b_{n1}}} \oplus \Bbb Z_{p_2^{b_{n2}}} \oplus \dots \oplus \Bbb Z_{p_k^{b_{nk}}}$

where, for example, the exponents $b_{11} \le b_{21} \le \dots \le b_{n1}$ are a rearrangement of the numbers $a_{11}, a_{21}, \dots, a_{n1}$ (in the first column) in increasing order.  Do the same for the other columns.

Now put together each of these rows into cyclic groups by multiplying their orders, thus

$\Large\ \ \Bbb Z_{N_1}$
$\Large \oplus \Bbb Z_{N_2}$
$\Large \vdots$
$\Large \oplus \Bbb Z_{N_n}$


$\large N_1 = p_1^{b_{11}} p_2^{b_{12}} \dots p_k^{b_{1k}}$,
$\large N_2 = p_1^{b_{21}} p_2^{b_{22}} \dots p_k^{b_{2k}}$,
$\large \vdots$
$\large N_n = p_1^{b_{n1}} p_2^{b_{n2}} \dots p_k^{b_{nk}}$.

In view of the fact that the $b_{1j} \le b_{2j} \le \dots \le b_{nj}$ is increasing for each $j$, we see that $N_1 | N_2 | \dots | N_n$, as required. $\blacksquare$

Latex on Blogger

LaTeX work here if you add the small script below:

Short exact sequence:
$\large 0 \to H \to G \to G/H \to 0$
and an integral
$\large \int \sqrt{x} dx$.

Each finite Abelian group is isomorphic to a direct sum of cyclic groups
$\large \Bbb Z_{m_1} \oplus \Bbb Z_{m_2} \oplus \dots \oplus \Bbb Z_{m_n}$
where $m_1|m_2|\dots | m_n$.

(One of my favorite results from group theory.)

Thanks to a gentle soul's responding at tex.stackexchange:

To get LaTeX to work on Blogger, go to Design, then to "edit HTML", then to "edit template". In the HTML file insert the following script right after where it says < head >:

<script type="text/javascript" src="">
 extensions: ["tex2jax.js","TeX/AMSmath.js","TeX/AMSsymbols.js"],
 jax: ["input/TeX", "output/HTML-CSS"],
 tex2jax: {
     inlineMath: [ ['$','$'], ["\\(","\\)"] ],
     displayMath: [ ['$$','$$'], ["\\[","\\]"] ],
 "HTML-CSS": { availableFonts: ["TeX"] }

Tuesday, June 17, 2014

Richard Feynman on Erwin Schrödinger

I thought it is interesting to see what the great Nobel Laureate physicist Richard Feynman said about Erwin Schrödinger's attempts to discover the famous Schrödinger equation in quantum mechanics:

When Schrödinger first wrote it [his equation] down,
He gave a kind of derivation based on some heuristic
Arguments and some brilliant intuitive guesses. Some
Of the arguments he used were even false, but that does
Not matter; the only important thing is that the ultimate
Equation gives a correct description of nature.
                                    -- Richard P. Feynman
(The Feynman Lectures on Physics, Vol. III, Chapter 16, 1965.)

It has been my experience in reading physics books that this sort of `heuristic' reasoning is part of doing physics. It is a very creative (sometimes not logical!) art with mathematics in attempting to understand the physical world. Dirac did it too when he obtained his Dirac equation for the electron.

Tuesday, June 3, 2014

Entangled and Unentangled States

Let's take a simple example of electron spin states. The reason it is 'simple' is that you only have two states to consider: either an electron's spin is 'up' or it is 'down'. We can use the notation ↑ to stand for the state that its spin is 'up' and the down arrow ↓ to indicate that its spin is down.

If we write, say, two arrows together like this ↑↓ it means we have two electrons, the first one has spin up and the second one spin down. So, ↓↓ means both particles have spin down.

Now one way in which the two particles can be arranged experimentally is in an entangled form. One state that would describe such a situation is a wavefunction state like this:

Ψ = ↑↓ - ↓↑.

This is a superposition state combining (in some mysterious fashion!) two basic states: the first one ↑↓ describes is a situation where the first particle has spin up and the second spin down, and the second state ↓↑ describes a situation where the first particle has spin down and the second particle has spin up. But when the two particles are in the combined superposition state Ψ (as above), it's in some sort of mix of those two scenarios. Like the case of the cat that is half dead and half alive! :-)

Why exactly is this state Ψ 'entangled' -- and what exactly do we mean by that? Well, it means that if you measure the spin of the first electron and you discover that its spin is down ↓, let's say, that picks out the part "↓↑" of the state Ψ! And this means that the second electron must have spin up! They're entangled! They're tied up together so knowing some spin info about one tells you the spin info of the other - instantly! This is so because the system has been set up to be in the state described by Ψ.

Now what about an unentangled state? What would that look like for our 2-electron example. Here's one:

Φ = ↑↑ + ↓↓ + ↑↓ - ↓↑.

Here this state is made up of two electrons that can have both their spins up (namely, ↑↑), both their spins down (↓↓), or they could be in the state Ψ (consisting of the opposite spins). In this wavefunction state Φ (called a "product state" which are generally not entangled), if you measure the spin, say, of the first electron and you find that it is up ↑, then what about the spin of the other one? Well, here you have two possibilities, namely ↑↑ and ↑↓ involved in Φ, which means that the second electron can be either in the up spin or the down spin. No entanglement, no correlation as in the Ψ case above. Knowing the spin state of one particle doesn't tell you what the other one has to be.

You can illustrate the same kind of examples with photon polarization, so you can have their polarizations entangeled or unentangled - depending on how the system is set up by us or by nature.

Thursday, May 29, 2014

Periodic Table of Finite Simple Groups

Chemistry has its well known and fantastic periodic table of elements. In group theory we have an analogous 'periodic table' that describes the classification of the finite simple groups (shown below). (A detailed PDF is available.) It summarizes decades worth of research work by many great mathematicians to determine all the finite simple groups. It is an amazing feat! And a work of great beauty.

Groups are used in studies of symmetry - in Math and in the sciences, especially in Physics and Chemistry.

A group is basically a set $G$ of objects that can be "combined", so that two objects $x, y$ in $G$ produce a third object $x \ast y$ in G. Loosely, we refer to this combination or operation as a 'multiplication' (or it could be an 'addition'). This operation has to have three basic rules:

1. The operation must associative, i.e. $(x\ast y)\ast z = x\ast(y\ast z)$ for all objects $x, y, z$ in $G$.
2. $G$ contains a special object $e$ such that $e\ast x = x = x\ast e$ for all objects $x$ in $G$.
3. Each object $x$ in $G$ has an associated object $y$ such that $x\ast y = e = y\ast x$.

Condition 2 says that the object $e$ has no effect on any other object - it is called the "identity" object. It behaves much like the real number 0 in relation to the addition + operation since $x + 0 = x = 0 + x$ for all real numbers. (Here, in this example, $\ast$ is addition + and e is 0.) As a second example, $e$ could also be the real number 1 if $ast$ stood for multiplication (in which case we take $G$ to be all real numbers except 0).

Condition 3 says that each object has an 'inverse' object. Or, that each object could be 'reversed'. It turns out that you can show that the $y$ in condition 3 is unique for each $x$ and is instead denote by $y = x^{-1}$.

The commutative property -- namely that $x\ast y = y\ast x$ -- is not assumed, so almost all of the groups in the periodic table do not have this property. (Groups that do have this property are called Abelian or commutative groups.) The Abelian simple groups are the 'cyclic' ones that appear in the right most column of the table. (Notice that their number of objects is a prime number $2, 3, 5, 7, 11, \dots$ etc.)

The periodic table lists all of the finite simple groups. So they are groups as we just described. And they are finite in that each group $G$ has finitely many elements. (There are many infinite groups used in physics but these aren't part of the table.)

But now what are 'simple' groups? Basically, they are ones that cannot be 'made up' of yet smaller groups or other groups. (More technically, a group $G$ is said to be simple when there isn't a nontrivial normal subgroup $H$ inside of $G$ -- i.e., $H$ is a subset of $G$ and is also a group under the same $\ast$ operation of $G$, and further $xHx^{-1}$ is contained in $H$ for any object $x \in G$.) So a simple group is like a basic object that cannot be "broken down" any further, like an 'atom', or a prime number.

One of the deepest results in the theory of groups that helped in this classification is the Feit-Thompson Theorem which says: each group with an odd number of objects is solvable. (The proof was written in the 1960s and is over 200 pages - published in Pacific Journal of Mathematics.)

Wednesday, May 28, 2014

St Augustine on the days of Creation

In his City of God, St Augustine said in reference to the first three days of creation:

"What kind of days these were it is extremely difficult, or perhaps impossible for us to conceive, and how much more to say!" (See Chapter 6.)

So a literal 24 hour day seemed not the plain meaning according to St Augustine. He does not venture to speculate but leaves the matter open. That was long long before modern science.

Sunday, May 25, 2014

Experiment on Dark Matter yields nothing

The most sensitive experiment to date by the Large Underground Xenon (LUX), which was designed to detect dark matter, has not registered any sign of the substance.

See Nature article:
No sign of dark matter in underground experiment
Eugenie Samuel Reich

As a result, some scientists are considering other (exotic) possibilities:

Dark-matter search considers exotic possibilities, by Clara Moskowitz

Saturday, May 17, 2014


This is my first (test) post on this blog. I'm just examining it, maybe to add a few thoughts and ideas now and then. Thank you for reading.