Wednesday, October 24, 2012


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This time, lets talk about a really simple equality, but it scares many high school kids. The equality is


Those kids will say, "I know logarithms, but this equality is too complicated."

I ask, "Tell me what is $\log_ax$?"

They answer, "It is the answer of $a$ to the power of something equals $x$."

I say, "Write it down?"

They write, "It is the answer of $a^{(\ )}=x$."

I ask, "What is the answer?"

They confuse, "$\log_ax$?"

I say, "Write the answer inside ( )."

They write, "$a^{\log_ax}=x$."

I say, "Now you know."

They reply, "Know what?"

I say, "The equality is nothing but just the definition of $\log_ax$."

They complain, "I still don't understand!"

I ask, "Sigh! Lets start again. Do you know logarithm?"

They say, "I know logarithms, but this equality is too complicated."

Thursday, October 18, 2012

Prove that 2n>n

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(I am wondering if I shall add a "level-indicator" on the page, because somehow the level of knowledge is not quite even among my posts...)

Preliminary concept: A set is just a collection of objects. A (possible empty) part of the set is called a subset. The set consists of all subsets of a specified set is called the power set of the specified set.

e.g. A={1,2,3,4} is a set. { }, {1}, {1 3}, and A itself, are subsets of A. The power set of A is
P(A) ={{ }, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4},
{3,4}, {1,2,3}, {1,2,4}, {1,3,4},{2,3,4},{1,2,3,4} }

The power set of A, can be denoted by P(A) or 2A. It is not hard to see that for a set with n elements, there is exactly 2n elements in the power set, hence we have the notation.

It looks obvious that 2n>n. Well, it is quite obvious when it is finite. 23>3, 24>4, how hard is that?

However, when n is infinite, 2n is also infinite. To compare two infinite sets is tricky. Well, it is easy to see that 2n≥n: If A={a,b,c,...} then the size of A is the same as {{a},{b},{c},...} which is a subset of P(A).

The hard part is: We have to make sure 2n≠n. In other words, we have to establish the fact that there is no one-to-one correspondence between A and P(A).

Consider a correspondence from A to P(A) such that any element aA corresponds to a subset S(a)∈P(A). Since A is infinite, we cannot write down all its elements, but we can have a local list:
α : ....α∈S(α)? β∈S(α)? γ∈S(α)? δ∈S(α)?
β : ....α∈S(β)? β∈S(β)? γ∈S(β)? δ∈S(β)? ....
γ : ....α∈S(γ)? β∈S(γ)? γ∈S(γ)? δ∈S(γ)? ....
δ : ....α∈S(δ)? β∈S(δ)? γ∈S(δ)? δ∈S(δ)? ....

It is a table of true and false. Now construct a new subset B containing only those element a such that the question a∈S(a)? is fales.

Note that if a∈S(a) then aB and so S(a)≠B; likewise if a∉S(a) then aB and so S(a)≠B. Hence the list

....α∈S(B)? β∈S(B)? γ∈S(B)? δ∈S(B)? ....

is different from any S(a) at exactly the a∈S(a)?-position. (We can see now it is exactly the Cantor's diagonal argument used in the previous two posts: Not enough names for numbers and  The are uncountably many subsets of natural numbers

Therefore no element of A corresponds to B and so this correspondence is not a one-to-one correspondence. As this is just any correspondence, so one-to-one correspondence cannot exist.

Sunday, October 14, 2012

The are uncountably many subsets of natural numbers

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In the post Not enough names for numbers, we prove that there are more real numbers between 0 and 1 than natural numbers, technically we say that the set of real numbers is uncountable. We now apply the same argument to prove that there are uncountably many real numbers made up entirely by 1 and 2 in the unit interval.

Consider a list of real numbers made up entirely by 1 and 2 in the unit interval:


construct a real number B=0.β1β2β3β4β5… such that βj=1 if αjj=2; 2 if αjj=1.

Again this number B is not in the list. Hence there is never a complete list of such real numbers.

Lets consider a somewhat different problem. The set of natural numbers is {1,2,3,…}. Each subset S of natural numbers is corresponding to a unique real number αS=0.α1α2α3α4α5… in a way that αj=1 if j∈S; 2 if j∉S, for example

Suppose there is a given list of subsets of natural numbers, then we convert it to a list of numbers in the unit interval made up entirely by 1 and 2, then we have a B which is not in the list, and finally we convert B to a subset SB={j βj=1} which is of course not in the given list of subsets of natural numbers.

Can we construct SB directly from the given list? Yes, we can!

SB={j : j∉ the j-th subset in the list}

In the first part of the previous section, we have a correspondence that each subset of natural numbers S maps to a real number αS in (0,1). Different subsets map to different real numbers in (0,1), and hence there is at least as many real numbers in (0,1) as the subsets of natural numbers.

Lets consider another correspondence. For each real number in (0,1), expressed in binary number system, for example 5/8=0.101000…. If α=0.α1α2α3α4α5…, then let Sα={j : αj=1}, for example S0.101000…={1,3}. Different real numbers in (0,1) maps to different subsets of natural numbers, and hence there is at least as many subsets of natural numbers as real numbers in (0,1).

Combining, we see that there are as many subsets of natural numbers as real numbers.

Monday, October 8, 2012

Not enough names for numbers

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Napier's constant, Pi, Planck constant, golden ratio, etc. We have given names to many special numbers. However, there are far many numbers than our human language can handle. It is not a philosophical statement, it is a true mathematical theorem which is surprisingly simple to prove.

A language consists of phrases which are finite sequences of a finite set of symbols. English phrases, for example, are finite sequences of 'a' to 'z' and the empty space. If we assign an ordering of the symbols, then we also assign an lexographical order of phrases. If we assign 'a'<'b'<...<'z'<' ', we will have 'ab cd'<'acbde'<'a bac'. In other words, all possible phrases in a language can be listed.

We are now going to prove that the set of all real numbers between 0 and 1 cannot be listed, and hence there are far more numbers than possible words:

Suppose on the contrary, the set of all real numbers between 0 and 1 can be listed as follows


Now, construct a real number 0.β1β2β3β4β5… such that βj≠αjj, 0 and 9.

Under this special construction, the new number is different from the j-th number in the list, as they have different j-th digits. (Note that two numbers with different digits can be the same, e.g. 0.10000… and 0.09999… but this will not happen in our construction.) Therefore, the new number is not in the list of all real numbers between 0 and 1, which is obviously a contradiction.

The above construction is the famous Cantor's diagonal argument by Georg Cantor (1845-1918).  This seemingly simple technique is indeed rather powerful, it is used in the arguments for Russell's paradox, the first of Gödel's incompleteness theorems, and Turing's answer to the Entscheidungsproblem.

Thursday, October 4, 2012

$\cos \theta=2$

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Question: Write down one solution to each of the following:

(a) $\cos \theta=0.5$
(b) $\cos \theta=1$
(c) $\cos \theta=-0.17$
(d) $\cos \theta=2$

Note that there is no ° in the question, so the answer should be in radian.

A high school student could do (a) and (b) without a calculator. Calculator will give a solution of (c) with $\pi/2<\theta<\pi$. For (d), school algebra tells us that there is no solution, as the value of cosine lies in between -1 and 1 -- however, it is not the case if we consider complex number.

There are three definitions of cosine:

1. As the ratio of the adjacent to the hypotenuse in a right angled triangle.  Under this definition, $\theta$ can only be $0<\theta<\pi/2$ and we have $0<\cos \theta<1$.

2. As the $x$-coordinate of point with polar coordinate $(1, \theta)$. Under this definition, $\theta$ can be any real number and it is still true that $-1\le \cos \theta\le 1$.

3. By the formula discovered by Leonhard Euler (1707-1783):
$$\cos \theta=1-\frac{1}{2!}\theta^2+\frac{1}{4!}\theta^4-\frac{1}{6!}\theta^6+\cdots$$
or equivalently
$$\cos \theta=\frac{e^{i\theta}+e^{-i\theta}}{2}.$$
Nowadays, the study of wave behaviour in physics and modern axiomatic approach in mathematics, cosine is no more an "angle" thing, and therefore Euler's formula serves as a better definition.

By using Euler's formula, we can have the cosine of a complex number.
One solution of the quadratic equation is $e^{-i\theta}=2+2\sqrt{3}$ and hence one solution of (d) is $\theta=i\log_e(2+\sqrt{3})$.

Tuesday, October 2, 2012

The two meanings of $\sqrt{4}$

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What is wrong with the following expression?

$$4=\sqrt{16}=\sqrt{-4 \times -4}=\sqrt{-4}\times \sqrt{-4}=-4.$$

It seems easy. $\sqrt{-4}$ doesn't make sense. Right? Well, yes and no.

If we just consider the usual real numbers, then square of any number is positive and $\sqrt{-4}$ simply does not exists.

However, there is something called complex numbers.  For complex numbers, square can be negative.  In this case, there are two ways of interpreting $\sqrt{-4}$.

1.  $\sqrt{-4}=2 i$, where $i$ is the imaginary unit.  In this case, the equality $\sqrt{-4 \times -4}=\sqrt{-4}\times \sqrt{-4}$ does not hold.

2. $\sqrt{-4}=\{ \pm 2 i \}$, i.e. it is a set.  We would also reinterpreting $\sqrt{16}=\{\pm 4\}$. In this case, we have
$$\{\pm 4\}=\sqrt{16}=\sqrt{-4 \times -4}=\sqrt{-4}\times \sqrt{-4}=\{\pm 2\}\times\{\pm 2\}=\{\pm 4\},$$
here we use element-wise multiplication of two sets.

In general, there are two different meanings of $\sqrt{x}$.

1. $\sqrt{x}$ is the unique positive square root of $x$, if $x>0$; $\sqrt{0}=0$; $\sqrt{x}=\sqrt{-x}i$ if $x<0$;  any one of the square root of $x$ if $x$ is not a real number.  Under this meaning, we have $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ only if at least one of $a$ and $b$ are nonnegative.

Under this meaning, the formula for the quadratic equation $ax^2+bx+c=0$ is $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.  As we have extended the definition of $\sqrt{\cdot }$, this formula now makes senses for any complex numbers $a\ne 0$, $b$ and $c$.

2. $\sqrt{x}$ is the set consists of the two roots of $x$.  Indeed, we can generalize it to $\sqrt[n]{x}$ being the set of all $n$-th roots of $x$.  For this meaning, we have $\sqrt[n]{ab}=\sqrt[n]{a}\times\sqrt[n]{b}$ for all complex numbers $a$ and $b$.

The formula for the quadratic equation is now simply $\frac{-b+\sqrt{b^2-4ac}}{2a}$, which is indeed the set of the two solutions.