bogus numbers?

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I have read an article who claims that complex numbers does not exist and any physics based on complex numbers are bogus science!

I am surprised that in 21st century there are still such idiots dare to write like that. Anyway, his key argument is the following:

Consider the system of two equations $$\begin{cases}(x-1)^2+(y-1)^2=1 \\ x+y=0\end{cases}$$ the first of which is a circle centered at (1,1) of radius 1 and the second of which is a straight line passing (0,0) and (1,-1). The system has complex roots, but the geometric objects have no intersection. Therefore, the complex roots are bogus.

Question: What is wrong with his argument?

Answer: Only when we consider $x$ and $y$ as real numbers, the two equations correspond to the two geometric objects.

There are three ways to understand complex numbers.

The simplest way is that $a+\sqrt{-1}b$ is the point $(a,b)$ on a plane. In this case $(x-1)^2+(y-1)^2=0$ and $x+y=0$ each describes a relation between two points and therefore both are equations for a two-dimensional figure in a four-dimensional space. We cannot visualize four-dimensional space!

The second way is to consider $a+\sqrt{-1}b$ as the polynomial $ax+b$ in the space of polynomials in such a way that we identify the polynomials $x^2$ and $-1$. The two equations are therefore relations among two polynomials.

The third way is to consider $a+\sqrt{-1}b$ as the matrix $\begin{pmatrix}a & b \\ -b & a\end{pmatrix}$. The sum and product of two complex numbers are the sum and prodcut of two matrices.

Why does the author write such article? Because Stephen Hawking talks about complex numbers in physics and he also argues that we do not need God in science!

multiple of 11

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Question: What are the smallest and largest multiples of 11 that are made up of the nine digits 1,2,3,4,5,6,7,8,9?

123456789
123456798
123456879
123456897
123457689
123457698
123457869
123457896
123457968
123457986
..................
..................
..................
987654312
987654321

Answer:

If $A=a_1a_2a_3a_4a_5a_6a_7a_8a_9$ is a multiple of 11, then we know the difference $(a_1+a_3+a_5+a_7+a_9)-(a_2+a_4+a_6+a_8)$ is also a multiple of 11, say $11x$.

Note that $a_1+a_2+\cdots+a_9=1+2+\cdots+9=45$ and hence $a_1+a_3+a_5+a_7+a_9=(45+11x)/2$ and $a_2+a_4+a_6+a_8=(45-11x)/2$.

Note that if $x=3$, then $a_2+a_4+a_6+a_8=6$ which is impossible. Likewise $x\ne -3$. Therefore $x$ can only be $1$ or $-1$.

Therefore either $a_1+a_3+a_5+a_7+a_9=28, a_2+a_4+a_8+a_9=17$ or $a_1+a_3+a_5+a_7+a_9=17, a_2+a_4+a_8+a_9=28$.

To get the smallest multiple of 11, we would like to see $A=12345\ldots$. Both $1+3+5+a_7+a_9=28$ and $1+3+5+a_7+a_9=17$ are impossible (the latter one is possible only if $\{a_7, a_9\}=\{2,6\}$ but we already have $a_2=6$.).

Try $A=1234\ldots$. Now $1+3+a_5+a_7+a_9=28$ is possible if $\{a_5,a_7,a_9\}=\{7,8,9\}$ (and so $\{a_6,a_8\}=\{5,6\}$) but $1+3+a_5+a_7+a_9=17$ is impossible. Therefore the smallest multiple of 11 is $A=123475869$.

To get the largest multiple of 11, we would like to see $A=98765\ldots$. Now $9+7+5+a_7+a_9=28$ is possible if $\{a_7,a_9\}=\{3,4\}$ (and so $\{a_6,a_8\}=\{1,2\}$). Thus the largest multiple of 11 is $A=987652413$.

Arithmetic sequence of primes

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Question: Suppose a, a+d, a+2d, ..., a+(n-1)d are primes. Find the largest possible n, under the condition that d cannot be a mulitple of 10.

    2     3     5     7    11    13    17    19    23    29
   31    37    41    43    47    53    59    61    67    71
   73    79    83    89    97   101   103   107   109   113
  127   131   137   139   149   151   157   163   167   173
  179   181   191   193   197   199   211   223   227   229
  233   239   241   251   257   263   269   271   277   281
  283   293   307   311   313   317   331   337   347   349
  353   359   367   373   379   383   389   397   401   409
  419   421   431   433   439   443   449   457   461   463
  467   479   487   491   499   503   509   521   523   541
  547   557   563   569   571   577   587   593   599   601
  607   613   617   619   631   641   643   647   653   659
  661   673   677   683   691   701   709   719   727   733
  739   743   751   757   761   769   773   787   797   809
  811   821   823   827   829   839   853   857   859   863
  877   881   883   887   907   911   919   929   937   941
  947   953   967   971   977   983   991   997

(Table of primes up to 1000)

Answer: Lets make a guess. We claim the largest possible n≥5.

a must be odd, otherwise a+2d is an even number greater than 2 and thus not a prime.

If d is odd, then a+d is an even prime greater than 2 and thus not a prime.

If d is even and not a multiple of 10, then one of a, a+d, a+2d, a+3d, a+4d have the last digit being 5 and hence it must be 5. Note that there is only one odd prime, i.e. 3, which is smaller than 5. Therefore it is only possible if a=5 or a+d=5.

If a+d=5 then a=3, d=2. The sequence is just 3,5,7.

The only case left is that a=5 and then a+5d is a multiple of 5. Therefore n is at most 5.

You can try to search using the table.

5, 11, 17, 23, 29 is a prime sequence with d=6.
5, 17, 29, 41, 53 is a prime sequence with d=12.

Therefore the largest possible n=5.

What if d is allowed to be a multiple of 10?

No matter how large n is, we can construct a sequence a, a+d, a+2d, ..., a+(n-1)d of primes. It is the famous Green-Tao theorem, proved by Ben Green and Terence Tao in 2004.

1, 11, 111, 1111 ....

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 (Easter Holiday gives me another excuse to be lazy....)

A simple number theory question: Show that there is at least one (and hence infinitely many) multiples of p among 1_b, 11_b, 111_b, 1111_b, 11111_b,... for all but finitely many prime numbers p. Here 11111_b etc are numbers written with base b numerals.

Answer after the good-looking easter eggs :P








Let n(k)=111...1_b denote the number with k 1's. Note that
n(k)=$b^{k-1}+b^{k-2}+b^{k-3}+\cdots+1$=(b^k-1)/(b-1).

We claim that when p is not a factor of b, then n(k) is a multiple of p for some k.

Use factor theorem, we know that $x^{k-1}+x^{k-2}+x^{k-3}+\cdots+1=(x-1)q(x)+k$. Consequently, n(k)=k mod (b-1). Therefore if p is a factor of b-1 (and hence not a factor of b), then p divides n(b-1).

By Fermat's little theorem, we know that if p is not a factor of b, then p divides b^p-1-1. Therefore if p is neither a factor of b nor a factor of b-1, then p divides n(p-1).

The claim is proved. Please note that if n(k) is a multiple of p, then so is n(2k), n(3k), and so on.

Permutation and Group

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A permutation of a permutation, is the product of the two permutations?
Right. The multiplication is not commutative, i.e in general ab≠ba. For example $$\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}$$ but $$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\1&3&2\end{pmatrix}=\begin{pmatrix}1&2&3\\2&1&3\end{pmatrix}.$$
What happens if I keep permuting using the same permutation?
If you keep permuting with same permutation, you will eventually return to the initial configuration. For example $$\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix} =\begin{pmatrix}1&2&3\\2&3&1\end{pmatrix}\begin{pmatrix}1&2&3\\3&1&2\end{pmatrix} =\begin{pmatrix}1&2&3\\1&2&3\end{pmatrix}.$$ The identity permutation, usually denoted by (1), is the permutation that changes nothing. We know that, say, there are n objects, then applying the same permutation for n! times will always end up to be the identity permuation, i.e. $$\tau^{n!}=(1)$$
Can I say τ^n!-1=1/τ?
Not quite. But you can write τ^n!-1=τ^-1, meaning that it is the multiplicative inverse of τ in the permutation multiplication.

Muliplication with an identity element and inverses. It is a group!?
Right. It is a group.
For every element α of a group G, we define a function A_α(g)=αg. It is not hard to see that A_α is a bijection and so it is a permutation. Indeed we have A_αA_β=A_αβ. Hence if we identify α with A_α, we see that each element of a group is indeed a permutation of the group!

Permutations

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Permutation of n objects is just a bijection on the set of these n objects?
Right.  If we don't specify the set, we means a permutation on the set {1,2,3,...,n}.

How to present a permutation?
Of course, we can present it as a bijection, e.g.
$$f(x)=\begin{cases}1 & x=1\\3 & x=2\\2 & x=3\end{cases}$$ However, traditionally we have a simpler presentation
$$\left(\begin{matrix}1&2&3\\1&3&2\end{matrix}\right)$$ which is, I believe, self-explanatory. The lower part is the new arrangement.

Why not just write (1 3 2)?
It is because we already use this symbol for a special type of permutations-the cycles. For instance (1 3 2) refers to the permutation f(1)=3, f(3)=2 and f(2)=1, i.e. $$\left(\begin{matrix}1&2&3\\3&1&2\end{matrix}\right).$$
It is easy to see cycles are interesting, mathematically?
First, we have to understand the multiplication of two permutations. If we consider two permutations $f$ and $g$ as two bijections on {1,2,3,...,n}, then their product is just the permutation $(fg)(x)=f\circ g(x)=f(g(x))$, that means if $g(r)=s$ and $f(s)=t$ then $(fg)(r)=s$. For instance, $$\left(\begin{matrix}1&2&3\\1&3&2\end{matrix}\right)\left(\begin{matrix}1&2&3\\3&1&2\end{matrix}\right)=\left(\begin{matrix}1&2&3\\2&1&3\end{matrix}\right).$$ It is known that every permutation is a product of cycles.

So cycles are the primes for permutations?
Not quite. The transpositions (i,j), i.e. the cycles of two elements, are the primes. Every permutation can be written as a product of transpositions. e.g. $$\left(\begin{matrix}1&2&3&4&5\\3&5&4&1&2\end{matrix}\right)=(1\, 3\, 4)(2\, 5)=(1\, 4)(1\, 3)(2\, 5).$$

1=0!?

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(Growing lazy, not a good sign....)

I have seen a proof, saying that 1=0!?
Which one do you mean '"1=0!"?' or '"1=0"!?', uh?

You got the trick. Why do we use "!" for factorial?
I am not a historian, I don't know why Christian Kramp choose n! to represent factorial.

Why does 0!=1?
Lets first consider the definition of factorial. The n! is the number of ways to permute n objects. In the set theory language, n! is the size of the set of all bijective maps on a set of n elements.
Let A be a set. A bijective map or a bijection or a permutation is a subset F of A×A such that

1. For any elements a∈A, there exists a unique F(a)∈A such that (a,F(a))∈F.
2. For any elements a∈A, there exists a unique F^-1(a)∈A such that (a,F^-1(a)∈F.

For example the map on {x,y,z} such that F(x)=y, F(y)=z, F(z)=x is a bijection (and hence a permutation), but G(x)=y, G(y)=x, G(z)=x is not.

On {x,y,z}, we have six permutations, right?
Yes. They are

• (F(x), F(y), F(z))=(x, y, z)
• (F(x), F(y), F(z))=(x, z, y)
• (F(x), F(y), F(z))=(y, x, z)
• (F(x), F(y), F(z))=(y, z, x)
• (F(x), F(y), F(z))=(z, x, y)
• (F(x), F(y), F(z))=(z, y, x)

Hence 3!=6.

On empty set ∅, ...?
Note that ∅ equals to and is the only subset of ∅×∅. There is no way we can violate the two conditions, and so ∅ is the bijection on ∅. Therefore 0!=1.