## Friday, April 26, 2013

### 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!

## Friday, April 19, 2013

### 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

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 $A=987652413$.

## Friday, April 12, 2013

### 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.

## Friday, April 5, 2013

### 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 1b, 11b, 111b, 1111b, 11111b,... for all but finitely many prime numbers p. Here 11111b etc are numbers written with base b numerals.

Answer after the good-looking easter eggs :P

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