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March 4

how to define anew way to count real number?

we know that numbers of the form a/b can be counted by find one to one mapping or one to one function defined on the natural number set as it`s domain ,a/b set as it`s range.

f(n):N→Q , f(1,n)= 1/n ,f(2,n) =2/n ,.....etc. n={1,2,3 }

abviously we can not find such function to count real number like SQR 2.

how to create this function?

let us consider the interval (0,1)

FIRST WAY, let us define the standared number p=123456789101112131415161718192021......etc. now let us define,

   F:N→R

f(1) = 0.p = 0.1234....etc  
f(0,1) = f(2)= 0.0p = 0.01234...etc 
f(1,1) = f(3)= 0.p+1 = 0.001234...etc +(1) 
f(0,1,1)=f(4)= 0.0p+1 = 0.1234..etc +(1)

now it mus occur that f(n) will match all the real numbers within the interval (0,1)

SECOND WAY, we consider an arbitrary real number like P=(2-SQR2)=0.1414...etc.and we follow the same above way to start counting real number.

MY QUESTION IS IF IT IS CORRECT TO FOLLOW THIS WAY??????

80.255.40.168 09:34, 4 March 2007 (UTC)MATH FREAK80.255.40.168 09:34, 4 March 2007 (UTC)[reply]

Without getting into the problems of your construct, one may simply note that the cardinality of the natural numbers is strictly less than the cardinality of the real numbers (i.e. the natural numbers are countable, while the real numbers are uncountable). Your statement that this function "f" would form a bijection between the sets of natural and real numbers contradicts this well known fact.

Word. I'm curious about the construct, though. What is it you're trying to say? I'm having some trouble following your notation. For instance, I'm fairly certain an function isn't used on simultaneously one, two, and three inputs. This "p" isn't a number as far as I can tell, although the "0.p" thing you give is. Then, you lose me. From f(1) to f(2), you shift it a decimal place. From f(2) to f(3), you shift it another, and add either 1, or what might be an infinitesimal. From f(3) to f(4), you shift it back over two decimal places. I don't get it. Black Carrot 20:52, 7 March 2007 (UTC)[reply]

Cardinality of the set of all functions

What would be the cardinality of the set of all functions? Clearly it is at least equal to that of the real numbers and it seems like it should be quite a very good deal larger, but I have no idea how to prove things have greater cardinality than the Reals. Any thoughts on such a question? Maelin (Talk | Contribs) 11:25, 4 March 2007 (UTC)[reply]

There is no set of all functions; such an object would lead to immediate paradoxes (like the Set of all sets). If you mean the set of all functions from the reals to the reals, it has cardinality equal to the cardinality of the power set of the reals, so it is indeed much larger than R. Algebraist 13:43, 4 March 2007 (UTC)[reply]

And since I have nothing better to do, I might as well prove that. The set F of all functions from R to R certainly contains all functions from R to {0,1}, and so is at least as large as this set which has the same cardinality as P(R). But by definition a function from R to R is just a subset of RxR, so |F|<=|P(RxR)|=|P(R)|. By Schroeder-Bernstein we have the result. Algebraist 14:17, 4 March 2007 (UTC)[reply]

Well, by definition a function from the reals to the reals is a special type of subset of R x R. –King Bee ^{(T • C)} 15:46, 4 March 2007 (UTC)[reply]

Sorry, that's what I meant (and what I used). Should be more explicit. Algebraist 16:22, 4 March 2007 (UTC)[reply]

Right, I just wanted to be precise, so that no one would get confused. =) –King Bee ^{(T • C)} 16:25, 4 March 2007 (UTC)[reply]

CORRECTION OF HOW TO DEFINE ANEW WAY TO COUNT REAL NUMBERS SET ?

We know that numbers of the form a/b can be counted by find one to one mapping or one to one function defined on the natural numbers set as it`s domain,a/b numbers set as it`s range,

F(n):N →Q , f(1,n)=1/n, f(2,n)=2/n,….etc. n= {1,2,3,.. }

Abviously we cannot find such function to count real number set like SQR2

Now how to create this function?

Let us consider the interval (0,1)

FIRST WAY, Let us define the standared number p=12345678910111213141516171819202122… etc.Now,

F:N→R,

F(1) = 0.p = 0.12345..etc. F(0,1) = f(2) = 0.0p = 0.012345…etc. F(1,1) = f(3) = 0.(p+1) = 0.12345…etc +1 ≠ 1.p = 1.1234…etc F(0,1,1) = f(4) = 0.0(p+1) = 0.012345…etc+1≠ 1.0p = 1.012345..etc. And so on. It must occur that f(n) will match all the real number within (0,1).

SECOND WAY, We may consider an arbitrary number like p=(2-sqr2) = 0.1414…etc. Applying the first way above to start counting real number..

My question is if it possible to adopt such kind of definition inorder to make the real number set countable? 80.255.40.168 12:41, 4 March 2007 (UTC)MATH FREAK[reply]

No. The place where you run into problems is that real numbers, assuming we are considering their decimal expansions, can have infinitely many digits, whereas all natural numbers have only a finite number of digits. You could construct such a 1-to-1 mapping between the set of all terminating decimals and the natural numbers, but that would only prove countability of a subset of the rational numbers and so would not be very interesting. Maelin (Talk | Contribs) 13:17, 4 March 2007 (UTC)[reply]

If you don't mind me asking. Why are you asking the same question twice on Reference Desk Mathematics? Do you really want us to answer the same question twice? 202.168.50.40 21:50, 4 March 2007 (UTC)[reply]

Trigonometric Identities

I'm stuck on two trigonometry questions:

1. How is this proof solved? $(cos(a-b))/(sin(a)sin(b))=cot(a)cot(b)+1$

2. How would I find the exact answer (in degrees) for: $(tan(40)-tan(10))/(1+tan(40)tan(10))$

Any help would be appreciated. Thanks. -- Sturgeonman 16:01, 4 March 2007 (UTC)[reply]

For number 1, are you allowed to use the identity that

\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)

? If so, it's completely trivial. –King Bee ^{(T • C)} 16:06, 4 March 2007 (UTC)[reply]

Oh, and for number 2, I would suggest writing

\tan x

as

\sin x/\cos x

and then applying identities similar to the one you have in number 1. –King Bee ^{(T • C)} 16:16, 4 March 2007 (UTC)[reply]

Thanks - You da man. -- Sturgeonman 16:28, 4 March 2007 (UTC)[reply]

For number 2, you should look at the identity for

\tan(x-y)

--Spoon! 22:09, 4 March 2007 (UTC)[reply]

Segre variety is irreducible : how to prove something about certain ideal

Hello,

I am trying to understand why the Segre variety (see Segre embedding) is irreducible. I had a problem with proving something about a certain ideal. It comes down to this :

Let $k$ be a field

let $R=k[Z_{i,j};i=0\cdots m,j=0\cdots n]$

Let $S=k[x_{0},\cdots ,x_{m},y_{0},\cdots ,y_{n}]$

Let $\phi$ be the unique ring morphism, mapping $Z_{i,j}$ onto $x_{i}y_{j}$ .

Show that the kernel is the ideal W generated by all element of type $Z_{a,b}Z_{c,d}-Z_{a,c}Z_{b,d}.$

Now the ideal I described is obviously a subset of the kernel, but how do I see all of the kernel is generated?

Thank you for any ideas, Evilbu 22:03, 4 March 2007 (UTC)[reply]

So, I don't know how far along you are, but just take an arbitrary element of the kernel and show you can write in the form you describe. Is this too difficult? –King Bee ^{(T • C)} 22:09, 4 March 2007 (UTC)[reply]

Well forgive me but...yes :( Evilbu 22:32, 4 March 2007 (UTC)[reply]

Oh, don't interpret my comment above as an insult. I was honestly asking if that was the difficult thing. (You asked for any ideas, so I didn't know if you had figured out what you were supposed to do). I should be able to help you once I understand the problem. What exactly is

Z_{i,j}

? –King Bee ^{(T • C)} 22:35, 4 March 2007 (UTC)[reply]

Okay thanks : each

Z_{i,j}

is just one of the variables in my polynomial ring (so there are

(m+1)(n+1)

of them. So my first ring

R

is just the polynomial ring over these variables over

k

. Evilbu 22:49, 4 March 2007 (UTC)[reply]

Alright, here's what I'm thinking. Can you use the fundamental homomorphism theorem somehow? Show first that the set you describe above is contained in the kernel, then show that that set is an ideal, and since the kernel is the smallest ideal in such and such, the two must be the same? Will that work here, or am I thinking incorrectly? –King Bee ^{(T • C)} 02:59, 5 March 2007 (UTC)[reply]

Thanks, but no, I have been thnking about that homomorphism theorem and I don't see how it can be applied here. But I have another idea : I think the problem I have can be attacked in a different way : prove that W is a prime ideal (please scroll upwards, and you will see I have come up with a name for that ideal). Evilbu 14:34, 5 March 2007 (UTC)[reply]

any help please?

Evaluate: lim as x approaches 0 from the positive side of quantity (sqrt(x) times cos(x)) divided by x^2

lim x->0+ ( $sqrt(x$ )*cosx) / x²—Preceding unsigned comment added by 65.110.234.70 (talk • contribs)

Please do your own homework. Consider simplifying the expression first. You have

{\dfrac {\sqrt {x}}{x^{2}}}

, and this can be simplified. –King Bee ^{(T • C)} 22:14, 4 March 2007 (UTC)[reply]