Question

The set of real numbers between 0 t0 1 should be finite or infinte and explain it by reason

Answer 1

infinite, and uncountable as well

Answer 2

how please explain me,

Answer 3

First let's look at the definition of countability:
"A set S is called countable if there exists an injective function f from S to the natural numbers N = {0, 1, 2, 3, ...}." (Copied from Wikipedia)

Answer 4

ok then

Answer 5

Let's suppose that the set is countable.

Answer 6

And define a function f(x) such that for every natural number x, f(x) refers to some real number between 0 and 1

Answer 7

@kc_kennylau i just started sets and i just find the definition of it just tell me that o and 1 between these two what the numbers are present

Answer 8

For example,
\[
f(0)=0.\color{red}3363684485...\\
f(1)=0.6\color{red}955102578...\\
f(2)=0.26\color{red}39483653...\\
f(3)=0.006\color{red}5093130...\\
f(4)=0.8864\color{red}533021...\\
f(5)=0.84447\color{red}42908...\\
f(6)=0.862720\color{red}1508...\\
f(7)=0.8665645\color{red}627...\\
f(8)=0.39569269\color{red}52...\\
f(9)=0.824607139\color{red}4...\\ \]
Now, let's take the first digit of the first number, second digit of the second number, third digit of the third number, etc... and form a new number:
\[0.3935541654...\]

Answer 9

Now add one to all digits of this (if it's 9, convert it to 0)
\[0.4046652765...\]

@ziqbal103 my modem was down :/

Answer 10

ziqbal, your question was just about the set being finite or infite. Don't get stuck on the countability therefore. :)
The set is inifite as already the rational numbers betwee 0 and 1 are infinite. Which you can prove very nicely by just writing them all down, each divisor in one line.

Answer 11

@techhelper @ziqbal103 oh i see sorry

Answer 12

To prove that the set of all real numbers between 0 and 1 is infinite:
The first digit has 10 probabilities.
The second digit has 10 probabilities.
The third digit has 10 probabilities.
And so on and so on.

Therefore, the number of all real numbers between 0 and 1
=10*10*10*.....
=\(\infty\)

Answer 13

@techhelper tell me how the rational numbers become infinte

Answer 14

and @kc_kennylau the difference b/w o and 1 there is more than 10 decimal digits i means infinite so am i right

Answer 15

@ziqbal103 sorry I fail to comprehend your last sentence.

Answer 16

Assuming they are finite, you can write them in a sequence that is always getting bigger: 0, a1, a2, a3, ... , 1. take any pair of them like a2 and a3 and have a look at a number between them: b=(a2+a3)/2. this is by definition a rational number again. however you just found a new one, that is not yet in your sequence, so your assumption of them being finite must be wrong.

Answer 17

@techhelper nicely done :D

Answer 18

For your real number question:

Assuming they are finite, you can write them in a sequence that is always getting bigger: 0, a1, a2, a3, ... , 1. take any pair of them like a2 and a3 and have a look at a number between them: b=(a2+a3)/2. this is by definition a real number again. however you just found a new one, that is not yet in your sequence, so your assumption of them being finite must be wrong.

Answer 19

@techhelper and @kc_kennylau now i get it thank you very much for helping

Answer 20

no problem :)

Answer 21

i wanna ask another question

Answer 22

\[\large z_1z_2=(a+ib)(c+id)\\\large\overline{z_1z_2}=(a-ib)(c-id)\\\large z_1z_2\overline{z_1z_2}\\\large=(a+ib)(a-ib)(c+id)(c-id)\\\large=(a^2+b^2)(c^2+d^2)\\\large=|z_1z_2|^2\]

Answer 23

I mean \(\large|z_1|^2|z_2|^2\)

Answer 24

@Directrix @ganeshie8 just wanna check if I'm correct thanks

Answer 25

i think same case will be with(z1+z2)(z1.z2whole bar)....?

Answer 26

Sorry I'm not familiarized with complex number :/

Answer 27

ok@kc-kennylau.and @techhelper you?

Answer 28

@ehuman

Answer 29

@CGGURUMANJUNATH

Answer 30

@kc_kennylau where are you help me i am sending you a link an attachement so that you make me clear that z1+z2whole modulus is less than z1modulus +z2 modulus
ATTACHEMENT

Answer 31

http://www.suitcaseofdreams.net/properties_modulus.htm
BY THIS LINK CAN YOU HELP ME @kc_kennylau

Answer 32

Sorry I'm not familiarized with complex number

Answer 33

This is my reasoning:

Take a number between 0 and 1. Divide it by 2. Divide the next number by 2. Keep dividing until you're convinced you will keep getting numbers no matter how much you divide.