Question

Let S = [0,1]. If x and y are in s with x ≠y. How can we show that there are m,n∈N such that x< m/2^n 9 years ago

Answer 1

what class is this for? I need to know what you can use...

Answer 2

Post Grad Class

Answer 3

sorry keep coming up with ideas and then they dont work....

Answer 4

so we can fin 1/2^n < y for some n for sure.

Answer 5

if we could then say 1/2^n-x>0 we would be done

Answer 6

you got any ideas? It has been a while since real analysis...

Answer 7

I was thinking of using the Density Theorem prove ....Real Analysis has not really been my thing...I just have little idea..Am mostly good with Numerical and Differential Equation

Answer 8

Let \(S = [0,1]\). If \(x\) and \(y\) are in \(S\) with \(x\ne y\).
How can we show that there are \(m,n\in \mathbb{N}\) such that \(x<\frac{m}{2^n}<y\).

Answer 9

Is that your statement?

Answer 10

I would asume you meant to say, \(\forall x,y\in\mathbb{N}: x<y\) .

Answer 11

Yes

Answer 12

(just adding that \(x<y\) has to be a condition as well )

Answer 13

Nope

Answer 14

The first statement is right

Answer 15

When I said "I assume" ... I meant to add to the statement

Answer 16

x, y are reals, not natural

Answer 17

Well, in the statement they are natural.

Answer 18

Yes they are natural in this case

Answer 19

m,n are naturals
x,y are reals in [0, 1]

Answer 20

oh, yes ...

Answer 21

Let S = [0,1]. If x and y are in S with x\(\color{red}{<}\)y.
Show that there are m,n∈N such that x< m/2^n <y.

I edited the statement

Answer 22

Here is an idea, not sure if it works all the way..

First we choose \(n\) such that \(1/2^n < |x-y|\).

Answer 23

That way we get a "measuring stick" whose length is less than the interval (x, y)

Answer 24

For example, \(n=\lfloor~~ -\log_2 |x-y| ~~\rfloor +1\) works just fine.

Answer 25

Since our measuring stick is less than the interval width (x,y),
there exists a multiple of the measuring stick \(m*1/2^n\) , which falls in the interval (x,y).

Does this argument sound any good ?

Answer 26

@ gameshie
I have been able to work on the hint given. Please does the below sounds gound? Then how can i complete this?
x<m2n<y ⟺ 2nx<m<2ny.x<m2n<y ⟺ 2nx<m<2ny. For x<yx<y such integers m,nm,n necessarily exist since, by the Archimedean property, 2n(y−x)>1 2n(y−x)>1 for large n,n, so said interval must contain an integer mm since the interval has length >1.

Answer 27

I have been able to work on the hint given. Please does the below sounds gound? Then how can i complete this?
x<m2^n<y ⟺ 2^nx<m<2^ny.x<m2n<y ⟺ 2^nx<m<2^ny. For x<y such integers m,n necessarily exist since, by the Archimedean property, 2^n(y−x)>1 for large n, so said interval must contain an integer mm since the interval has length >1.

Answer 28

I think you are missing some division signs

Answer 29

Hello Guys....can anyone provide me with any useful textbook on real analysis that i can have an idea to solve this? I have been on it for a while without any success