Question

Challenge: Show that any positive natural number \(a\) can be written as the sum of numbers in the form \(\dfrac{1}{2^n}\) where \(n \in \mathbb{N}\) and a number can be used any number of times, but ALL numbers -- \(\left\{\dfrac{1}{2^n}:n\in\mathbb{N}\right\}\) -- are to be used.

Answer 1

a= 1/2 +1/2 +1/2 + ...

Answer 2

I have a solution, but it may just be because I'm interpreting the problem incorrectly.

Answer 3

where there are 2a numbers of 1/2

Answer 4

Right. Same thought here, Suaravshakya.

Answer 5

n=1

Answer 6

lol, I'd add something to my problem.

Answer 7

But there are many solutions. Instead of summing up halves, you could sum up quarters or eighths or sixteenths, or some random combination of those.

Answer 8

yep

Answer 9

Well show that any integer cannot be expressed as sum of 1/2^n where n cannotbe repeated

Answer 10

Same logic will help

Answer 11

Ah, much more interesting question =)
It's fairly intuitive, but to prove it rigorously may be difficult.

Answer 12

It's not really rigorous.

Answer 13

Well, actually Suar, it's not just that any integer cannot be expressed if you disallow n to be repeated. In fact, the sum cannot reach or exceed 1.

Answer 14

Actually, if you read the question, I've written "positive number". =/

Answer 15

Oh hey. Did you edit it? I could have sworn this said integers before.

Answer 16

Heh, edited this a minute before you commented that. Any solutions?

Answer 17

I'm confused by "A number can be used any number of times but all numbers are to be used"

How can all numbers be used? If you want all numbers to be used, then I suppose this is not a finite sum, but an infinite sum.

Answer 18

And I believe this calls not for a proof but for a disproof.

Answer 19

So now is the final edit. Take a look at the question.

Answer 20

@Hero: Because my homework problems look like\[2x + 3x = 10, \text{ solve for }x.\]

Answer 21

lololol
truefacts

Answer 22

So assuming you are allowing infinite sums, this is not difficult.
\[\Large \sum_{n=1}^{\infty} \frac{1}{2^n} = 1\]
Therefore,
\[\Large a\sum_{n=1}^{\infty} \frac{1}{2^n} = a\]

Answer 23

\[a=\frac{1}{2}k+0\left(1+\sum_{i >1}\frac{1}{2^i}\right), \quad \text{ where } k: a=\frac{1}{2}k.\]

Answer 24

@SmoothMath Infinite sums are stated explicitly to be used in the definition of all numbers. Your answer is on par with mine in terms of Smart Alec material. This is a compliment.

Answer 25

Yes! Just a little note: Use\[\sum_{n =1}^{\infty} \dfrac{1}{2^n} + \sum_{n =1}^{\infty} \dfrac{1}{2^n}+ \sum_{n =1}^{\infty} \dfrac{1}{2^n} \cdots(a \ \text{times}) = a\]because you aren't allowed to use the number \(a\) since it is strictly not in the form shown above...

Answer 26

\[\sum_{n =1}^{\infty} \dfrac{1}{2^n} + \sum_{n =1}^{\infty} \dfrac{1}{2^n}+ \sum_{n =1}^{\infty} \dfrac{1}{2^n} \cdots(a \ \text{times}) = a\sum_{n \ge 1}2^{-n}\]

Answer 27

It is a good thing to point out, and I almost did fix it but was too lazy.
Another way to satisfy the requirement is by the property of sums:
\(\Large a\sum_{n=1}^{\infty} \frac{1}{2^n} = \sum_{n=1}^{\infty} a*\frac{1}{2^n}\)

Answer 28

Ah, and Limitless said the same thing =)

Answer 29

@ParthKohli Why don't you prove that this works for _only_ \(x=\dfrac{1}{2}\) in \[\sum_{n \ge 1}x^{n}?\]

Answer 30

or something similar.

Answer 31

@SmoothMath To be rough with you, Smooth, I simply demonstrated that I am a bit of a smart alec for the second time in this thread in an attempt to be humorous.

Answer 32

@Limitless: What does?

Answer 33

@ParthKohli Show that
\[a=a\sum_{n \ge 1}x^n \iff x=\frac{1}{2}.\]

Answer 34

Hmm, that's just another example of infinite geometric series.

Answer 35

\[S_n = \dfrac{a}{1 - r} ~ \iff r > |1| \]

Answer 36

Oops, I meant\[r < |1|\]

Answer 37

That's close. Be specific.

Answer 38

OK, do you need the derivation of it?