can every real number be represented by binary floating point number?

Question

I've got to prove that every real x, greater than 0, can be represented as $$x = n*2^c$$ where $$n\in <0,5;1)$$ $$n=\frac12\sum_{i=2}^\infty d_{i}2^{-i}$$ $$c \in \Bbb Z,\ \ d_{-2,-3,...}=\{0,1\}$$ What I got is that if we take (1) and divide it by 2^(n+1) we will get (2). It is now clear that I have to show that n can be every real number between <0,5; 1). Well, it is logical, that if we divide by 2 multiple times we have greater "resolution" every time so that sum (up to infinity) can be every real number in assumed interval. But I have no clue how to write it in mathematical way. Any help would be appreciated. $$2^c\le x<2^{c+1}\ (1)$$ $$\frac12\le \frac{x}{2^{c+1}} < 1(2)$$

Answer 1

For a real postive $x > 0$ there is $-\infty < c < +\infty$ such that $2^{c-1} \lt x \le 2^{c}$
Why is this true ? Because the sequence of half open intervals $ ((2^{c-1}, 2^{c}])_{c = -\infty, \infty}$ partitions the postitve real line $x > 0$. (To be more precise, the sequences $ ((2^{c-1}, 2^{c}])_{c = 1, \infty}$ and $ ((2^{-c}, 2^{1-c}])_{c = 1, \infty}$)

Then $x = 2^c * (x/2^c)$. Put $y = (x/2^c)$ so $x = 2^c * y$ and $1/2 \lt y \le 1$, i.e. $2^{-1}\lt y \le 2^0$

Now we just need to show that any $1/2 \lt y \le 1$ has a binary representation. Do this by recursively constructing a sequence of binary representations that converges to $y$.....
Put $r_1 = y$ then $2^{-1}\lt r_1 \le 2^0$
If $r_n \gt 2^{-n} $ put $b_n = 1$ else $b_n = 0$ and put $r_{n+1} = r_{n} - 2^{-n}b_n$
And (by induction) $y = (\Sigma _{i = 1, n} 2^{-i} b_i) + r_{n+1}$

Claim that for all $n$, $ 0 \le r_{n} \le 2^{1-n}$
Certainly true for n = 1.
If true for n,
then $r_n > 2^{-n} $ (and $r_{n} \le 2^{1-n}) \implies 0 \le r_{n+1} \le 2^{-n}$
while $r_n \le 2^{-n} \implies r_{n+1} = r_n \le 2^{-n}$.
So, by induction the claim is true for all n.

Then the sequence $((s_n = \Sigma _{i = 1, n} 2^{-i} b_i))_{n = 1, \infty}$ converges to $y$ (since for any $\epsilon > 0$ there is N with $r_N \le 2^{1-N} \lt \epsilon$)
I.e. $y = \Sigma _{i = 1, \infty} 2^{-i} b_i$ and then $0.b_1b_2....$ is a binary representation for y.

Note: this construction always generates an infinite sequence. If $y= 0$ then an infinite sequence of zeroes, but otherwise any remainder which is exactly $2^{-n}$ will be represented as $\Sigma _{i = n+ 1, \infty}2^{-n}$. This is the same as saying $1 = 0.9999999...$.
I think a better binary representation is achieved by changing the starting inequalities so that for $x > 0$ there is $-\infty < c < +\infty$ such that $2^{c-1} \le x \lt 2^{c}$ .
Or is this what you neant by $n\in <0,5;1)$ ? If so you can make the necessary adjustments.

Answer 2

Let $\frac12 \leq n < 1$ and let $c$ be any integer. Then since $2^c > 0,$

$$ \tfrac12 \times 2^c \leq n \times 2^c < 1 \times 2^c,$$ that is, $$ 2^{c-1} \leq n \times 2^c < 2^c.$$

(This is also true for any real number $c,$ but since you want $c$ to be an integer, I've added the assumption that it is an integer.)

So if you want to set $x = n\times 2^c,$ where $c$ is an integer (therefore also a real number!) and $\frac12 \leq n < 1,$ then you want

$$ 2^{c-1} \leq x < 2^c.$$

The mathematical statement you want is, "For any positive real number $x,$ there exists an integer $c$ such that $ 2^{c-1} \leq x < 2^c.$"

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If you need a proof of that fact, you could proceed like this, provided that you are allowed to use the fact that every set of integers bounded above by a real number has a greatest member.

For the case $x \geq 1,$ let $A = \{k \in \mathbb N \mid 2^k \leq x\}$ (that is, $A$ is the set of integers $k$ such that $2^k \leq x$). The set $A$ is a set of integers, and it is non-empty because $0$ is a member (because $2^0=1\leq x$), therefore it has a greatest member.

Let $m$ be the greatest member of $A.$ Then $m+1 \not\in A,$ so $x < 2^{m+1},$ but since $m\in A,$ $2^m \leq x.$ That is, $2^m \leq x < 2^{m+1}.$ Now let $c = m +1.$ then $$ 2^{c-1} \leq x < 2^c,$$ proving the fact for the case $x \geq 1.$

For $x < 1,$ then $\frac1x \geq 1,$ and therefore (by the case we have just proved) there exists an integer $d$ such that $ 2^{d-1} \leq \frac1x < 2^d.$ Therefore $$ 2^{-d} < x \leq 2^{1-d} .$$ If $x =2^{1-d}$ exactly, then set $c = 2 - d$ (hence $2^{c-1} = 2^{1-d} = x$); otherwise set $c = 1 - d.$ Then $$ 2^{c-1} \leq x < 2^c,$$ proving the fact for the case $x < 1.$