To clarify the idea I shall explain what I mean.

The current number systems deals with numbers in the form of tens (10s) , meaning that every 10 units in the ones unit equal 1 unit in the tens unit. I shall talk about a system that is a bit different, a system that deals with numbers in the form of twoes (2s) , every 2 units in the ones unit equal 1 unit in the twoes unit. So the system looks like the binary system. But we shall extend this system to include decimal numbers, each decimal after the point is half the value of the previous unit. For example the number 0.1 in our binary-like system is equal to 0.5 in the tens system. And 0.01 is equal to 0.25, and so on.

My question is , is a number like 0.251 possible to be reached in the binary-like number system and be represented by a finite number of digits ? if not, then what is the resulted number that we get from trying to reach it, is it a continuous fraction repeating itself, or a transcendental number not repeating itself ?

  • $\begingroup$ To clarify to the reader: this is the ordinary binary real number expansion. $\endgroup$ – Jam Mar 1 '20 at 20:52
  • $\begingroup$ To OP: you know from your definition that $0.251 = \ldots + a_1\cdot 2^1 + a_0\cdot 2^0+ a_{-1}\cdot \frac12+ a_{-2}\cdot \frac14 +\ldots$. Have you tried finding these coefficients? To put it another way, how many $0.5$'s are in $0.251$, and how many $0.25$'s, or $0.125$'s, etc. $\endgroup$ – Jam Mar 1 '20 at 20:54
  • $\begingroup$ @Jam No , I don't know how to , I need help $\endgroup$ – Amr S. Mar 1 '20 at 20:57
  • $\begingroup$ Find the largest multiple of each negative power of $2$ (i.e., $\frac12, \frac14, \frac18, \ldots$) that fits into $0.251$. To start you off, there are $0$ $\frac12$'s since $0.5$ doesn't fit. You can eventually find that you get a recurring expansion since $0.251$ is not a finite sum of powers of $2$. $\endgroup$ – Jam Mar 1 '20 at 21:00
  • $\begingroup$ The binary expansion in question is $0.010\overline{0000001000001100010010011011101001011110001101010011111101111100111011011001000101101000011100101011}$ where the overbar denotes the recurring digits. The recurring section is longer than I thought so it's not very quick to do by hand but you can easily convert numbers to their binary expansions with Wolfram Alpha. $\endgroup$ – Jam Mar 1 '20 at 21:09

A finite decimal number can be expressed as $x/10^m$, for some integers $x$ and $m\ge0$.

Similarly, a (rational) number has a finite $b$-adic expansion if and only if it can be expressed as $y/b^n$, for some integers $y$ and $n\ge0$.

Now, suppose $$ \frac{251}{1000}=\frac{y}{2^n} $$ Then $$ 251\cdot2^n=1000y $$ which is a contradiction, because the right-hand side is divisible by $5$ and the left-hand side isn't.

The converse, however, is true: every finite binary expansion can be converted to a finite decimal expansion because $$ \frac{y}{2^n}=\frac{y\cdot5^n}{10^n} $$

A rational number always has a repeating (or finite) expansion in every (integer) basis.

  • $\begingroup$ So would trying to represent it in the binary system result in a a decimal that doesn't repeat itself/ transcendental number ? $\endgroup$ – Amr S. Mar 1 '20 at 21:06
  • $\begingroup$ @AmrS. See addition. You can easily find a proof. $\endgroup$ – egreg Mar 1 '20 at 21:07
  • $\begingroup$ so the answer to the latter question is yes , right ? just making sure $\endgroup$ – Amr S. Mar 1 '20 at 21:11
  • $\begingroup$ @AmrS. The binary expansion of $0.251$ is repeating, because $0.251$ is a rational number. $\endgroup$ – egreg Mar 1 '20 at 21:17

Let $x$ be a number with finite binary representation. Then, if the leftmost one is $k$ positions from the fractional point, $z=x \times 2^k $ is an odd integer. Then $x$ can be expressed as a irredudible fraction $z/2^k$.

In your case $x=\frac{251}{1000}=\frac{251}{2^3 5^3}$ cannot be written in such form.


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