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I'm having some trouble in trying to crack the logic behind this pattern. Up to binary 7 (111) I can see that the '1' holds the positions that can be occupied by moving from right to left. However, it becomes increasingly complex and can't see the pattern, specially from 10000 onwards. I've 'cheated' and used a decimal>binary conversion chart, however, I still don't see a clear pattern in the movements of the 1s. Exercise as appears in the book

Thanks in advance.

PS. The source is Geldand's Algebra.

LocoVoco
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    This is simple binary counting. What's the problem? In the $1$s position it alternates $101010....$. In the $2$s position it alternates $00110011....$ in the $4$s position it alternates $000011110000....$ etc. – David G. Stork Jul 12 '19 at 15:19
  • The problem lies in that I can't seem to follow the logic behind the positioning of the 1s. – LocoVoco Jul 12 '19 at 15:20
  • What's the logic behind the positioning of the 1s in the following sequence? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, ... – Magma Jul 12 '19 at 15:22
  • It's a decimal sequence n+1. Well, I can see that. But how do I know where to move the 1 when counting in binary? Do I just move it to the left or...? Just a general rule or key would be great since I myself can't seem to crack it. Thanks. – LocoVoco Jul 12 '19 at 15:25
  • You don't move any digits to the left or right when you count in decimal, so why would you do so in binary? Decimal is just binary with ten digits instead of two. – Magma Jul 12 '19 at 15:27
  • I asked a friend as well and it's from 'smaller' to 'bigger'. 110 (6) is bigger than 101 (5); 111 (7) is bigger than 110 (6) and so on. I just couldn't see it. Is that right? – LocoVoco Jul 12 '19 at 15:28
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    I think you're stuck in the wrong mindset. What is the decimal number after 45516923999999? Which process do you use to find it? – Magma Jul 12 '19 at 15:30
  • Thanks very much guys. You'll see me asking more often than not around here :) – LocoVoco Jul 12 '19 at 15:32
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    You could probably use some practice with number bases. See https://www.mathsisfun.com/numbers/bases.html – John Douma Jul 12 '19 at 15:52
  • Thanks, John. I only knew of decimal and binary bases up until now. – LocoVoco Jul 12 '19 at 16:01

2 Answers2

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Not giving any chance to a cheat you can simply understand what it meant to be binary. So for a known number system a.k.a ' decimal number system ' we have 10 unique numbers and the numbers above 10 can be formulated using 1-10 numbers, regarding 10 as base . Likely in 'binary' system we have just two numbers 0,1 and we have to formulate every number by them and we will call them " the numbers having base 2" . And the meaning of it is more clear in the case conversion.

Supposed you have decimal number and you wanna make it binary. So u can proceed in a way that first you write it in a form $2^{n}$+p for the highest $n\in\Bbb{N}$ . And then go for long drive with p and break it in the manner. And continue it untill u see 1 at the end . For an example, Let's have 37. And we want to break it by 2 . So, here it will be like

37= $2^{5}$ +5,

5= $2^{2}$+1. And we will stop here . So resembling it all we get,

37= $2^{5}$ + $2^{2}$ +1

Now , comes the next part . As we have broken down a decimal and it looks like , `$2^{p}+2^{q}+.....+1$ . Thus we can proceed further assuming it a polynomial of degree p . And if 'D' is the decimal number then , f(x)=D is the polynomial equation whose root is 2 . And if we just collect it's coefficients and order them we will get a binary transportation. For example, we have the number 73 . And

73= $2^{6}$+ $2^{3}$+1 And , thus it can be written as , 73= $2^{6}$+ 0× $2^{5}$ + 0×$2^{4}$ +1× $2^{3}$+0× $2^{2}$+0×$2^{1}$+1× $2^{0}$`

And it is a polynomial equation of degree 6 . And as it is mentioned, the binary number (transformed) will be $1001001_{2}$ .

Now ,if you have a binary number which u have to transform into decimal , then just count how many 1 and 0 s are there , and then subtracting 1 from them , u get the order of the polynomial and the work is over . For an example if the number is ,`$100101_{2}$

Thus there is a polynomials of degree 5 and thus the decimal number will be ,

$2^{5}$+0×$2^{4}$+0×$2^{3}$+1×$2^{2}$ +0×$2^{1}$+1×$2^{0}$ i.e 37.

And u got it ! Just remember a binary number can't be started with 0 as it violates the condition of having a ' fix' degree of a polynomial.

R29nb2xlIFBp
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Each position from right to left is a power of $2$ using conventions $m_j$ meaning $m$(base)j and $n^k$ meaning $n$(to the power of)$k$.

So $1_2=2^0=1_{10}\quad 10_2=2^1=2_{10}\quad 100_2=2^2=4_{10}$ and so on. Using this pattern we can see that $11_2=2_{10}+1_{10}=3_{10}\qquad 110_2=4_{10}+2_{10}=6_{10}\qquad 111_2=4_{10}+2_{10}=1_{10}=7_{10}.$

In the example $10000$, the $1$ is in position $5$ so, letting $p=$position, we have $10000_2=2^{p-1}=2^{5-1}=2^4=16_{10}$

Now, let's take the sequence you showed from the last number $1100=2^3+2^2=8_{10}+4_{10}=12_{10}$

The next number is simply $1100_2+1_2=1101_2=13_{10}$

The number after that has to move the rightmost $1$ to the left and leave $0$ in the last position because of a binary carry. (Think how, in base $10,\quad 9+1=10$). So we have $1101_2+1_2=1110_2=14_{10}.$ Likewise, $1110_2+1_2=1111_2=15_{10}.$

Now pay attention:$\quad 1111_2+1_2=10000_2=16_{10}\quad $ because all of the digits had to carry.

You can figure this stuff out by hand, in most cases because the numbers are small.

Example: $10101=2^4+2^2+2^0=16_{10}+4_{10}+1_{10}=21_{10}$

poetasis
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