Hello world!
I'm Lilia living on another planet full of adventures and fantasies. While living on Earth I believe that there are 10 digits because we have 10 fingers, therefore we can count all single digits from 0-9. And after 9 why do we write 10?
Similarity 1: Here 1 in 10 represents how many times we use all our fingers in the example of 21 we use them twice and so on. The same happens on the planet where I used to live, but because here all creatures have 2 fingers they use only 0s and 1s to count. 0=0, 1=1, 2=10 ( here again 1 represents the time they use their fingers), 3=11... and how to represent 300 in a two-based system. Every time we need to divide it by 2 and write the remainder, the same process with the quotient till the quotient will be 1. So the 300 in 10 based system will be equal to 100101100 in 2 based system.
Similarity 2: The formula (base)^n-1 represents the greatest number, where n is the number of digits. In a 10-based system, 9 is the biggest digit 99 is the biggest 2-digit number .... each of them can be represented as 10^n -1. The same is true for numbers in two-based systems 1, 11, 111.... the formula is 2^n -1 which gives the value of these numbers in 10 based system.
Similarity 3: We can do the same operations ( sum, difference, multiplication, division)in a two-based system as we used to in decimal (10-based system) ones.
1. Addition (+)
We have 2 numbers 15 and 4.
15+4=19
15 in 2 based system will be (1111)
15:2=7( r.1)
7:2=3(r.1)
3:2=1(r.1)
1:2=0(r.1)
(Read the remainders from bottom to top)
So 15 is 1111 in a two-based system
4:2=2(r.0)
2:2=1(r.0)
1:2=0(r.1)
|=>4(10)=100(2)
(10). 15+4 <=> (2). 1111
+ 100
________
10011
2. Subtraction (-)
We have two numbers 20 and 5.
20:2=10( r.0)
10:2=5(r.0)
5:2=2(r.1)
2:2=1(r.0)
1:2=0(r.1)
(10) 20 <=> (2) 10100
5:2=2(r.1)
2:2=1(r.0)
1:2=0(r.1)
(10) 5 <=> (2) 101
(10) 20-5=15 <=> (2) 10100
- 101
_______
1111
3. Multiplication (*)
(10) 20*5=100 <=> (2) 10100
* 101
______
10100
+ 00000
+ 10100
__________
1100100
Checking: 1 1 0 0 1 0 0
(6)(5)(4)(3)(2)(1)(0)
(2^0)*0+(2^1)*0+(2^2)*1+(2^3)*0+(2^4)*0+(2^5)*1+(2^6)*1= 0+0+4+0+0+32+64= 100
I hope it was helpful:)
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