Guys First with formula Step by Step (1000)2
binary to decimal
Answers
Answer:The decimal expansion of a number is its representation in base-10 (i.e., in the decimal system). In this system, each "decimal place" consists of a digit 0-9 arranged such that each digit is multiplied by a power of 10, decreasing from left to right, and with a decimal place indicating the 10^0=1s place. For example, the number with decimal expansion 1234.56 is defined as
1234.56 = 1×10^3+2×10^2+3×10^1+4×10^0+5×10^(-1)+6×10^(-2)
(1)
= 1×1000+2×100+3×10+4+5×1/(10)+6×1/(100).
(2)
Expressions written in this form sum_(k)b_k10^k (where negative k are allowed as exemplified above but usually not considered in elementary education contexts) are said to be in expanded notation.
Other examples include the decimal expansion of 25^2 given by 625, of pi given by 3.14159..., and of 1/9 given by 0.1111.... The decimal expansion of a number can be found in the Wolfram Language using the command RealDigits[n], or equivalently, RealDigits[n, 10].
The decimal expansion of a number may terminate (in which case the number is called a regular number or finite decimal, e.g., 1/2=0.5), eventually become periodic (in which case the number is called a repeating decimal, e.g., 1/3=0.3^_), or continue infinitely without repeating (in which case the number is called irrational).
The following table summarizes the decimal expansions of the first few unit fractions. As usual, the repeating portion of a decimal expansion is conventionally denoted with a vinculum.
fraction decimal expansion fraction decimal expansion
1 1 1/(11) 0.09^_
1/2 0.5 1/(12) 0.083^_
1/3 0.3^_ 1/(13) 0.076923^_
1/4 0.25 1/(14) 0.0714285^_
1/5 0.2 1/(15) 0.06^_
1/6 0.16^_ 1/(16) 0.0625
1/7 0.142857^_ 1/(17) 0.0588235294117647^_
1/8 0.125 1/(18) 0.05^_
1/9 0.1^_ 1/(19) 0.052631578947368421^_
1/(10) 0.1 1/(20) 0.05
If r=p/q has a finite decimal expansion (i.e., r is a regular number), then
r = (a_1)/(10)+(a_2)/(10^2)+...+(a_n)/(10^n)
(3)
= (a_110^(n-1)+a_210^(n-2)+...+a_n)/(10^n)
(4)
= (a_110^(n-1)+a_210^(n-2)+...+a_n)/(2^n·5^n).
(5)
Factoring possible common multiples gives
r=p/(2^alpha5^beta),
(6)
where p≢0 (mod 2, 5). Therefore, the numbers with finite decimal expansions are fractions of this form. The first few regular numbers are 1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, ... (OEIS A003592).
Any nonregular fraction m/n is periodic, and has a decimal period lambda(n) independent of m, which is at most n-1 digits long. If n is relatively prime to 10, then the period lambda(n) of m/n is a divisor of phi(n) and has at most phi(n) digits, where phi is the totient function. It turns out that lambda(n) is the multiplicative order of 10 (mod n) (Glaisher 1878, Lehmer 1941). The number of digits in the repeating portion of the decimal expansion of a rational number can also be found directly from the multiplicative order of its denominator.
When a rational number m/n with (m,n)=1 is expanded, the period begins after s terms and has length t, where s and t are the smallest numbers satisfying
10^s=10^(s+t) (mod n).
(7)
When n≢0 (mod 2, 5), s=0, and this becomes a purely periodic decimal with
10^t=1 (mod n).
(8)
As an example, consider n=84.
10^0=1 10^1=10 10^2=16 10^3=-8; 10^4=4 10^5=40 10^6=-20 10^7=-32; 10^8=16,
(9)
so s=2, t=6. The decimal representation is 1/84=0.01190476^_. When the denominator of a fraction m/n has the form n=n_02^alpha5^beta with (n_0,10)=1, then the period begins after max(alpha,beta) terms and the length of the period is the exponent to which 10 belongs (mod n_0), i.e., the number x such that 10^x=1 (mod n_0). If q is prime and lambda(q) is even, then breaking the repeating digits into two equal halves and adding gives all 9s. For example, 1/7=0.142857^_, and 142+857=999. For 1/q with a prime denominator other than 2 or 5, all cycles n/q have the same length (Conway and Guy 1996).
If n is a prime and 10 is a primitive root of n, then the period lambda(n) of the repeating decimal 1/n is given by
lambda(n)=phi(n),
(10)
where phi(n) is the totient function. Furthermore, the decimal expansions for p/n, with p=1, 2, ..., n-1 have periods of length n-1 and differ only by a cyclic permutation. Such numbers n are called full reptend primes.
To find denominators with short periods, note that
10^1-1 = 3^2
(11)
10^2-1 = 3^2·11
(12)
10^3-1 = 3^3·37
(13)
10^4-1 = 3^2·11·101
(14)
10^5-1 = 3^2·41·271
(15)
10^6-1 = 3^3·7·11·13·37
(16)
10^7-1 = 3^2·239·4649
(17)
10^8-1 = 3^2·11·73·101·137
(18)
10^9-1 = 3^4·37·333667
(19)
10^(10)-1 = 3^2·11·41·271·9091
(20)
10^(11)-1 = 3^2·21649·513239
(21)
10^(12)-1 = 3^3·7·11·13·37·101·9901.
(22)
period primes
1 3
2 11
3 37
4 101
5 41, 271
6 7, 13
7 239, 4649
8 73, 137
9 333667
10 9091
11 21649, 513239
12 9901
13 53, 79, 265371653
14 909091
15 31, 2906161
16 17, 5882353
17 2071723, 5363222357
18 19, 52579
19 1111111111111111111
20 3541, 27961
Answer:
decimal number:
8
Decimal from signed 2's complement:
N/A
Hex number:
8
Decimal calculation steps:
(1000)₂ = (1 × 2³) + (0 × 2²) + (0 × 2¹) + (0 × 2⁰) = (8)₁₀
Explanation:
hope so it will help you