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Chapter 2: Problem 27

Write \((100)_{10}\) in the binary and ternary systems,

Short Answer

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Question: Convert the decimal number 100 to binary and ternary systems. Answer: In the binary system, 100 can be represented as 1100100. In the ternary system, 100 can be represented as 10201.

Step by step solution

01

Convert to Binary System (Base 2)

First, we will convert \((100)_{10}\) to the binary system. We will perform successive divisions by 2 and record the remainders. 1. \(100 \div 2 = 50\) with a remainder of \(0\) 2. \(50 \div 2 = 25\) with a remainder of \(0\) 3. \(25 \div 2 = 12\) with a remainder of \(1\) 4. \(12 \div 2 = 6\) with a remainder of \(0\) 5. \(6 \div 2 = 3\) with a remainder of \(0\) 6. \(3 \div 2 = 1\) with a remainder of \(1\) 7. \(1\) is our last non-zero quotient. Now, read the remainders backwards (from bottom to top): \(1100100\). Thus, \((100)_{10} = (1100100)_{2}\) in the binary system.
02

Convert to Ternary System (Base 3)

Next, we will convert \((100)_{10}\) to the ternary system. We will perform successive divisions by 3 and record the remainders. 1. \(100 \div 3 = 33\) with a remainder of \(1\) 2. \(33 \div 3 = 11\) with a remainder of \(0\) 3. \(11 \div 3 = 3\) with a remainder of \(2\) 4. \(3 \div 3 = 1\) with a remainder of \(0\) 5. \(1\) is our last non-zero quotient. Now, read the remainders backwards (from bottom to top): \(10201\). Thus, \((100)_{10} = (10201)_{3}\) in the ternary system.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binary Conversion
Converting a number from the decimal system (also known as base 10) to the binary system (base 2) involves dividing the number by 2 and recording the remainders. Each remainder represents a digit in the binary system. The process stops when the quotient becomes zero.
To perform this conversion, follow these steps:
  • Divide the number by 2.
  • Record the remainder.
  • Repeat with the quotient until it is zero.
  • The binary number is formed by reading the remainders in reverse order.
For example, let's convert 100 from base 10 to base 2:
  • 100 梅 2 = 50 remainder 0
  • 50 梅 2 = 25 remainder 0
  • 25 梅 2 = 12 remainder 1
  • 12 梅 2 = 6 remainder 0
  • 6 梅 2 = 3 remainder 0
  • 3 梅 2 = 1 remainder 1
  • Finally, note down 1 鈥� this is your last nonzero quotient.
The binary representation of 100 is therefore 1100100, by reading the remainders bottom to top. This method allows us to precisely convert any decimal to binary, one bit at a time.
Ternary Conversion
Ternary conversion follows a similar iterative process as binary conversion, but with division by three. This process will translate a decimal number into a ternary number. In the ternary system, each digit can only be 0, 1, or 2, reflecting its base-3 nature. This step-by-step method ensures accurate translation into the base 3 system.
  • Divide the number by 3.
  • Record the remainder.
  • Continue dividing the quotient by 3 until it becomes zero.
  • Read the remainders in reverse order to acquire the ternary number.
An example is converting the decimal number 100 into ternary:
  • 100 梅 3 = 33 remainder 1
  • 33 梅 3 = 11 remainder 0
  • 11 梅 3 = 3 remainder 2
  • 3 梅 3 = 1 remainder 0
  • Finally, note down 1 鈥� the last nonzero quotient.
Reading from the last remainder to the first, the number 100 in base 10 converts to 10201 in the ternary system. This systematic approach can be used for any base 10 conversion to base 3, ensuring an effective calculation of each digit.
Base 10
Our everyday number system is the decimal system, or base 10, which uses ten digits ranging from 0 to 9. Each position in a number represents a power of ten. This system is intuitive and has been historically widespread, largely due to humans having ten fingers, making it practical for daily counting and computations.In the decimal system, numbers are structured as follows:
  • Each digit in a number has a positional value, increasing as you move left.
  • The rightmost position holds the value of \(10^{0} = 1\), and the value increases to the left as \(10^1 = 10\), \(10^2 = 100\), and so forth.
  • Addition and multiplication are straightforward due to the ten-based structure.
To convert a number from another base to base 10, multiply each digit by its base raised to the power of its position. Add up these values to get the number in base 10. This step-by-step arithmetic makes it simple to understand and convert numbers between different systems. Its versatility spans many applications in everyday life, science, and engineering.

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Most popular questions from this chapter

Verify that the rational numbers \(\mathbb{Q}\) satisfy all the axioms for real numbers except the axiom of completeness.

Show that a) from a system of closed intervals covering a closed interval it is not always possible to choose a finite subsystem covering the interval; b) from a system of open intervals covering an open interval it is not always possible to choose a finite subsystem covering the interval; c) from a system of closed intervals covering an open interval it is not always possible to choose a finite subsystem covering the interval.

Show that a) every infinite set contains a countable subset; b) the set of even integers has the same cardinality as the set of all natural numbers; c) the union of an infinite set and an at most countable set has the same cardinality as the original infinite set; d) the set of irrational numbers has the cardinality of the continuum; e) the set of transcendental numbers has the cardinality of the continuum.

Using the principle of induction, show that a) the sum \(x_{1}+\cdots+x_{n}\) of real numbers is defined independently of the insertion of parentheses to specify the order of addition; b) the same is true of the product \(x_{1} \cdots x_{n}\) c) \(\left|x_{1}+\cdots+x_{n}\right| \leq\left|x_{1}\right|+\cdots+\left|x_{n}\right|\) d) \(\left|x_{1} \cdots x_{n}\right|=\left|x_{1}\right| \cdots\left|x_{n}\right|\) e) \(((m, n \in \mathbb{N}) \wedge(m-1\) and \(n \in \mathbb{N}\), equality holding only when \(n=1\) or \(x=0\) (Bernoulli's inequality); g) \((a+b)^{n}=a^{n}+\frac{n}{1 !} a^{n-1} b+\frac{n(n-1)}{2 !} a^{n-2} b^{2}+\cdots+\frac{n(n-1) \cdots 2}{(n-1) !} a b^{n-1}+b^{n}(\) Newton's binomial formula).

Show that a) if \(I\) is any system of nested closed intervals, then, $$ \sup \\{a \in \mathbb{R} \mid[a, b] \in I\\}=\alpha \leq \beta=\inf \\{b \in \mathbb{R} \mid[a, b] \in I\\} $$ and $$ [\alpha, \beta]=\bigcap_{[a, b] \in I}[a, b] $$ b) if \(I\) is a system of nested open intervals \(] a, b\left[\right.\) the intersection \(\left.\bigcap_{] a, b[\in I}\right] a, b[\) may happen to be empty. Hint : \(] a_{n}, b_{n}[=] 0, \frac{1}{n}[.\)
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