Chapter 12: Number Systems, Binary Conversions, and Base Changes Study Guide

Overview of Number Systems and Base 10

• The study of number systems begins with our standard base 10 system, which has been utilized throughout the course.

• Our system is a place value system, which includes the following positions:

  • Ones place

  • Tens place

  • Hundredths place (likely intended as hundreds place in the context of whole numbers)

  • Thousands place, etc.

• In base 10, we utilize the digits $0$ through $9$.

• Once you reach $10$ objects, you must move into the next place value. For example, once you count ten individual items, you move into the tens place and are left with a zero in the ones place.

Historical Context of Different Bases

• Different civilizations have utilized alternative bases throughout history:

  • Babylonians: Utilized a base 60 system. Once you reached the number $59$, you had to move to the next value, which was the $60^1$ column.

  • Mayans: Utilized a positional system based on $20$. Once you reached $20$, you moved into position two.

Binary Numbers (Base 2)

• Binary numbers operate in base 2.

• They exclusively use the digits $0$ and $1$.

• The Conversion Algorithm: To convert from base 10 to a different base (such as binary), follow these steps:

  1. Divide the base 10 number by the target base (in this case, $2$).

  2. Perform long division and record the remainder.

  3. Continue dividing the resulting quotient by the base.

  4. Repeat this process until the remaining number is smaller than the base (i.e., you cannot divide by the base anymore).

  5. The remainders obtained form the new number. The first remainder found corresponds to the ones place, the second remainder to the next position, and so on. The final quotient (the last number that could not be divided) becomes the leading digit.

Examples of Base 10 to Binary Conversion

• Converting $43_{10}$ to Binary ($Base\,2$):

  • $43 \div 2 = 21$ remainder $1$ (This $1$ goes in the ones place: $2^0$)

  • $21 \div 2 = 10$ remainder $1$ (Next position: $2^1$)

  • $10 \div 2 = 5$ remainder $0$ (Next position: $2^2$ - Note: The zero is vital and cannot be omitted)

  • $5 \div 2 = 2$ remainder $1$ (Next position: $2^3$)

  • $2 \div 2 = 1$ remainder $0$ (Next position: $2^4$)

  • Final quotient is $1$, which cannot be divided by $2$ further. This goes into the final slot ($2^5$).

  • Result: $101011_2$

• Checking the Answer for $43_{10}$:

  • To check, expand the binary result using powers of $2$:

  • $1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 43$

• Converting $13_{10}$ to Binary ($Base\,2$):

  • $13 \div 2 = 6$ remainder $1$

  • $6 \div 2 = 3$ remainder $0$

  • $3 \div 2 = 1$ remainder $1$

  • The final $1$ cannot be divided further and moves to the leading position.

  • Result: $1101_2$

• Checking the Answer for $13_{10}$:

  • $1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 8 + 4 + 0 + 1 = 13$

Binary Arithmetic (Addition)

• When adding in base 2, recuerd the rule: $2_{10}$ is equal to $10_2$. You can only use digits $0$ and $1$. If a sum reaches $2$, you must carry over to the next place value.

• Example 1: $101_2 + 10_2$

  • Ones column: $1 + 0 = 1$

  • Tens column: $0 + 1 = 1$

  • Hundreds column: $1$

  • Result: $111_2$

• Example 2: Adding where carrying is required

  • Sum of $0 + 1 = 1$

  • Sum of $0 + 1 = 1$

  • Sum of $1 + 1 = 2$. Since the digit $2$ does not exist in base 2, write $0$ and carry the $1$ to the next column.

  • The next calculation involves adding the carried $1$ to the existing digits. If you have $1 + 1$, the result is again $10$.

  • Resulting values discussed in the example: $10011_2$.

• Example 3: $111_2 + 111_2$ (conceptualized)

  • $1 + 1 = 0$, carry $1$

  • $1 + 1 + 1 = 11$ (which is $3_{10}$), write $1$, carry $1$

  • Final result discussed in sequence: $1010_2$.

Converting to Other Bases

• The same division algorithm applies regardless of the base.

• Base 4 Conversion (Digits: $0$ through $3$):

  • Example: Convert $10_{10}$ to Base 4

    • $10 \div 4 = 2$ remainder $2$

    • The quotient $2$ cannot be divided by $4$ further.

    • Result: $22_4$

    • Check: $2 \times 4^1 + 2 \times 4^0 = 8 + 2 = 10$

  • Example: Convert $86_{10}$ to Base 4

    • $86 \div 4 = 21$ remainder $2$

    • $21 \div 4 = 5$ remainder $1$

    • $5 \div 4 = 1$ remainder $1$

    • The final $1$ moves to the front.

    • Result: $1112_4$

• Base 8 Conversion (Digits: $0$ through $7$):

  • Example: Convert $489_{10}$ to Base 8

    • $489 \div 8 = 61$ remainder $1$

    • $61 \div 8 = 7$ remainder $5$

    • The final $7$ moves to the front.

    • Result: $751_8$

Converting Back to Base 10

• To convert any base back to base 10, use the place value expansion method (the check step).

• Example 1: Convert $1010_2$ to Base 10

  • $1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 8 + 0 + 2 + 0 = 10_{10}$

• Example 2: Convert $143_5$ to Base 10

  • Note: The transcript corrects an error where a digit "5" was used in base 5; digits in base 5 must be $0$ through $4$.

  • $1 \times 5^2 + 4 \times 5^1 + 3 \times 5^0$

  • $25 + 20 + 3 = 48_{10}$

• Example 3: Convert $5254_6$ to Base 10

  • $5 \times 6^3 + 2 \times 6^2 + 5 \times 6^1 + 4 \times 6^0$

  • $5 \times 216 + 2 \times 36 + 5 \times 6 + 4 \times 1$

  • $1080 + 72 + 30 + 4 = 1186_{10}$