Leading zeros and fixed-width binary
Name:
Date:
Class:
A
Quick Questions
Shade in the A or B circle for each question.
1. Which shows 101 padded to 8 bits?
2. Does adding leading zeros change the value of a binary number?
3. What is 11 written as a 4-bit binary number?
4. How many bits are in 00001111?
5. What is the value of 00000101?
B
Match the Pairs
Draw a line connecting each item on the left to its match on the right.
| Left column | Right column | |
|---|---|---|
| 100000101 | AThe binary number 101 padded with leading zeros to fill 8 bits. Its value is still 5. | |
| 20110 | BThe zeros were added on the right, which changes the value to 160. Padding zeros must go on the left: 00000101. | |
| 3Writing 11 as a 4-bit binary group by adding leading zeros | C0011 | |
| 4Writing 1001 as an 8-bit binary number by padding with zeros on the left | DA 4-bit group holding the binary number 110, which is 6 in denary. | |
| 5101 padded to 8 bits: 10100000 | E00001001 | |
| 60011 is bigger than 11 because it has more digits | FLeading zeros add nothing. 0011 and 11 are both 3, so the value is the same. |
C
Definitions
00000101
D
Fill in the Blanks
Complete each line of code by filling in the missing parts.
1.
101 as 8 bits: _____101
2.
=DEC2BIN(5,_)
3.
3 in 4 bits is __11
4.
00000101 has the same value as ___
5.
=_______(5,8)
E
Spot the Error
Shade in the letter of the line that contains an error.
Question 1
A
n = 5
B
bits = bin(n)[2:]
C
padded = bits.zfill(8)
D
print(len(padded) == 4)
Question 2
A
n = 5
B
bits = format(n, '08')
C
print(bits)
D
print(int(bits, 2))
Question 3
A
n = 6
B
nibble = format(n, '04b')
C
print(int(nibble, 10))
Question 4
A
a = int('101', 2)
B
b = int('00010101', 2)
C
print(a == b)
Question 5
A
bits = 101
B
padded = bits.zfill(8)
C
print(padded)
Answer Key
Leading zeros and fixed-width binary
Section A: Quick Questions
Q1
B — 00000101
Q2
A — No, 0101 and 101 are both 5
Q3
A — 0011
Q4
B — 8 bits
Q5
A — 5
Section B: Match the Pairs
Q1
00000101 → The binary number 101 padded with leading zeros to fill 8 bits. Its value is still 5.
Q2
0110 → A 4-bit group holding the binary number 110, which is 6 in denary.
Q3
Writing 11 as a 4-bit binary group by adding leading zeros → 0011
Q4
Writing 1001 as an 8-bit binary number by padding with zeros on the left → 00001001
Q5
101 padded to 8 bits: 10100000 → The zeros were added on the right, which changes the value to 160. Padding zeros must go on the left: 00000101.
Q6
0011 is bigger than 11 because it has more digits → Leading zeros add nothing. 0011 and 11 are both 3, so the value is the same.
Section C: Definitions
Q1
00000101 — 00000101 is the binary number 101 padded with leading zeros to fill 8 bits. Zeros added on the left do not change the value, so it is still 5.Section D: Fill in the Blanks
Q1
101 as 8 bits: 00000101Q2
=DEC2BIN(5,8)Q3
3 in 4 bits is 0011Q4
00000101 has the same value as 101Q5
=DEC2BIN(5,8)Section E: Spot the Error
Q1
D — zfill(8) pads the binary number to 8 bits, so the length check should compare with 8, not 4.
Q2
B — The format code needs the letter b to give binary. '08b' gives 00000101 (8 bits, padded with zeros). '08' would just give a decimal number.
Q3
C — The nibble 0110 is binary, so int() must use base 2. Base 10 reads it as the decimal number 110 instead of 6.
Q4
B — The 8-bit version of 101 is 00000101. Writing 00010101 changes a bit to a 1 (it is 10101, which is 21), so it is no longer equal to 5.
Q5
A — The binary number must be a string such as '101' so zfill can add zeros on the left. 101 without quotes is a plain integer and has no zfill.