Write Integers Greater Than 1 in Standard Form
Atoms of Knowledge
Before we start, make sure you understand these building blocks:
CATEGORICAL
Identifying place value of digits: Can you identify which position each digit occupies (units, tens, hundreds, thousands, etc.)?
TRANSFORMATION
Moving the decimal point: Can you move a decimal point left or right and count how many places it moved?
FACT
Standard form definition: Standard form is written as \(A \times 10^n\) where \(1 \leq A < 10\) and \(n\) is an integer.
CATEGORICAL
Recognizing significant digits: Can you identify which digits are significant (non-zero or zeros between non-zero digits) versus trailing zeros?
TRANSFORMATION
Counting decimal place movements: Can you count accurately how many places the decimal point needs to move?
FACT
Powers of 10: \(10^1 = 10\), \(10^2 = 100\), \(10^3 = 1000\), \(10^4 = 10000\), etc.
I Do – Worked Examples
Example 1: Write 3500 in standard form
3500
Step 1: Identify where the decimal point currently sits
The decimal point is after the last digit: 3500.
The decimal point is after the last digit: 3500.
Step 2: Move the decimal point to get a number between 1 and 10
Moving the decimal 3 places to the left: 3500 → 3.500 → 3.5
We get \(A = 3.5\)
Moving the decimal 3 places to the left: 3500 → 3.500 → 3.5
We get \(A = 3.5\)
Step 3: Count how many places the decimal moved
The decimal moved 3 places to the left, so \(n = 3\)
The decimal moved 3 places to the left, so \(n = 3\)
Step 4: Write in standard form
\(3500 = 3.5 \times 10^3\)
\(3500 = 3.5 \times 10^3\)
Why this works: Standard form requires a number between 1 and 10 multiplied by a power of 10. Moving the decimal 3 places left means we multiply by \(10^3\) to get back to the original value. The trailing zeros don’t appear in our final answer because \(3.5 \times 10^3\) automatically represents 3500.
Example 2: Write 30,500 in standard form
30,500
Step 1: Identify where the decimal point currently sits
The decimal point is after the last digit: 30500.
The decimal point is after the last digit: 30500.
How this relates to Example 1: In Example 1, we worked with 3500, which had all its zeros at the end. Now we’re looking at 30,500, which has a zero in the middle (between 3 and 5). This is important because the zero in the middle is significant and must be preserved in our answer, unlike trailing zeros which disappear in standard form.
Step 2: Move the decimal point to get a number between 1 and 10
Moving the decimal 4 places to the left: 30500 → 3.0500 → 3.05
We get \(A = 3.05\) (the zero between 3 and 5 is significant)
Moving the decimal 4 places to the left: 30500 → 3.0500 → 3.05
We get \(A = 3.05\) (the zero between 3 and 5 is significant)
Step 3: Count how many places the decimal moved
The decimal moved 4 places to the left, so \(n = 4\)
The decimal moved 4 places to the left, so \(n = 4\)
Step 4: Write in standard form
\(30500 = 3.05 \times 10^4\)
\(30500 = 3.05 \times 10^4\)
Comparing the two answers: Example 1 gave us \(3.5 \times 10^3\) and Example 2 gives us \(3.05 \times 10^4\). Notice how Example 2’s answer has an extra zero (3.05 instead of 3.5) because the original number had a zero in the middle that’s significant. Also, Example 2 has a higher power (\(10^4\) instead of \(10^3\)) because it’s a larger number. The same process works for both – we just need to be careful to preserve zeros that appear between non-zero digits.
We Do – Guided Practice
Write 8200 in standard form
8200
Step 1: Identify where the decimal point currently sits
The decimal point is after the last digit: 8200.
The decimal point is after the last digit: 8200.
Step 2: Move the decimal point to get a number between 1 and 10
Moving the decimal 3 places to the left: 8200 → 8.200 → 8.2
We get \(A = 8.2\)
Moving the decimal 3 places to the left: 8200 → 8.200 → 8.2
We get \(A = 8.2\)
Step 3: Count how many places the decimal moved
The decimal moved 3 places to the left, so \(n = 3\)
The decimal moved 3 places to the left, so \(n = 3\)
Step 4: Write in standard form
\(8200 = 8.2 \times 10^3\)
\(8200 = 8.2 \times 10^3\)
Why this works: We moved the decimal 3 places left to create 8.2, which is between 1 and 10. Multiplying by \(10^3\) brings us back to the original value of 8200.
You Do – Independent Practice
Question 1: Write 4300 in standard form
\(4300 = 4.3 \times 10^3\)
Working: Decimal moves 3 places left: 4300 → 4.3
Working: Decimal moves 3 places left: 4300 → 4.3
Question 2: Write 67000 in standard form
\(67000 = 6.7 \times 10^4\)
Working: Decimal moves 4 places left: 67000 → 6.7
Working: Decimal moves 4 places left: 67000 → 6.7
Question 3: Write 20300 in standard form
\(20300 = 2.03 \times 10^4\)
Working: Decimal moves 4 places left: 20300 → 2.03
Working: Decimal moves 4 places left: 20300 → 2.03
Question 4: Write 850 in standard form
\(850 = 8.5 \times 10^2\)
Working: Decimal moves 2 places left: 850 → 8.5
Working: Decimal moves 2 places left: 850 → 8.5
Question 5: Write 401000 in standard form
\(401000 = 4.01 \times 10^5\)
Working: Decimal moves 5 places left: 401000 → 4.01
Working: Decimal moves 5 places left: 401000 → 4.01
Question 6: Write 9200 in standard form
\(9200 = 9.2 \times 10^3\)
Working: Decimal moves 3 places left: 9200 → 9.2
Working: Decimal moves 3 places left: 9200 → 9.2
Question 7: Write 56000 in standard form
\(56000 = 5.6 \times 10^4\)
Working: Decimal moves 4 places left: 56000 → 5.6
Working: Decimal moves 4 places left: 56000 → 5.6
Question 8: Write 10500 in standard form
\(10500 = 1.05 \times 10^4\)
Working: Decimal moves 4 places left: 10500 → 1.05
Working: Decimal moves 4 places left: 10500 → 1.05
Question 9: Write 730 in standard form
\(730 = 7.3 \times 10^2\)
Working: Decimal moves 2 places left: 730 → 7.3
Working: Decimal moves 2 places left: 730 → 7.3
Question 10: Write 208000 in standard form
\(208000 = 2.08 \times 10^5\)
Working: Decimal moves 5 places left: 208000 → 2.08
Working: Decimal moves 5 places left: 208000 → 2.08
Challenge Questions
1. Factual Recall
Write 12400 in standard form.
Answer: \(1.24 \times 10^4\)
Working: Decimal moves 4 places left: 12400 → 1.24
Explanation: This is a straightforward application of the standard form routine. We identify that the decimal must move 4 places to position 1.24, which is between 1 and 10, giving us the power of 4.
Working: Decimal moves 4 places left: 12400 → 1.24
Explanation: This is a straightforward application of the standard form routine. We identify that the decimal must move 4 places to position 1.24, which is between 1 and 10, giving us the power of 4.
2. Carry Out a Routine
Convert these numbers to standard form:
(a) 34000
(b) 90700
(c) 505000
Answers:
(a) \(3.4 \times 10^4\)
(b) \(9.07 \times 10^4\)
(c) \(5.05 \times 10^5\)
Working:
(a) 34000 → 3.4 (4 places)
(b) 90700 → 9.07 (4 places, zero is significant)
(c) 505000 → 5.05 (5 places, zero between digits is significant)
Explanation: Each conversion follows the same routine but demonstrates variety. Part (a) has all zeros at the end. Part (b) shows a zero between non-zero digits that must be preserved. Part (c) shows zeros both in the middle and at the end, testing careful identification of significant digits.
(a) \(3.4 \times 10^4\)
(b) \(9.07 \times 10^4\)
(c) \(5.05 \times 10^5\)
Working:
(a) 34000 → 3.4 (4 places)
(b) 90700 → 9.07 (4 places, zero is significant)
(c) 505000 → 5.05 (5 places, zero between digits is significant)
Explanation: Each conversion follows the same routine but demonstrates variety. Part (a) has all zeros at the end. Part (b) shows a zero between non-zero digits that must be preserved. Part (c) shows zeros both in the middle and at the end, testing careful identification of significant digits.
3. Classify
Which of these numbers are correctly written in standard form?
(a) \(12.5 \times 10^3\)
(b) \(4.6 \times 10^2\)
(c) \(0.8 \times 10^4\)
(d) \(7.03 \times 10^5\)
Answers: (b) and (d) are correct
Working:
(a) Incorrect: 12.5 is not between 1 and 10
(b) Correct: 4.6 is between 1 and 10 ✓
(c) Incorrect: 0.8 is not between 1 and 10
(d) Correct: 7.03 is between 1 and 10 ✓
Explanation: Standard form requires \(1 \leq A < 10\). This means \(A\) must be at least 1 but less than 10. Option (a) fails because 12.5 ≥ 10, and option (c) fails because 0.8 < 1. Options (b) and (d) both satisfy the condition, with (d) demonstrating that zeros within the decimal number are perfectly acceptable.
Working:
(a) Incorrect: 12.5 is not between 1 and 10
(b) Correct: 4.6 is between 1 and 10 ✓
(c) Incorrect: 0.8 is not between 1 and 10
(d) Correct: 7.03 is between 1 and 10 ✓
Explanation: Standard form requires \(1 \leq A < 10\). This means \(A\) must be at least 1 but less than 10. Option (a) fails because 12.5 ≥ 10, and option (c) fails because 0.8 < 1. Options (b) and (d) both satisfy the condition, with (d) demonstrating that zeros within the decimal number are perfectly acceptable.
4. Interpret
A calculator displays \(6.4 \times 10^3\) as the answer to a calculation. What is the ordinary number this represents? Explain what the power of 3 tells us.
Answer: 6400
Working: \(6.4 \times 10^3 = 6.4 \times 1000 = 6400\)
Explanation: The power of 3 tells us how many places to move the decimal point to the right to convert back to ordinary form. Starting with 6.4, we move the decimal 3 places right: 6.4 → 64 → 640 → 6400. The power essentially tells us the magnitude or size of the number – in this case, it’s in the thousands. Each increase in the power of 10 makes the number ten times larger.
Working: \(6.4 \times 10^3 = 6.4 \times 1000 = 6400\)
Explanation: The power of 3 tells us how many places to move the decimal point to the right to convert back to ordinary form. Starting with 6.4, we move the decimal 3 places right: 6.4 → 64 → 640 → 6400. The power essentially tells us the magnitude or size of the number – in this case, it’s in the thousands. Each increase in the power of 10 makes the number ten times larger.
5. Prove/Justify
Explain why \(4500 = 4.5 \times 10^3\) and why this cannot be written as \(45 \times 10^2\) in standard form, even though \(45 \times 100 = 4500\).
Answer: \(45 \times 10^2\) is not standard form because 45 is not between 1 and 10.
Working:
Standard form requires \(1 \leq A < 10\)
In \(4.5 \times 10^3\): \(A = 4.5\) ✓ (satisfies condition)
In \(45 \times 10^2\): \(A = 45\) ✗ (45 is not less than 10)
Explanation: While both expressions equal 4500 mathematically, only \(4.5 \times 10^3\) follows the standard form convention. Standard form has a strict definition requiring the first number to be between 1 and 10. This ensures every number has exactly one correct standard form representation, making it easier to compare sizes of numbers and work with very large or very small values consistently. The form \(45 \times 10^2\) would be rejected in any formal mathematical context even though the arithmetic is correct.
Working:
Standard form requires \(1 \leq A < 10\)
In \(4.5 \times 10^3\): \(A = 4.5\) ✓ (satisfies condition)
In \(45 \times 10^2\): \(A = 45\) ✗ (45 is not less than 10)
Explanation: While both expressions equal 4500 mathematically, only \(4.5 \times 10^3\) follows the standard form convention. Standard form has a strict definition requiring the first number to be between 1 and 10. This ensures every number has exactly one correct standard form representation, making it easier to compare sizes of numbers and work with very large or very small values consistently. The form \(45 \times 10^2\) would be rejected in any formal mathematical context even though the arithmetic is correct.
6. Extend
If \(3.2 \times 10^4 = 32000\), what would \(3.2 \times 10^5\) equal? Explain the relationship between increasing the power by 1 and the resulting ordinary number.
Answer: \(3.2 \times 10^5 = 320000\)
Working:
\(3.2 \times 10^4 = 32000\)
\(3.2 \times 10^5 = 3.2 \times 10 \times 10^4 = 10 \times 32000 = 320000\)
Explanation: Increasing the power of 10 by 1 multiplies the ordinary number by 10. This is because \(10^5 = 10 \times 10^4\). You can think of it as moving the decimal point one additional place to the right. More generally, if you increase the power by \(n\), you multiply the ordinary number by \(10^n\). This relationship shows why standard form is so useful for very large numbers – we can easily see how much bigger one number is than another by comparing their powers of 10.
Working:
\(3.2 \times 10^4 = 32000\)
\(3.2 \times 10^5 = 3.2 \times 10 \times 10^4 = 10 \times 32000 = 320000\)
Explanation: Increasing the power of 10 by 1 multiplies the ordinary number by 10. This is because \(10^5 = 10 \times 10^4\). You can think of it as moving the decimal point one additional place to the right. More generally, if you increase the power by \(n\), you multiply the ordinary number by \(10^n\). This relationship shows why standard form is so useful for very large numbers – we can easily see how much bigger one number is than another by comparing their powers of 10.
7. Construct
Create three different numbers in standard form that would round to \(5 \times 10^3\) when rounded to 1 significant figure. Write them in both standard form and ordinary form.
Example answers:
\(4.7 \times 10^3 = 4700\)
\(5.2 \times 10^3 = 5200\)
\(5.4 \times 10^3 = 5400\)
Working:
Numbers that round to \(5 \times 10^3\) must be in the range from \(4.5 \times 10^3\) to less than \(5.5 \times 10^3\) (in ordinary form: 4500 to 5499).
Explanation: This question tests understanding of both standard form and rounding. When rounding to 1 significant figure, we look at the first digit after the decimal point in the \(A\) value. Any number from \(4.5 \times 10^3\) up to (but not including) \(5.5 \times 10^3\) will round to \(5 \times 10^3\). The power of 10 must be 3 to stay in the thousands range. Many correct answers are possible – the key is ensuring \(A\) is between 4.5 and 5.5 (not including 5.5) and the power is 3.
\(4.7 \times 10^3 = 4700\)
\(5.2 \times 10^3 = 5200\)
\(5.4 \times 10^3 = 5400\)
Working:
Numbers that round to \(5 \times 10^3\) must be in the range from \(4.5 \times 10^3\) to less than \(5.5 \times 10^3\) (in ordinary form: 4500 to 5499).
Explanation: This question tests understanding of both standard form and rounding. When rounding to 1 significant figure, we look at the first digit after the decimal point in the \(A\) value. Any number from \(4.5 \times 10^3\) up to (but not including) \(5.5 \times 10^3\) will round to \(5 \times 10^3\). The power of 10 must be 3 to stay in the thousands range. Many correct answers are possible – the key is ensuring \(A\) is between 4.5 and 5.5 (not including 5.5) and the power is 3.
8. Criticize
A student converts 20050 to standard form and writes: \(2.005 \times 10^4\). Another student says this is wrong and writes: \(2.05 \times 10^4\). Who is correct and why?
Answer: The first student is correct.
Working:
20050 → move decimal 4 places: 2.0050
Simplified: \(2.005 \times 10^4\)
Check: \(2.005 \times 10000 = 20050\) ✓
Second student’s answer: \(2.05 \times 10000 = 20500\) ✗
Explanation: The second student made a critical error by treating the zero in the middle as if it were at the end. The number 20050 has the zero between the 2 and the 5 as significant – it represents “twenty thousand and fifty”, not “twenty thousand and five hundred”. When we move the decimal point 4 places, we must preserve all digits including that middle zero: 2.0050, which simplifies to 2.005. The second student’s answer of \(2.05 \times 10^4 = 20500\) is a completely different number. This demonstrates the importance of carefully tracking all significant digits, including zeros that appear between non-zero digits.
Working:
20050 → move decimal 4 places: 2.0050
Simplified: \(2.005 \times 10^4\)
Check: \(2.005 \times 10000 = 20050\) ✓
Second student’s answer: \(2.05 \times 10000 = 20500\) ✗
Explanation: The second student made a critical error by treating the zero in the middle as if it were at the end. The number 20050 has the zero between the 2 and the 5 as significant – it represents “twenty thousand and fifty”, not “twenty thousand and five hundred”. When we move the decimal point 4 places, we must preserve all digits including that middle zero: 2.0050, which simplifies to 2.005. The second student’s answer of \(2.05 \times 10^4 = 20500\) is a completely different number. This demonstrates the importance of carefully tracking all significant digits, including zeros that appear between non-zero digits.
