Writing in Standard Form

Starting from 0.00047, we move the decimal point 4 places to the right to get 4.7. Each rightward shift is a multiplication by 10. Because we made the decimal value \( 10^4 \) times bigger, we must divide the exponent by \( 10^4 \) (multiply by \( 10^{-4} \)) to compensate and keep the value balanced. Therefore \( 0.00047 = 4.7 \times 10^{-4} \). We can verify: \( 4.7 \div 10 = 0.47 \), \( \div 10 = 0.047 \), \( \div 10 = 0.0047 \), \( \div 10 = 0.00047 \) ✓

Alternatively, think of it as place value: the 4 is in the ten-thousandths column (\( 10^{-4} \)) and the 7 is in the hundred-thousandths column (\( 10^{-5} \)), so \( 0.00047 = 4 \times 10^{-4} + 7 \times 10^{-5} = 4.7 \times 10^{-4} \).

Standard form requires a number to be written as \( a \times 10^n \) where \( 1 \leq a < 10 \). Here, \( a = 56 \), which is not between 1 and 10. Although \( 56 \times 10^3 \) does equal 56,000, the notation doesn’t satisfy the definition. The correct standard form is \( 5.6 \times 10^4 \), since 5.6 is between 1 and 10.

The constraint on \( a \) is what makes standard form useful — it guarantees every number has exactly one standard form representation, and the power of 10 immediately tells you the order of magnitude. Without the constraint, you could write 56,000 as \( 560 \times 10^2 \), \( 0.56 \times 10^5 \), etc., which defeats the purpose.

Converting both to ordinary numbers: \( 3.5 \times 10^4 = 35{,}000 \) and \( 8.9 \times 10^3 = 8{,}900 \). Clearly \( 35{,}000 > 8{,}900 \), so \( 3.5 \times 10^4 \) is greater. The key insight is that the power of 10 determines the order of magnitude: \( 10^4 = 10{,}000 \) while \( 10^3 = 1{,}000 \), a tenfold difference.

Students may instinctively compare the \( a \) values (\( 8.9 > 3.5 \)) and conclude the second is larger. But this ignores the power of 10. When comparing standard form numbers, always compare the powers first — a higher power means a larger number, regardless of the \( a \) value (since \( a \) is always between 1 and 10).

A number is in standard form if it can be written as \( a \times 10^n \) where \( 1 \leq a < 10 \) and \( n \) is an integer. If the number \( a \) itself satisfies \( 1 \leq a < 10 \), we must write it as \( a \times 10^0 \), since \( 10^0 = 1 \), and multiplying by 1 doesn’t change the value. For example, to write 7 in standard form, it becomes \( 7 \times 10^0 \).

This surprises students who associate standard form only with “very big” or “very small” numbers. But the definition doesn’t require a large or small power — \( 10^0 = 1 \) is perfectly valid. It shows that standard form is a continuous format, not just an inherent property of extreme numbers.

Example: \( 4 \times 10^3 = 4{,}000 \)

Another: \( 2.5 \times 10^2 = 250 \)

Creative: \( 9.99 \times 10^3 = 9{,}990 \) — just under 10,000, showing how close the boundary can get.

Trap: \( 25 \times 10^2 = 2{,}500 \) — this equals a whole number less than 10,000, but it’s not written in standard form because 25 is not between 1 and 10.

Example: \( 5 \times 10^{-1} = 0.5 \)

Another: \( 3.2 \times 10^{-3} = 0.0032 \)

Creative: \( 9.9 \times 10^{-1} = 0.99 \) — just under 1. Many students don’t realise numbers this close to 1 still need a negative power.

Trap: \( 0.4 \times 10^2 = 40 \) — this looks like it might be between 0 and 1 because of the 0.4, but it’s actually 40. And it’s not in standard form because 0.4 is less than 1.

Example: \( 9.5 \times 10^2 = 950 \) and \( 1.01 \times 10^3 = 1{,}010 \) — difference of 60.

Another: \( 9.8 \times 10^1 = 98 \) and \( 1.02 \times 10^2 = 102 \) — difference of just 4.

Creative: \( 9.99 \times 10^3 = 9{,}990 \) and \( 1.001 \times 10^4 = 10{,}010 \) — difference of only 20, despite one being in the thousands and the other in the ten-thousands.

Trap: \( 5 \times 10^2 = 500 \) and \( 6 \times 10^3 = 6{,}000 \) — difference of 5,500, which is not less than 100. Students who don’t convert to ordinary numbers might think “the numbers look close.”

Example: \( (2 \times 10^3) \times (3 \times 10^5) = 6 \times 10^8 \)

Another: \( (4 \times 10^4) \times (1.5 \times 10^4) = 6 \times 10^8 \)

Creative: \( (8 \times 10^{-2}) \times (7.5 \times 10^9) = 60 \times 10^7 = 6 \times 10^8 \) — using a negative power is unexpected, and the intermediate result needs adjusting back into standard form.

Trap: \( (20 \times 10^3) \times (3 \times 10^4) = 60 \times 10^7 = 6 \times 10^8 \) in value, but \( 20 \times 10^3 \) is not in standard form because 20 ≥ 10. The question requires both numbers to be in standard form.

Example: \( -4 \times 10^3 = -4{,}000 \)

Another: \( -2.5 \times 10^5 = -250{,}000 \)

Creative: \( -1.01 \times 10^{-4} = -0.000101 \) — combining a negative coefficient with a negative power!

Trap: \( 4.5 \times 10^{-3} \) — this looks “negative” because of the minus sign in the power, but it’s actually just a small positive decimal (\( 0.0045 \)). To make a number negative in standard form, the minus sign must go in front of the \( a \) value.

\( 3 \times 10^2 = 300 \), which is greater than 10. But consider \( 1 \times 10^1 = 10 \), which is equal to 10, not greater than it. This is the only counterexample — any \( a > 1 \) with power 1 gives a result greater than 10, and any power ≥ 2 gives at least 100.

The edge case \( 1 \times 10^1 = 10 \) catches out students who assume “positive power” automatically means “bigger than 10.” The precise wording matters: greater than is not the same as greater than or equal to.

True case: \( (2 \times 10^3) \times (3 \times 10^4) = 6 \times 10^7 \) — already in standard form since \( 1 \leq 6 < 10 \). False case: \( (5 \times 10^3) \times (4 \times 10^2) = 20 \times 10^5 \) — not in standard form since 20 ≥ 10. This must be rewritten as \( 2 \times 10^6 \).

The result is in standard form only when the product of the two \( a \) values stays between 1 and 10. If the product of the \( a \) values is 10 or more, you need to adjust by moving one factor of 10 into the power.

In standard form, \( a \times 10^n \) requires \( 1 \leq a < 10 \). Since \( a \) is always positive and \( 10^n \) is always positive (for any integer \( n \)), the product \( a \times 10^n \) is always positive. For example, \( 3 \times 10^{-2} = 0.03 \), which is small but positive.

This is the “negative power = negative number” misconception. A negative power makes the number small (between 0 and 1), not negative. The negative in the power affects the size, not the sign. To write a negative number like \( -4{,}500 \) you would need \( -4.5 \times 10^3 \) — the negative sign goes in front of the \( a \) value, not in the power.

If \( 0 < x < 1 \) and \( x = a \times 10^n \) where \( 1 \leq a < 10 \), then we need \( a \times 10^n < 1 \). Since \( a \geq 1 \), we need \( 10^n < 1 \), which requires \( n < 0 \) (a negative power). If \( n = 0 \), then \( a \times 10^0 = a \geq 1 \), which is not between 0 and 1.

This is the logical flip side of ASN Q3 above. A negative power doesn’t make a number negative — it makes it less than 1. And conversely, any number between 0 and 1 must have a negative power when written in standard form. The two ideas reinforce each other.

The student has correctly identified that \( 45 \times 10^3 = 45{,}000 \), but this is not standard form. This is the “any number times a power of 10” misconception — the student treats standard form as simply factoring out powers of 10, without applying the constraint that the first number (\( a \)) must satisfy \( 1 \leq a < 10 \). Since 45 is not between 1 and 10, this isn’t standard form.

The correct answer is \( 4.5 \times 10^4 \). The decimal point in 45,000 must be moved 4 places to the left to give 4.5, and we compensate with \( 10^4 \). A quick check: \( 45 \times 10^3 = 4.5 \times 10 \times 10^3 = 4.5 \times 10^4 \).

The student correctly identified that the decimal moves 3 places and that \( a = 3.6 \), but used a positive power instead of a negative one. This is the “positive power regardless of direction” misconception — the student counts the number of places correctly but forgets (or doesn’t understand) that moving the decimal to the right for a small number requires a negative power.

The correct answer is \( 3.6 \times 10^{-3} \). A quick check reveals the error: \( 3.6 \times 10^3 = 3{,}600 \), which is clearly not 0.0036. The negative power indicates the number is less than 1 — every number between 0 and 1 has a negative power in standard form.

The answer 4,000 is correct, but the reasoning — “the power tells you how many zeros to write after the number” — only works because \( a = 4 \) happens to be a whole number with no decimal digits. This is the “add zeros” misconception.

If the student applied this method to \( 4.5 \times 10^3 \), they would write 4.5000 = 4.5, not the correct answer of 4,500. The correct understanding is that the power tells you how many places to move the decimal point to the right (for positive powers). Starting from 4.0, moving the decimal 3 places right gives 4,000. Starting from 4.5, moving 3 places right gives 4,500. The “add zeros” shortcut fails whenever the \( a \) value has digits after its decimal point.

The student has only compared the \( a \) values (\( 9.2 > 1.8 \)) and completely ignored the powers of 10. This is the “compare the front number” misconception — a very common error when students first encounter standard form. It echoes the “longer is larger” misconception from decimals.

Converting both: \( 9.2 \times 10^4 = 92{,}000 \) and \( 1.8 \times 10^5 = 180{,}000 \). The second number is almost twice as large. When comparing numbers in standard form, the power of 10 must be considered first — a higher power means a larger order of magnitude, and since the \( a \) values are always between 1 and 10, a difference of one in the power always outweighs any difference in the \( a \) values.

The student has confused the rules for addition with the rules for multiplication. This is the “adding powers during addition” misconception.

When multiplying, you would indeed add the powers. But when adding, the power of 10 dictates the place value column. \( (3 \times 10^4) + (2 \times 10^4) \) is simply \( 30{,}000 + 20{,}000 = 50{,}000 \), which is \( 5 \times 10^4 \). The power does not change, just as adding 3 thousands and 2 thousands gives 5 thousands, not 5 millions.