Adding and Subtracting in Standard Form

A common error is to add the coefficients directly (3 + 2 = 5) and write \( 5 \times 10^4 \). You cannot add \( 3 \times 10^4 \) and \( 2 \times 10^3 \) to get 5, just like you cannot add \( 3x^4 \) and \( 2x^3 \) to get \( 5x^4 \). They are different mathematical ‘units’. It is like trying to add 3 kilograms and 200 grams to get 303 kilograms.

To add in standard form abstractly, first convert to matching powers: \( 2 \times 10^3 = 0.2 \times 10^4 \). Then \( 3 + 0.2 = 3.2 \), giving \( 3.2 \times 10^4 \). The student error of getting \( 5 \times 10^4 = 50{,}000 \) is over 18,000 too large, showing why matching powers is essential.

Students who confuse addition with multiplication may calculate \( 6 \times 6 = 36 \) for the coefficients and \( 10^3 \times 10^3 = 10^6 \) for the powers, getting \( 36 \times 10^6 = 36{,}000{,}000 \) — absurdly large. This is the “applying multiplication rules to addition” misconception. The correct method is to add the coefficients since the powers match: \( 6 + 6 = 12 \), giving \( 12 \times 10^3 \).

But \( 12 \times 10^3 \) is not yet in standard form because the coefficient must satisfy \( 1 \leq \text{coefficient} < 10 \). Converting: \( 12 \times 10^3 = 1.2 \times 10^4 \). Checking: \( 6{,}000 + 6{,}000 = 12{,}000 = 1.2 \times 10^4 \). This question illustrates two steps: add the coefficients (don’t multiply them), then convert to standard form if necessary.

Since both numbers have the same power of 10, we add the coefficients: \( 7.6 + 5.1 = 12.7 \), giving \( 12.7 \times 10^4 \). Some students stop here, but \( 12.7 \times 10^4 \) is not in standard form because the coefficient must be between 1 and 10. This is the “forgetting to convert to proper standard form” error. We need to adjust: \( 12.7 \times 10^4 = 1.27 \times 10^5 \).

Checking: \( 7.6 \times 10^4 = 76{,}000 \) and \( 5.1 \times 10^4 = 51{,}000 \). The sum is \( 127{,}000 = 1.27 \times 10^5 \). Standard form always requires \( 1 \leq \text{coefficient} < 10 \). Many students forget this final adjustment, especially when the addition itself is straightforward.

To subtract, we first convert to the same power. The common error is the “incorrect power adjustment” misconception — converting \( 5 \times 10^4 \) incorrectly when changing to \( 10^6 \). Since we’re increasing the power by 2 (from \( 10^4 \) to \( 10^6 \)), we must divide the coefficient by 100, giving \( 0.05 \times 10^6 \). Then \( 2 \; – \; 0.05 = 1.95 \), so the answer is \( 1.95 \times 10^6 \). A student who divides by only 10 instead of 100 would get \( 0.5 \times 10^6 \), leading to \( 1.5 \times 10^6 \) — an incorrect answer.

Checking with ordinary numbers: \( 2 \times 10^6 = 2{,}000{,}000 \) and \( 5 \times 10^4 = 50{,}000 \). The difference is \( 2{,}000{,}000 \; – \; 50{,}000 = 1{,}950{,}000 = 1.95 \times 10^6 \). The key is recognising that each increase of 1 in the power of 10 means dividing the coefficient by 10.

When dealing with negative powers, a common misconception is that a “larger” negative number means a larger value. Here, \( 10^{-3} \) is ten times larger than \( 10^{-4} \). To add them, we must match the powers. Converting the smaller power (\( 10^{-4} \)) to the larger power (\( 10^{-3} \)): \( 8 \times 10^{-4} = 0.8 \times 10^{-3} \).

Then \( 4 + 0.8 = 4.8 \), giving \( 4.8 \times 10^{-3} \). Students often incorrectly convert the other way, or simply add the numbers to get \( 12 \times 10^{-7} \), entirely ignoring that these numbers sit in different, microscopic decimal places.

Example: \( 6 \times 10^4 + 4 \times 10^4 \) (since \( 60{,}000 + 40{,}000 = 100{,}000 = 1 \times 10^5 \))

Another: \( 9.5 \times 10^4 + 5 \times 10^3 \) (since \( 95{,}000 + 5{,}000 = 100{,}000 = 1 \times 10^5 \))

Creative: \( 9.9999 \times 10^4 + 1 \times 10^0 \) (since \( 99{,}999 + 1 = 100{,}000 = 1 \times 10^5 \) — using very different powers)

Trap: \( 5 \times 10^2 + 5 \times 10^3 \). A student might add the coefficients (\( 5 + 5 = 10 \)) and add the powers (\( 2 + 3 = 5 \)), getting “\( 10 \times 10^5 = 1 \times 10^6 \)” or just writing “\( 1 \times 10^5 \).” But actually \( 500 + 5{,}000 = 5{,}500 = 5.5 \times 10^3 \), nowhere near \( 1 \times 10^5 \). The student has confused the addition rule with the multiplication rule for powers.

Example: \( 6 \times 10^3 + 5 \times 10^3 = 11 \times 10^3 = 1.1 \times 10^4 \) (power increases from \( 10^3 \) to \( 10^4 \))

Another: \( 8.5 \times 10^5 + 3.2 \times 10^5 = 11.7 \times 10^5 = 1.17 \times 10^6 \) (power increases from \( 10^5 \) to \( 10^6 \))

Creative: \( 9.99 \times 10^9 + 2 \times 10^7 = 10.01 \times 10^9 = 1.001 \times 10^{10} \) (barely tips over into \( 10^{10} \))

Trap: \( 4 \times 10^3 + 5 \times 10^4 \). A student might think “I’m adding, so the power must go up.” But \( 4{,}000 + 50{,}000 = 54{,}000 = 5.4 \times 10^4 \). The power (\( 10^4 \)) is the same as the larger original — it doesn’t increase. The power only increases when the sum of coefficients (at the same power) reaches 10 or more.

Example: \( 2 \times 10^3 \; – \; 5 \times 10^3 = -3 \times 10^3 \) (since \( 2{,}000 \; – \; 5{,}000 = -3{,}000 \))

Another: \( 1.5 \times 10^4 \; – \; 2.8 \times 10^4 = -1.3 \times 10^4 \) (since \( 15{,}000 \; – \; 28{,}000 = -13{,}000 \))

Creative: \( 4 \times 10^2 \; – \; 3 \times 10^5 = -2.996 \times 10^5 \) (since \( 400 \; – \; 300{,}000 = -299{,}600 \) — the result has a much larger magnitude than the first number)

Trap: \( 3 \times 10^5 \; – \; 4 \times 10^4 \). A student sees \( 3 < 4 \) and thinks the answer must be negative. But \( 3 \times 10^5 = 300{,}000 \) is much larger than \( 4 \times 10^4 = 40{,}000 \), so \( 300{,}000 \; – \; 40{,}000 = 260{,}000 = 2.6 \times 10^5 \), which is positive. The student has compared the coefficients without considering the powers.

Example: \( 5 \times 10^6 \; – \; 4.99 \times 10^6 = 0.01 \times 10^6 = 1 \times 10^4 \) (power drops from \( 10^6 \) to \( 10^4 \), a drop of 2)

Another: \( 3.1 \times 10^5 \; – \; 3.09 \times 10^5 = 0.01 \times 10^5 = 1 \times 10^3 \) (power drops from \( 10^5 \) to \( 10^3 \), a drop of 2)

Creative: \( 1.001 \times 10^8 \; – \; 1 \times 10^8 = 0.001 \times 10^8 = 1 \times 10^5 \) (power drops from \( 10^8 \) to \( 10^5 \), a drop of 3)

Trap: \( 6 \times 10^7 \; – \; 2 \times 10^5 \). A student might think “the powers differ by 2, so the answer’s power must drop by 2.” But \( 60{,}000{,}000 \; – \; 200{,}000 = 59{,}800{,}000 = 5.98 \times 10^7 \). The answer keeps power \( 10^7 \) — the power only drops significantly when the two numbers are very close in value, not simply when the original powers are far apart.

This is sometimes true and sometimes false — it depends on whether the sum of the coefficients stays below 10. True case: \( 2 \times 10^3 + 3 \times 10^3 = 5 \times 10^3 \). The coefficients add to 5, which is between 1 and 10, so the power stays at \( 10^3 \). False case: \( 6 \times 10^3 + 7 \times 10^3 = 13 \times 10^3 = 1.3 \times 10^4 \). The coefficients add to 13, which is ≥ 10, so the result must be converted, raising the power to \( 10^4 \).

The power changes whenever the coefficient sum reaches 10 or more. This is the “assuming the power always stays the same” misconception.

This is always true. Because both numbers in standard form are positive (coefficients between 1 and 10), their sum is greater than the larger number. A number greater than \( a \times 10^n \) (where \( a \geq 1 \)) must, when written in standard form, have a power of at least \( n \). Therefore the power of the sum is always ≥ the larger original power.

Students might think a counterexample exists, but they won’t find one: adding a positive number to another positive number can never make it smaller. The power can stay the same (e.g. \( 5 \times 10^4 + 2 \times 10^3 = 5.2 \times 10^4 \)) or increase (e.g. \( 8 \times 10^4 + 5 \times 10^4 = 1.3 \times 10^5 \)), but it can never decrease.

This is sometimes true and sometimes false — it depends on the size of the gap between the coefficients. True case: \( 8 \times 10^4 \; – \; 3 \times 10^4 = 5 \times 10^4 \). The coefficient 5 is between 1 and 10, so the result is already in standard form. False case: \( 3 \times 10^4 \; – \; 2.5 \times 10^4 = 0.5 \times 10^4 \). The coefficient 0.5 is less than 1, so this is not in standard form — it needs converting to \( 5 \times 10^3 \).

Whenever the coefficients are close in value, the difference will be less than 1 and the result will need adjusting. This is the “assuming subtraction always gives standard form” misconception.

This is never true. This method confuses the rules for adding in standard form with the rules for multiplying. When you multiply, you multiply the coefficients and add the powers: \( (a \times 10^m) \times (b \times 10^n) = ab \times 10^{m+n} \). But addition doesn’t work this way.

Example: \( 2 \times 10^4 + 3 \times 10^2 = 20{,}000 + 300 = 20{,}300 = 2.03 \times 10^4 \). The incorrect method gives \( (2 + 3) \times 10^{4+2} = 5 \times 10^6 = 5{,}000{,}000 \) — wildly wrong. The correct method is to convert to the same power first, then add the coefficients. This is the “confusing addition rules with multiplication rules” misconception, which persists because the multiplication rule is superficially similar.

Note: Evaluating these reveals a beautiful structural feature: the first and third options evaluate to the exact same value!

The student has applied the “multiplication rule instead of the addition rule”. When multiplying standard form numbers, you multiply the coefficients and add the powers. But when adding, you must first match the powers of 10. Converting: \( 2 \times 10^4 = 0.2 \times 10^5 \). Then \( 6 + 0.2 = 6.2 \), giving \( 6.2 \times 10^5 \).

Checking: \( 6 \times 10^5 = 600{,}000 \) and \( 2 \times 10^4 = 20{,}000 \), so the sum is \( 620{,}000 = 6.2 \times 10^5 \). The student’s answer of \( 12 \times 10^9 = 12{,}000{,}000{,}000 \) is almost 20,000 times too large, showing how mixing up the operations leads to wildly incorrect results.

The student has the correct answer — \( 3 \times 10^5 + 7 \times 10^4 = 300{,}000 + 70{,}000 = 370{,}000 = 3.7 \times 10^5 \) — but the method of “putting the digits together” is not mathematically valid. It happens to work here only because the powers differ by exactly 1, so the second coefficient slots into the first decimal place.

If the question were \( 3 \times 10^5 + 7 \times 10^3 \) instead, the student’s method would give \( 3.7 \times 10^5 = 370{,}000 \). But the correct answer is \( 300{,}000 + 7{,}000 = 307{,}000 = 3.07 \times 10^5 \). The proper method is to convert to the same power first: \( 7 \times 10^4 = 0.7 \times 10^5 \), then \( 3 + 0.7 = 3.7 \). Getting the right answer by a flawed method is dangerous because it fails on different numbers.

The student correctly identified that the powers match and correctly added the coefficients, but stopped too early. The answer \( 17.7 \times 10^3 \) is not in proper standard form because the coefficient 17.7 is not between 1 and 10. This is the “forgetting to adjust to standard form” error. The final step is to convert: \( 17.7 \times 10^3 = 1.77 \times 10^4 \) (dividing the coefficient by 10 and increasing the power by 1).

Checking: \( 9{,}400 + 8{,}300 = 17{,}700 = 1.77 \times 10^4 \). Standard form always requires the coefficient to satisfy \( 1 \leq \text{coefficient} < 10 \). Many students forget this final adjustment, especially when the addition itself is straightforward.

The student has applied the “subtract everything separately” misconception — subtracting the coefficients (\( 8 \; – \; 4 = 4 \)) and subtracting the powers (\( 7 \; – \; 5 = 2 \)) as if each part of the standard form can be dealt with independently. This doesn’t correspond to any valid standard form operation. (Even division, which does involve subtracting powers, requires dividing the coefficients, not subtracting them.)

The correct method is to match the powers first. Converting: \( 4 \times 10^5 = 0.04 \times 10^7 \). Then \( 8 \; – \; 0.04 = 7.96 \), giving \( 7.96 \times 10^7 \). Checking: \( 80{,}000{,}000 \; – \; 400{,}000 = 79{,}600{,}000 = 7.96 \times 10^7 \). The student’s answer of \( 4 \times 10^2 = 400 \) is absurdly small compared to the correct answer of nearly 80 million, which shows that treating each part of standard form independently produces nonsensical results.

The student has a fundamental misunderstanding of negative numbers and powers. \( 10^{-2} \) (which is 0.01) is ten times larger than \( 10^{-3} \) (which is 0.001). The student’s logic that “3 is bigger than 2” completely ignores the negative sign in the index.

To subtract correctly, we should convert the smaller power (\( 10^{-3} \)) to match the larger power (\( 10^{-2} \)): \( 2 \times 10^{-3} = 0.2 \times 10^{-2} \). Then \( 5 \; – \; 0.2 = 4.8 \), giving the correct answer of \( 4.8 \times 10^{-2} \).