Scientific Notation: Free Response

Take the large number first. Its decimal point sits after the final zero, and exactly one nonzero digit has to end up in front of it, so slide the point left until it rests between the $2$ and the $8$. Counting the hops from the far right to that spot gives $8$ places. Each leftward hop divides by $10$, so multiplying back by $10^{8}$ restores the value:

$284{,}000{,}000 = 2.84 \times 10^{8}.$

The six trailing zeros are not significant digits. They were only holding the size, and the power of ten now holds it instead.

The small number moves the other way. Its first nonzero digit is the $9$ in the fifth decimal place, so the point slides $5$ places right to sit just after it. Each rightward hop multiplies by $10$, so dividing back five times, which is multiplying by $10^{-5}$, leaves the value unchanged:

$0.0000905 = 9.05 \times 10^{-5}.$

The zero between the $9$ and the $5$ stays, because it sits among the significant digits, while the four zeros in front of the $9$ were placeholders and disappear with the move. Both coefficients, $2.84$ and $9.05$, are at least $1$ and below $10$, which is what the form demands.

An exponent is an instruction for the decimal point: its sign gives the direction and its size gives the number of places.

The first exponent is positive, so the number is large and the point moves right, $6$ places. After the $4$ there are only two digits to pass, so four zeros fill the places the digits do not reach:

$4.06 \times 10^{6} = 4{,}060{,}000.$

The second exponent is negative, so the number is small and the point moves left, $5$ places, padding with zeros as it goes:

$9.3 \times 10^{-5} = 0.000093.$

The check is that the direction matches the size in each case. Multiplying by $10^{6}$ has to grow the coefficient, and $4{,}060{,}000$ is far larger than $4.06$. Multiplying by $10^{-5}$ has to shrink it, and $0.000093$ is far smaller than $9.3$, while still being a positive number: the minus sign lives on the exponent, not on the value.

The sign is decided before any counting, because it only records which way the point has to travel, and the size of the number already says that.

If the number is $10$ or more, its first nonzero digit sits at least one place to the left of where the coefficient needs it, so the point moves left. Leftward moves shrink the number, and the power of ten has to pay that back by growing it, so the exponent is positive.

If the number is between $0$ and $1$, its first nonzero digit sits to the right of the point, so the point moves right. Rightward moves grow the number, so the power of ten has to shrink it back, and the exponent is negative.

The remaining case is a number from $1$ up to but not including $10$. It is already a legal coefficient, so the point does not move at all, and there is nothing to pay back:

$7.5 = 7.5 \times 10^{0}.$

That is why the three ranges line up with the three signs, and the boundaries of the coefficient range are exactly where the sign changes.

The size of the exponent is settled by counting, and it counts places. Every place the point moves is one factor of ten gained or lost, so moving it $k$ places has to be compensated by $k$ factors of ten in the other direction. That is what the exponent records, which is why an exponent of $8$ and an exponent of $-5$ describe journeys of eight places and five places, differing only in which way the point walked.

Regroup so that the coefficients stand together and the powers of ten stand together, which multiplication allows:

$(7 \times 10^{5}) \times (8 \times 10^{6}) = (7 \times 8) \times (10^{5} \times 10^{6}).$

The coefficients give $56$, and the two powers of ten combine by adding their exponents:

$56 \times 10^{11}.$

That is the right value in the wrong form, because $56$ is not below $10$. Peel one factor of ten off the coefficient and hand it to the power:

$56 \times 10^{11} = 5.6 \times 10^{1} \times 10^{11} = 5.6 \times 10^{12}.$

The coefficient is now $5.6$, safely between $1$ and $10$. Nothing changed value along the way: the coefficient was divided by ten and the power was multiplied by ten in the same step, and those two cancel each other exactly.

Split the quotient the same way, coefficients with coefficients and powers of ten with powers of ten:

$\frac{3.5 \times 10^{2}}{7 \times 10^{6}} = \frac{3.5}{7} \times \frac{10^{2}}{10^{6}}.$

The coefficients divide to $0.5$, and the powers of ten divide by subtracting exponents, $2 - 6 = -4$:

$0.5 \times 10^{-4}.$

This time the coefficient has landed too low, since $0.5$ is below $1$. So borrow a factor of ten from the power instead of giving one to it, which multiplies the coefficient by ten and lowers the exponent by one:

$0.5 \times 10^{-4} = 5 \times 10^{-1} \times 10^{-4} = 5 \times 10^{-5}.$

As a check, $3.5 \times 10^{2}$ is $350$ and $7 \times 10^{6}$ is $7{,}000{,}000$, and $350 \div 7{,}000{,}000 = 0.00005$, which is exactly what $5 \times 10^{-5}$ says.

The first line is sound. The coefficients give $6 \times 4 = 24$, and the powers of ten add their exponents to give $10^{7}$:

$(6 \times 10^{4}) \times (4 \times 10^{3}) = 24 \times 10^{7}.$

The student is also right that $24$ cannot stand as a coefficient. The fault is in the exchange that follows. Rewriting $24$ as $2.4$ divides one factor of the product by ten, and dividing one factor by ten makes the whole product ten times smaller unless the other factor is multiplied by ten to compensate. So the power of ten has to grow, not shrink:

$24 \times 10^{7} = 2.4 \times 10^{1} \times 10^{7} = 2.4 \times 10^{8}.$

The student's version is off by a factor of $100$: one factor of ten lost by shrinking the coefficient, and a second lost by shrinking the power as well instead of growing it.

The cheapest check is to leave scientific notation for a moment. The two numbers are $60{,}000$ and $4{,}000$, and their product is $240{,}000{,}000$, which is $24$ followed by seven zeros:

$240{,}000{,}000 = 2.4 \times 10^{8}.$

The student's answer stands for $2{,}400{,}000$, which is nowhere near it. Converting one line to ordinary form, or even counting the zeros roughly, would have caught this slip, and a misplaced factor of ten is the mistake this notation is most prone to.

The number of pages is the room available divided by the room one page takes:

$\frac{4.8 \times 10^{14}}{6 \times 10^{6}} = \frac{4.8}{6} \times \frac{10^{14}}{10^{6}}.$

The coefficients give $4.8 \div 6 = 0.8$, and the powers of ten subtract their exponents, $14 - 6 = 8$:

$0.8 \times 10^{8}.$

The coefficient is below $1$, so this is not yet proper form. Borrowing a factor of ten from the power multiplies the coefficient by ten and lowers the exponent by one:

$0.8 \times 10^{8} = 8 \times 10^{7}.$

So about $8 \times 10^{7}$ pages fit, which is $80$ million pages. Multiplying back is the check: $8 \times 6 = 48$ and $7 + 6 = 13$, so the pages would take $48 \times 10^{13}$ bytes, and renormalising that gives $4.8 \times 10^{14}$ bytes, the capacity the system started with.

Only the size of one item has changed, so the shape of the calculation is the same:

$\frac{4.8 \times 10^{14}}{3 \times 10^{4}} = \frac{4.8}{3} \times \frac{10^{14}}{10^{4}}.$

The coefficients give $4.8 \div 3 = 1.6$, and the exponents subtract to $14 - 4 = 10$:

$1.6 \times 10^{10}.$

This coefficient needs no repair, since $1.6$ is already at least $1$ and below $10$.

For the ordinary form, the exponent $10$ sends the decimal point $10$ places right. There is one digit after the point in $1.6$, so nine zeros fill the places the digits do not reach:

$1.6 \times 10^{10} = 16{,}000{,}000{,}000.$

That is sixteen billion images, and the written-out version shows why the notation earns its keep: eleven digits and three separators say what four symbols said above.

Compare the two file sizes directly, which is a division of one size by the other:

$\frac{6 \times 10^{6}}{3 \times 10^{4}} = 2 \times 10^{2}.$

So one scanned page takes $200$ times as much room as one lower-resolution image.

Now think about what filling the system means. The capacity is a fixed amount of room, and it is being cut into pieces of one size or the other. Pieces that are $200$ times smaller fit into that same room $200$ times as often, so the count rises by exactly the factor the size fell by. The capacity never enters the comparison, because it is the same room in both cases: double it and both counts double, leaving the factor between them untouched. That is why a single division of the two file sizes answers the question, and why the answer would be identical for a library with a system ten times larger or ten times smaller.

One caution about counting whole items. The factor of $200$ is exact for the quotients themselves, and here both quotients come out whole, since this capacity divides evenly by both file sizes. A capacity that left a few bytes unused would round each count down to a whole item, which can shift a count by one without changing the factor between the quotients.

The counts from the earlier parts agree with this, which is a check rather than the reason. Multiplying $8 \times 10^{7}$ by $200$ doubles the coefficient to $16$ and raises the exponent by $2$, giving $16 \times 10^{9}$, which renormalises to $1.6 \times 10^{10}$.

Sort by exponent first. Three of the four carry $10^{-5}$ and one carries $10^{-7}$. Since $-7$ lies further left on the number line than $-5$, $10^{-7}$ is the smaller power of ten, and with every coefficient in the range from $1$ up to $10$ no coefficient can make up a whole factor of ten. So the gold leaf is thinnest outright:

$1.2 \times 10^{-7} < 1.3 \times 10^{-5}.$

It is worth seeing why the narrow gap between the coefficients $1.2$ and $1.3$ cannot rescue that comparison. Written out, the two thicknesses are $0.00000012$ and $0.000013$, and the second is over a hundred times the first. The coefficients were never in the running.

The other three tie on the exponent, so their powers of ten are identical and the comparison falls to the coefficients alone, where $1.3 < 1.7 < 9$:

$1.3 \times 10^{-5} < 1.7 \times 10^{-5} < 9 \times 10^{-5}.$

So the order from thinnest to thickest is gold leaf, plastic wrap, aluminium foil, printer paper. The exponent settled every comparison involving the gold leaf; the coefficients settled the rest, and no comparison needed both.

The rule the technician is using is a real one, but it carries a condition, and the condition is not met here.

Comparing exponents first is valid only when both coefficients sit in the range from $1$ up to $10$. That range is what stops a coefficient being large enough, or small enough, to cross a whole power of ten on its own. The supplier's figure has coefficient $0.14$, which is below $1$, so the figure is not in scientific notation at all and its exponent cannot be set against the exponents in the table.

Put it into proper form first. Write $0.14$ as $1.4 \times 10^{-1}$ and merge that spare factor of ten into the power:

$0.14 \times 10^{-4} = 1.4 \times 10^{-1} \times 10^{-4} = 1.4 \times 10^{-5}.$

Now the comparison is legitimate. The film shares the exponent $-5$ with three of the four materials, so their coefficients decide, and $1.3 < 1.4 < 1.7 < 9$. The film is thicker than the plastic wrap and thinner than both the aluminium foil and the printer paper, which is a long way from thickest on the list.

The general point is that an exponent reports size honestly only once the coefficient has been pinned to its range. An unnormalised coefficient can hide a whole factor of ten, and this one hid exactly one: the $-4$ on the supplier's sheet was doing the work of a $-5$.

One pair is enough to retire a claim of this shape, and such a pair is easy to build on purpose: take a large coefficient with a small exponent, and a small coefficient with a large one.

$7.5 \times 10^{2} = 750, \qquad 1.1 \times 10^{5} = 110{,}000.$

Both are properly written, since $7.5$ and $1.1$ are each at least $1$ and below $10$, so the shortcut has been given every chance to work. Yet the coefficient $7.5$ is the larger of the two coefficients while $750$ is far smaller than $110{,}000$. The shortcut is wrong.

The reason it fails is that the coefficient carries only the significant digits, and digits alone say nothing about size. Two coefficients both confined to the same window, $7.5$ and $1.1$, can belong to numbers that differ by well over a hundred, because size is the job of the power of ten.

The working rule reverses the order of the checks. Compare the exponents first: with both coefficients between $1$ and $10$, the number with the larger exponent is the larger number, since the smaller-exponent number cannot climb as high as the next power of ten while the larger-exponent number is already at least that high. Only when the exponents are equal do the coefficients decide, and then they decide on their own, because the two powers of ten are identical.

Evaluate them one at a time, moving the decimal point as many places right as each exponent names.

$725 \times 10^{2} = 72{,}500.$

The second needs four places, and $7.25$ has only two digits after the point, so two zeros fill the rest:

$7.25 \times 10^{4} = 72{,}500.$

The third starts further left and needs six places, landing in the same spot:

$0.0725 \times 10^{6} = 72{,}500.$

So all three are true statements about one number, and being true is not the same as being in scientific notation. The form asks one thing of these three beyond correctness: the coefficient must be at least $1$ and below $10$. Test the three coefficients against that. The coefficient $725$ fails at the top, since it is not below $10$. The coefficient $0.0725$ fails at the bottom, since it is not at least $1$. Only $7.25$ passes at both ends, so only $7.25 \times 10^{4}$ is in scientific notation.

Notice what the two failures have in common. Each has the decimal point parked somewhere other than just after the first nonzero digit, with the exponent adjusted to compensate. The compensation is what keeps the value right while the form stays wrong.

Write the list out, dividing the coefficient by ten at each step and raising the exponent by one to keep the value fixed:

$72{,}500 \times 10^{0}, \quad 7{,}250 \times 10^{1}, \quad 725 \times 10^{2},$

$72.5 \times 10^{3}, \quad 7.25 \times 10^{4}, \quad 0.725 \times 10^{5}.$

The list runs on in both directions, and every step multiplies or divides the coefficient by exactly ten, because that is what shifting one place along the power of ten costs.

Now lay the allowed range over the list. It starts at $1$ and stops just short of $10$, so its top end is exactly ten times its bottom end: the window is one factor of ten wide. The coefficients march past that window in steps of ten, so as soon as one of them is inside, the next one up is at least $10$ and out the top while the next one down is below $1$ and out the bottom: at most one candidate can be inside. And because the list runs on forever in both directions while the window is a full factor of ten wide, no candidate can step clean over the window without landing in it, so at least one must be inside. Exactly one candidate is inside, then, and here it is $7.25$:

$72{,}500 = 7.25 \times 10^{4}.$

The width is what does the work. Suppose the rule allowed any coefficient from $1$ up to $100$, a window a hundred times wide. Then $7.25$ and $72.5$ would both qualify, so $7.25 \times 10^{4}$ and $72.5 \times 10^{3}$ would both count as correct and two people converting the same number could disagree with each other and both be right. A window narrower than a factor of ten fails the opposite way: the candidates would step straight over it and some numbers would have no expression at all. One factor of ten, with one end included and the other excluded so that no candidate is counted twice, is the width that gives every positive number exactly one form.

Test the rival range the way the standard one was tested. It runs from just above $1$ up to and including $10$, so its top end is again exactly ten times its bottom end, and again one end is included and the other is not. Those are the only two properties the uniqueness argument used, so the argument survives the swap unchanged: candidates stepping by tens put exactly one member inside any window one factor of ten wide, wherever that window is placed.

So the rival convention is not wrong. Every positive number would still have exactly one expression under it. What changes is which expression that is, and for most numbers it changes nothing at all. For the number on the board, $7.25$ is greater than $1$ and at most $10$, so both conventions choose the same expression:

$72{,}500 = 7.25 \times 10^{4}.$

The two conventions part company only at the powers of ten themselves. Under the standard rule a coefficient of $1$ is allowed and a coefficient of $10$ is not, giving

$1{,}000 = 1 \times 10^{3},$

while under the rival rule $1$ is barred and $10$ is allowed, forcing $1{,}000 = 10 \times 10^{2}$.

That is what the standard convention buys. Its coefficients always have exactly one digit in front of the decimal point, which is what makes the notation quick to read and quick to line up: the digits sit in the same place every time and the exponent alone carries the size. Under the rival rule a coefficient could have two digits before the point, and the tidy description of the form would need an exception written into it. Uniqueness is not the reason the standard range was chosen, since both ranges have it and both get it from the width. Readability is.