Study Companion: Programme F.1 — Arithmetic

Almost everything in this programme is a rule you already obey. You carry digits, you flip the sign twice, you find a common denominator, you round to three figures. This file is about why those rules are the only ones that could have worked — in nearly every case they are forced, not chosen, and seeing what forces them is the difference between remembering arithmetic and understanding it.

The rules, procedures and drill are in the companion reference sheet, which follows the book's topic order. This file tells the story in eight units instead, because the programme's many topics are really eight ideas.

How to use this companion

Each unit is an argument: a real situation and a question, the attempt a sensible person would make first, the picture that settles it, a check with numbers, and only then the rule. Read it once front to back. The Narration spine at the end of each unit summarises what the video shows.

Unit 01: Place Value Is a Choice

The problem

A car's odometer reads

Drive one more kilometre and the last wheel turns from 5 to 6. Drive five more and it reaches 9 and then rolls over to 0, nudging the wheel to its left on by one. Nothing about the shape of the digit 4 says "forty thousand" — its value comes entirely from which wheel it is on. So what is a numeral actually saying?

First attempt

"Each digit is worth ten times the one to its right." True, but it describes the odometer without explaining it — and it does not say what happens to the right of a decimal point, or why ten.

The picture

Each wheel counts how many times the wheel to its right has gone all the way round. The rightmost wheel counts kilometres; the next counts full turns of ten; the next counts full turns of those, so hundreds; and so on. Reading the odometer is therefore a sum:

A numeral is a polynomial in its base, with the digits as the coefficients. Positions to the right of the point continue the pattern downwards, as negative powers: 3.125=3+1(10^{-1})+2(10^{-2})+5(10^{-3}).

Now the question "why ten" answers itself: nothing in the picture needed ten. Ten is a fact about hands. An odometer whose wheels had only the digits 0 and 1 would roll over at every second click — that is binary, base 2:

The subscript names the base. Base ten is also called denary.

Check it with numbers

Duodecimal needs single symbols for ten and eleven; the book uses X for ten and Λ for eleven. Hexadecimal uses the letters A to F for ten to fifteen.

Adding up a long numeral faster — nested multiplication. Rolling the wheels in by hand gives a quicker way to convert: start with the leftmost digit, and repeatedly multiply by the base and add the next digit. For 2753_8: 2\to2\times8+7=23\to23\times8+5=189\to189\times8+3=1515. Each multiplication by 8 promotes everything so far up one wheel.

The rule it gives

To convert a whole number into base b, divide repeatedly by b and read the remainders from the bottom up. The reason is in the picture: the last digit is the units wheel, the only part not multiplied by a power of b, so it is exactly the remainder on dividing by b; dividing strips that wheel off and leaves the rest of the number shifted down one place. Repeat, and the digits peel off from the right — so they arrive in reverse order.

To convert a fractional part, multiply repeatedly by b and read the whole-number parts from the top down. Multiplying by the base moves the first digit after the point across the point, into the units position, where it can be read off; then discard it and repeat. For 0.8125 in binary: 1.625\to1, 1.25\to1, 0.5\to0, 1.0\to1, so 0.8125_{10}=0.1101_2 (check: \frac12+\frac14+\frac1{16}=0.8125).

Worked example

Convert 245_{10} to hexadecimal, using octal as a stepping stone. Dividing by 8: 245\to30 r5, 30\to3 r6, 3\to0 r3, so 245_{10}=365_8. Each octal digit is exactly three binary digits, because 8=2^3: 3\to011, 6\to110, 5\to101, giving 011\,110\,101_2. Regroup in fours from the point, because 16=2^4: 1111\,0101_2, which is \text{F}5_{16}. Check: 15\times16+5=245.

Where this shows up

A byte is eight binary digits, and reading them in binary is painful. Group them in fours and each group becomes one hexadecimal digit — one hex digit holds exactly four bits, with no remainder and no ambiguity. That is the only reason hexadecimal exists: it is binary, compressed by a factor of four, for humans. A web colour #FF8000 is three bytes — red full, green half, blue off — and every memory dump and register map you will read is written this way.

Narration spine. Roll an odometer until a wheel carries, and read the display as a sum of powers of ten. Replace the ten-digit wheels with two-digit ones and watch the same machinery count in binary. Peel the digits of 245 off by division, one wheel per step, so the upward read is visibly the order they came off in. Close by grouping the bits in fours into hex.

Unit 02: Signed Numbers, and Why a Negative Times a Negative Is Positive

The problem

A freezer thermometer reads -2\,{}^\circC and the display must show how much warmer that is than -4\,{}^\circC. Numbers now have a direction as well as a size. On a line, further right is greater: -2>-4, and -4<3. The integers are the whole numbers, zero, and their negatives.

Addition and subtraction are motion along the line. Adding a positive moves right; adding a negative moves left; subtracting a negative reverses the reversal:

(Write brackets around a negative number when it follows an operation sign: 5-(-3), never 5--3.)

Multiplication is where belief runs out. Everyone can recite (-3)(-2)=6. Almost nobody can say why it is not -6.

First attempt

Continue a pattern, multiplying -2 by a decreasing number:

Each step down raises the product by 2, so the next should be (-1)(-2)=2. Suggestive — but four terms of a pattern are a reason to expect something, not a reason it must hold.

The picture

Think of multiplying by a number as doing something to the whole number line. Multiplying by 3 stretches it by a factor of 3 about zero. Multiplying by -1 turns it round: every point swaps to the other side of zero, so the line is flipped end over end. Multiplying by -3 is "stretch by 3 and flip".

Now (-3)(-2): start at -2, which is already on the far side. Flip it and it lands on +2; stretch by 3 and it lands on 6. Two flips cancel, because turning something round twice brings it back to where it started.

This picture shows what the rule does. It does not yet show that arithmetic had to work this way — for that we need an argument.

Check it with numbers

Insist that the distributive law, the law connecting multiplication with addition, keeps working. Start from something indisputable: 2-2=0, and anything times 0 is 0.

So (-3)(-2) is the number you must add to -6 to get 0. There is exactly one:

Nothing was chosen. Once (-3)(2)=-6 and the distributive law holds, the value is forced.

The rule it gives

Like signs give a positive product or quotient; unlike signs give a negative one:

Worked example

(-18)\div(-3)=6 and 18\div(-3)=-6. And (-2)^3=(-2)(-2)(-2)=4(-2)=-8: an odd number of flips leaves the result flipped.

Where this shows up

A sign is a direction, and physics has to survive reversing it twice. A current of -2 A through a resistor is current flowing the other way; the power it dissipates is I^2R, and (-2)^2=4 makes the heating positive either way — a resistor warms up regardless of which way the current runs. If a negative times a negative were negative, the algebra would predict a resistor that cools itself.

Narration spine. Walk the number line for adding and subtracting signed numbers. Show the multiplication pattern and say plainly that it is not yet a proof. Show multiplication by -1 as a flip of the whole line, and two flips cancelling. Then run the distributive argument line by line, and let the forced value land.

Unit 03: Conventions and Laws Are Different Things

The problem

Type 14-3\times4 into a spreadsheet and it answers 2. Work it left to right by hand and you get 44. The expression as written does not have a meaning until something decides the order.

First attempt

"It's just BODMAS." But that phrase hides two quite different kinds of rule, and mixing them up leads to the belief that arithmetic is a pile of arbitrary conventions. Most of it is not.

The picture

• Precedence is a convention. It is an agreement about how to read a written expression, like reading English left to right. The book's precedence rules: brackets first (the innermost first, when nested); then powers; then multiplication and division, working from the left, as they are met; then addition and subtraction, from the left. It could have been otherwise, which is why brackets exist.
• Commutativity, associativity and distributivity are laws. They are facts about numbers, true whatever the notation, and they are what license you to rearrange a calculation.

Distributivity has a picture. A rectangle of height a and width b+c can be cut by one vertical line into rectangles of area ab and ac, and cutting does not change area:

and the same cut, with one piece removed, gives a(b-c)=ab-ac.

Check it with numbers

The convention: 34+10\div(2-3)\times5=34+10\div(-1)\times5=34+(-10)\times5=34-50=-16. Nested brackets, innermost first: 4(10-2[7-4])=4(10-2\times3)=4(10-6)=16.

The laws: 3(4+5)=27 and 3\times4+3\times5=12+15=27.

Division distributes from the right but not from the left. Share 24+12 sweets among 6 children and each gets (24+12)\div6=6 — the same as sharing each bag separately, 24\div6+12\div6=4+2. But dividing 24 among a group of 4+2 children is not the same as dividing among 4 and among 2 separately: 24\div(4+2)=4, while 24\div4+24\div2=18.

Subtraction and division obey neither commutativity nor associativity: 8-3\ne3-8, (8-4)-2\ne8-(4-2).

The rule it gives

Division distributes over + and - from the right only: (b\pm c)\div a=b\div a\pm c\div a.

Worked example

(4+5\times6)\div2-12\div4\times2-1: inside the brackets, 5\times6=30, so 34\div2=17; then 12\div4\times2=3\times2=6; so 17-6-1=10.

Where this shows up

Every rearrangement you make before substituting numbers — taking a common factor out of a sum of stiffnesses, grouping resistances before inverting — is one of these laws being used. The convention only decides how a formula is punctuated; the laws decide what you may do to it. When a formula is typed into a spreadsheet or a program, the brackets you add are what make the convention irrelevant.

Narration spine. Evaluate 14-3\times4 two ways to two answers. Separate convention from law. Cut the rectangle on screen for distributivity, then share sweets to show division distributing from the right but not the left.

Unit 04: The Atoms of the Integers

The problem

Two gears mesh, one with 12 teeth and one with 18. Mark one tooth on each. How many tooth-steps before the same two teeth meet again? And why do gear designers sometimes add a tooth on purpose?

First attempt

Turn the gears and count. It works, but it gives no way to predict the answer for other gears — and it hides why the answer is what it is.

The picture

Break each number into its prime factors. A factor divides a number exactly; a prime is a whole number with exactly two factors, 1 and itself: 2,3,5,7,11,13,\dots (1 is not prime.)

126=2\times3^2\times7. Split it a different way — 6\times21 first — and you land on exactly the same primes. That is the Fundamental Theorem of Arithmetic: every whole number above 1 factors into primes in exactly one way, apart from order. It is also why 1 is excluded from the primes: if 1 counted, you could insert as many 1s as you liked and the factorization would no longer be unique.

Uniqueness turns questions about divisibility into questions about exponents. Line up the prime powers of two numbers in columns:

A common factor can use only primes both numbers have, and only up to the smaller power — take more and one of them runs out. A common multiple must contain enough of every prime to cover both, so it takes the larger power.

Check it with numbers

\operatorname{HCF}(144,66)=2\times3=6 and \operatorname{LCM}(144,66)=2^4\times3^2\times11=1584. Check: 144/6=24, 66/6=11; 1584/144=11, 1584/66=24.

Because each column takes one smaller and one larger power, and together those are both powers, \operatorname{HCF}\times\operatorname{LCM} uses every prime power of both numbers exactly once: 6\times1584=9504=144\times66.

The rule it gives

• HCF (highest common factor): primes common to both, each to its smaller power.
• LCM (lowest common multiple): every prime present, each to its larger power.
• \operatorname{HCF}(a,b)\times\operatorname{LCM}(a,b)=ab.

Worked example

84=2^2\times3\times7 and 512=2^9. Only 2 is common, so \operatorname{HCF}=2^2=4 and \operatorname{LCM}=2^9\times3\times7=10\,752. Check: 4\times10\,752=43\,008=84\times512.

Where this shows up

The marked teeth meet again after \operatorname{LCM}(12,18)=36 tooth-steps — only 3 turns of the small gear — so the same pairs of teeth keep grinding against each other and wear matched flats.

Give the small gear 13 teeth instead. Now \operatorname{HCF}(13,18)=1, the LCM is the full 13\times18=234, and every tooth meets every other tooth before any pair repeats, so wear spreads evenly. That extra tooth is called a hunting tooth, and the design decision is made entirely by a highest common factor.

Narration spine. Mesh the 12- and 18-tooth gears and turn them until the marked teeth meet again. Factor 126 two ways and land on the same primes. Line up the prime powers of 144 and 66 as columns and read the HCF and LCM off the exponents. Swap in the 13-tooth gear and watch every tooth take its turn.

Unit 05: Fractions Are One Idea Wearing Six Hats

The problem

A concrete mix is specified as half gravel and a third sand, with cement making up the rest. What is the ratio of gravel to sand to cement, and how much of each goes into 120 kg? Answering needs equivalent fractions, a common denominator, subtraction, ratios and division — and they are all one idea: a fraction is a number of equal parts, and the only thing that ever matters is whether the parts are the same size.

A fraction is one integer, the numerator, divided by another, the denominator; that makes it a rational number. It is proper if the numerator is smaller (\frac47), improper if larger (\frac{12}5), and mixed when written as a whole number and a fraction (2\frac25).

First attempt

\frac12+\frac13=\frac25? Add tops and bottoms. But \frac25 is less than \frac12 on its own — adding sand made the mixture smaller. Wrong.

The picture

Equivalence: refine the partition. Shade 2 of 3 equal parts of a bar. Cut every part in half: nothing moved, but there are now 4 shaded parts out of 6.

\frac ab=\frac{ak}{bk} is a relabelling of a partition that did not change. Run it backwards — cancel common factors — and you reduce to lowest terms: \frac{84}{108}=\frac79.

Addition: count only equal pieces. Halves and thirds are different sizes, so they cannot be counted together. Refine both until they agree: cutting halves into thirds and thirds into halves both give sixths. The common denominator is the LCM of the denominators (Unit 04):

Multiplication is area. Take \frac12 of a rectangle, then \frac23 of that. Shading in two directions cuts it into 3\times2=6 cells, of which 2 are shaded twice: \frac23\times\frac12=\frac26=\frac13. "Of" means multiply.

Division asks how many fit. \frac23\div\frac57: how many \frac57-sized pieces fit into \frac23? A whole \frac57 is 5 sevenths, so one seventh fits \frac75 times as often as \frac57 does. Dividing by \frac57 is multiplying by its reciprocal, \frac75: \frac23\times\frac75=\frac{14}{15}.

Check it with numbers

\frac12+\frac13=\frac56=0.8333, and 0.5+0.3333=0.8333. The concrete: gravel \frac12=\frac36, sand \frac13=\frac26, so cement is the remainder, 1-\frac56=\frac16. Over the common denominator 6 the numerators are 3:2:1 — the ratio — and they add up to the denominator, 6 parts. In 120 kg one part is 20 kg: 60 kg gravel, 40 kg sand, 20 kg cement.

The rule it gives

A ratio lists the numerators of fractions over a common denominator. A percentage is a fraction with denominator 100: \frac{12}{25}=\frac{48}{100}=48\%, and 15\% of 240 is \frac{15}{100}\times240=36.

Percentage change multiplies. An 8\% increase gives the original plus 0.08 of it: V+0.08V=1.08V. Each later increase applies to the new value, so repeated change multiplies: 500(1.08)^3=629.86, not 500+3(40)=620.

Worked example

\frac59+\frac16: the LCM of 9 and 6 is 18, so \frac{10}{18}+\frac3{18}=\frac{13}{18}. And the reciprocal of -5 is -\frac15.

Where this shows up

A drive train with stages of 95\%, 92\% and 88\% efficiency does not lose 5+8+12=25\%. Each stage passes on a fraction of what it receives, so the fractions multiply: 0.95\times0.92\times0.88=0.769 — 76.9\% gets through.

Narration spine. Show the half-and-third concrete spec and the wrong "add tops and bottoms" answer, struck out. Shade a bar and refine the partition. Show halves and thirds failing to add, then refine both to sixths and read off the 3:2:1 ratio. Shade a rectangle two ways for multiplication, then count how many \frac57 pieces fit into \frac23.

Unit 06: Decimals, Rounding, and Numbers That Never End

The problem

A fuel pump divides the cost by the price and shows 31.428571\ldots litres. How many digits should it display — and is \frac17 really a number whose decimal goes on forever?

First attempt

"Just show what the calculator shows." But a display has a fixed number of places, and a measurement has a fixed precision; ten digits on a pump are a claim nobody can back up. And a pattern like 0.142857142857\ldots must mean something exact.

The picture

Rounding picks the nearest mark. Rounding 4550 to the nearest 100 asks which of 4500 and 4600 it is nearer. Exactly halfway, the book's rule is to round up — 4600. For a negative number, round the size and keep the sign: -4550\to-4600.

Significant figures are counted from the first non-zero digit; decimal places are counted after the point. Round by looking at the first digit dropped: 5 or more rounds the last kept digit up. 7.846\to7.85 (2 dp); 0.004736\to0.00474 (3 sf); 2.345\to2.35 (3 sf). Zeros added to meet the requirement are trailing zeros: 13.1 to 3 dp is 13.100.

Why some decimals never end. Long division of 1 by 7 can only ever leave a remainder from 0 to 6. Once a remainder repeats, every digit after it repeats too — so every fraction's decimal either stops (endless zeros) or cycles. In dot notation, \frac13=0.\dot3 and \frac17=0.\dot14285\dot7, the dots over the first and last digit of the repeating block.

Check it with numbers

A repeating decimal can be turned back into a fraction by shifting it against itself. Let x=0.\dot1\dot8=0.181818\ldots; then 100x=18.181818\ldots, and subtracting kills the tail:

If the repeat starts later, shift the non-repeating part out of the way first. x=0.4\dot6=0.4666\ldots: 10x=4.666\ldots, so 9x=4.2 and x=\frac{4.2}9=\frac{42}{90}=\frac7{15}.

The rule it gives

• A terminating decimal is a fraction over a power of ten: 0.52=\frac{52}{100}=\frac{13}{25}.
• A reduced fraction terminates exactly when its denominator's only primes are 2 and 5 — the primes that divide the base (Unit 01 meets Unit 04): \frac7{20}=0.35 since 20=2^2\times5, but \frac13 recurs.
• Rational numbers are exactly those whose decimals end or repeat. Irrational numbers, such as \sqrt2, \pi and e, have decimals that never repeat, so they can only be rounded or written as symbols. Together they form the real numbers.

Worked example

\frac38=0.375 (terminates: 8=2^3). \frac5{12}=0.41\dot6 (recurs: 12 contains a 3). And 0.\dot2\dot1=\frac{21}{99}=\frac7{33}.

Where this shows up

The pump's 31.428571\ldots litres is \frac{220}7; displayed to 2 dp it is 31.43, and the customer is billed on that. Carry exact values — fractions, \sqrt2, \pi — as symbols through a calculation and round once at the end; rounding early compounds the error through every later step.

Narration spine. Put 4550 on a number line between 4500 and 4600 and round it. Divide 1 by 7 with the remainders on display until one repeats and the digits start cycling. Shift 0.1818\ldots against 100 times itself and cancel the tails. Sort denominators by whether their primes are only 2 and 5.

Unit 07: Powers, and What a Fractional One Could Possibly Mean

The problem

A scale model of a bridge beam holds its load easily. Build the real beam at twice the size in every direction, and it may fail. Why? Doubling every length multiplies the cross-section, which carries load, by 2^2=4 — but the volume, and so the weight, by 2^3=8. The weight outruns the strength. Powers are how physical scaling works, so they need to be understood exactly.

First attempt

A power is repeated multiplication: 2^3=2\times2\times2. The number repeated is the base; the count is the index or power. Then 2^3\times2^4 is three copies of 2 followed by four more — seven copies, 2^7=128.

But what could 2^{1/2} mean? You cannot multiply 2 by itself half a time. The counting story breaks.

The picture

Write the copies out, and the laws of powers are just bookkeeping on how many there are:

Now refuse to give up the first law, and let it define the powers that counting cannot:

• a^1\times a^0=a^{1+0}=a^1, so a^0 must be 1.
• a^n\times a^{-n}=a^0=1, so a^{-n}=\frac1{a^n} — a negative power is a reciprocal.
• 2^{1/2}\times2^{1/2}=2^{1/2+1/2}=2^1=2, so 2^{1/2} must be a number whose square is 2.

That last step pins 2^{1/2} down up to its sign: both \sqrt2 and -\sqrt2 square to 2, and the law cannot choose between them.

Check it with numbers

2^3\times2^4=8\times16=128=2^7. 5^6\div5^2=15\,625\div25=625=5^4. (2^3)^2=8^2=64=2^6 — which is not 2^{(3^2)}=2^9=512. Different bases combine only under the same power: 2^3\times5^3=(2\times5)^3=1000, but 2^3\times3^2 does not simplify.

The rule it gives

Roots. a^{1/n} is a number whose nth power is a, and a^{m/n}=\left(a^{1/n}\right)^m. The sign needs a decision:

• Odd roots are unique, and odd roots of negatives are negative: (-27)^{1/3}=-3.
• Even roots of positive numbers come in pairs, \pm. The surd sign \sqrt{\ } means the positive one by convention. These notes also read a^{1/n} as the positive root, so 16^{1/4}=2; where both are wanted, write \pm. Stroud writes \pm for even fractional powers, e.g. 16^{1/4}=\pm2 — expect that form in the book's answers.
• Even roots of negative numbers have no real value: (-16)^{1/2} must wait for complex numbers.

A surd is a root left in root form because it is irrational, such as \sqrt2=1.41421\ldots. Keep it exact until the end: \frac1{\sqrt2}=\frac{\sqrt2}2.

Worked example

\frac{3^5\times3^{-2}}{3^2}=3^{5-2-2}=3^1=3. And 8^{2/3}=\left(8^{1/3}\right)^2=2^2=4.

Where this shows up

The beam: doubling its lengths quadruples its strength but multiplies its weight by eight, so a design that works as a small model can fail full-size. The same exponent mismatch is why large animals have disproportionately thick bones, and why a structure cannot be validated by testing a miniature and multiplying up.

Narration spine. Grow a beam to twice its size and show the 4\times cross-section against the 8\times volume. Write out copies of 2 and collapse them into exponents. Ask what half a copy could mean, then impose the addition law and let the square root be forced — up to its sign. Show \sqrt2 and -\sqrt2 both squaring to 2, and state the convention that picks one.

Unit 08: Standard Form, and Precision Is a Claim About Reality

The problem

A capacitor is labelled 47\ \muF, a radio station broadcasts at 98.4 MHz, and a machinist's drawing says \varnothing25.0 mm. The first two need a compact way to write very large and very small numbers; the third makes a claim about precision. Will a \varnothing25.0 mm shaft fit a \varnothing25.1 mm hole?

First attempt

Write the numbers out: 0.000047 F and 98\,400\,000 Hz. Counting zeros invites mistakes. And "25.0 fits in 25.1, there is 0.1 mm to spare" treats measurements as exact.

The picture

Standard form. Multiplying by 10 moves every digit one place left, so a number can be written as a mantissa between 1 and 10 times a power of ten, the exponent: 52\,674=5.2674\times10^4, 0.000047=4.7\times10^{-5}. Preferred standard form restricts the exponent to multiples of 3, so up to three digits sit before the point — matching the SI prefixes on real instruments: 4.7\times10^{-5}\ \text{F}=47\times10^{-6}\ \text{F}=47\ \mu\text{F}.

A measurement is an interval. A diameter reported as 25.0 mm to the nearest 0.1 mm is not the number 25; it is everything that rounds to 25.0:

Check it with numbers

Working in standard form. Multiply mantissas, add exponents: (3.2\times10^4)(5\times10^{-7})=16\times10^{-3}=1.6\times10^{-2}. To add, first make the exponents equal: 4.1\times10^3+2.5\times10^4=0.41\times10^4+2.5\times10^4=2.91\times10^4.

Checking a calculation. \frac{41.8\times0.0213}{7.92}\approx\frac{(4\times10)(2\times10^{-2})}{8}=0.1. A calculator reading 0.1124 is plausible; 1.124 or 0.01124 means a slipped decimal point.

The fit. Shaft and hole are each known to \pm0.05 mm. In the worst case the shaft is 25.05 and the hole 25.05: zero clearance, and the parts may not assemble.

The rule it gives

Accuracy of a result (the book's rule). A calculation from measured values is no more accurate than its least accurate input, so give the result to the smallest number of significant figures among the measurements. Exact numbers — counts, a factor of 2, the \pi in a formula — do not limit it. A plate 12.6 m by 4.1 m has area 51.66 m², reported as 52 m² because 4.1 has only 2 sf.

Bounds, for the worst case. Push the interval ends through the formula: that plate's area lies between 12.55\times4.05=50.83 and 12.65\times4.15=52.50 m². Multiplying quantities approximately adds their relative uncertainties — here about 0.4\%+1.2\% — so the least precise measurement dominates.

Worked example

A rectangle measures 10.0 mm by 5.0 mm, each to the nearest 0.1 mm. Nominal area 50 mm², reported as 50 mm² (2 sf). Bounds: 9.95\times4.95=49.25 to 10.05\times5.05=50.75 mm², about \pm1.5\% — the sum of 0.5\% and 1\%, to first order.

Where this shows up

Tolerance stacking is the shaft-and-hole calculation repeated across every dimension in an assembly, and it is done with bounds, never with nominal values. The instrument side of the same idea: a meter reading 47.3\ \muA is claiming three significant figures, and a result calculated from it cannot honestly claim more.

Narration spine. Show the 47\ \muF label and slide the decimal point with each power of ten. Multiply two numbers in standard form by combining mantissas and exponents. Estimate a calculation before revealing the calculator's result. Draw 25.0 mm as a band, put the hole's band beside it, and slide both to their worst case until the clearance vanishes.