[{"data":1,"prerenderedAt":1063},["ShallowReactive",2],{"layer:properties-of-numbers:deepen":3},{"layer":4,"contentHash":1037,"dependencyHashes":1038,"approval":1057,"releaseId":1062},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":40,"sourceIds":1032,"reviewStatus":1033,"authoring":1034},1,"properties-of-numbers","en","deepen","Why the rules must be true","Proofs with arrays and boxes, the distributive law behind every method, zero through history, and the road to algebra","Prove the commutative, associative and distributive laws for every whole number, see why long multiplication and divisibility tests work, show why division by zero would make 0 = 1, prove parity facts with letters, and meet the properties as the rules of algebra.",[13,14,15,16,17],"Prove the commutative, associative and distributive laws using arrays, boxes and rectangles, and notice what each picture assumes.","Use (a + b)², (a − b)² and (a + b)(a − b) to calculate squares and products quickly.","Explain why long multiplication, short division and the tests for 3, 9 and 11 work.","Prove that giving 1 ÷ 0 a value leads to 0 = 1, and find hidden divisions by zero.","Decide which properties an invented operation has, and link each algebra step to a property.",50,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Go deeper",{"label":26,"value":27},"Reading time","≈ 50 minutes",{"label":29,"value":30},"Prior knowledge","Properties stated with letters",{"label":32,"value":33},"Chapters","11",{"label":35,"value":36},"Labs","Expansion lab, operation sort, algebra match",{"label":38,"value":39},"Stretch","Class 8 algebra identities",[41,45,66,72,75,81,86,104,109,112,137,141,153,158,161,176,179,203,208,223,235,278,296,301,312,322,327,332,335,365,374,386,391,394,408,441,445,450,453,474,479,501,504,515,520,523,557,561,574,655,660,663,675,697,732,736,773,778,831,977,981,997,1003,1007,1011,1015],{"id":42,"type":43,"markdown":44},"intro-deepen","prose","In the earlier layers you *used* the properties and *tested* them. This layer asks the deepest question: **how do we know they are true for every number, even numbers nobody will ever write down?** And then: what do these properties let us build?\n\nYou will meet short, real proofs, some using pictures (arrays and boxes of cubes) and some using letters. You will see that long multiplication, divisibility tests and the whole of school algebra are built on the distributive property. You will find out why defining a value for 1 ÷ 0 would break arithmetic, trace how Indian mathematicians argued about zero 1,400 years ago, and invent your own operations to see which properties they keep. Take it slowly; the ideas here are the ones that make mathematics feel solid.",{"id":46,"type":47,"tone":48,"items":49},"spec-deepen","spec","neutral",[50,54,58,62],{"label":51,"big":52,"value":53},"Proof","covers every case","An argument that works for every number at once, not just the ones you tried.",{"label":55,"big":56,"value":57},"Tool 1","pictures","Arrays, boxes of cubes and rectangles prove the commutative, associative and distributive laws.",{"label":59,"big":60,"value":61},"Tool 2","letters","Writing even as 2m and odd as 2m + 1 turns parity rules into one-line proofs.",{"label":63,"big":64,"value":65},"Tool 3","contradiction","Assume the opposite and show it leads to nonsense, like 0 = 1.",{"id":67,"type":68,"title":69,"eyebrow":70,"navLabel":71},"ch1","chapter","Why a × b = b × a for every pair of numbers","Chapter 01","1 Proving commutativity",{"id":73,"type":43,"markdown":74},"comm-proof","We checked 4 × 6 = 6 × 4 with chairs. But to be sure that 3,417 × 98,006 = 98,006 × 3,417 without calculating, we need an argument that does not depend on the particular numbers.\n\n**The array proof.** Take any whole numbers a and b. Build an array of dots with a rows and b dots in each row. By the meaning of multiplication, it holds a × b dots. Now rotate the whole array a quarter turn. No dot is added or removed, so it still holds a × b dots. But now it has b rows with a dots in each row, which is b × a dots by the meaning of multiplication. The same set of dots was counted two ways, so **a × b = b × a**. Nothing in the argument depended on what a and b were, so it covers every pair.\n\n**The addition proof** is even simpler: a red beads followed by b blue beads on a string. Read left to right: a + b. Turn the string round: b + a. Same beads.\n\nThis style of proof, *counting the same collection in two ways*, is one of the most powerful ideas in mathematics. You will meet it again in combinatorics and in the handshake problem in Extend.",{"id":76,"type":77,"variant":78,"title":79,"markdown":80},"nuance-zero-arrays","callout","nuance","What about zero?","The array proof quietly assumes a and b are at least 1, because an array with 0 rows is hard to draw. The cases with zero are checked separately: a × 0 = 0 (a rows of no dots) and 0 × a = 0 (no rows at all), so a × 0 = 0 × a. A careful proof always asks, \"did my picture really cover every case?\"",{"id":82,"type":77,"variant":83,"title":84,"markdown":85},"aha-commute-rare","aha","Most operations are not commutative","Among all the ways to combine two numbers, commutativity is rare. Powers: 2³ = 8 but 3² = 9. Subtraction, division, \"the first number\", \"twice the first plus the second\": none are commutative. Addition and multiplication are the lucky exceptions, and the proofs above show exactly where the luck comes from: both are about counting a collection, and counting does not care about direction.",{"id":87,"type":88,"prompt":89,"options":90,"explanation":103},"predict-power-swap","prediction","Powers are not commutative: 2³ = 8 but 3² = 9. Is there **any** pair of different whole numbers a and b (both at least 1) with a to the power b equal to b to the power a?",[91,94,97,100],{"id":92,"label":93},"a","No, never",{"id":95,"label":96},"b","Yes, exactly one pair",{"id":98,"label":99},"c","Yes, infinitely many pairs",{"id":101,"label":102},"d","Only pairs that include 1","**Yes, exactly one pair: 2 and 4.** 2⁴ = 2 × 2 × 2 × 2 = 16 and 4² = 4 × 4 = 16. Pairs with 1 fail (1⁵ = 1 but 5¹ = 5). Mathematicians have proved that 2 and 4 is the only pair of different whole numbers that works. So \"never commutative\" would be slightly wrong; the honest statement is \"not commutative in general\". One lucky pair does not make an operation commutative, just as one failure makes it not commutative.",{"id":105,"type":68,"title":106,"eyebrow":107,"navLabel":108},"ch2","Why grouping does not matter: a box of cubes","Chapter 02","2 Proving associativity",{"id":110,"type":43,"markdown":111},"assoc-proof","Build a box out of small cubes: **2 cubes deep, 3 wide and 4 tall**. How many cubes?\n\n- Count by horizontal layers: each layer is 2 × 3 = 6 cubes, and there are 4 layers: (2 × 3) × 4 = 24.\n- Count by vertical slices from front to back: each slice is 3 × 4 = 12 cubes, and there are 2 slices: 2 × (3 × 4) = 24.\n\nSame box, same cubes, so (2 × 3) × 4 = 2 × (3 × 4) = 24. With a cubes by b cubes by c cubes, the same two counts show **(a × b) × c = a × (b × c)** for all whole numbers. Combined with commutativity, it means a product of many numbers can be rearranged in **any** order and grouping. A box of a × b × c cubes can be counted from any side.",{"id":113,"type":114,"caption":115,"columns":116,"rows":121},"box-table","table","Counting a 2 × 3 × 4 box of cubes in three different ways",[117,118,119,120],"Count by","Each piece has","Number of pieces","Total",[122,127,132],[123,124,125,126],"horizontal layers","2 × 3 = 6","4","6 × 4 = 24",[128,129,130,131],"front-to-back slices","3 × 4 = 12","2","12 × 2 = 24",[133,134,135,136],"side-to-side slices","2 × 4 = 8","3","8 × 3 = 24",{"id":138,"type":77,"variant":78,"title":139,"markdown":140},"nuance-general-assoc","From three numbers to any number","The associative law talks about three numbers. Why can we then add a list of 100 numbers in any grouping? Because any two groupings can be changed into each other by moving one pair of brackets at a time, and each move is one use of the associative law. Mathematicians call this **general associativity**. It is why we can write 2 + 7 + 9 + 1 without any brackets at all: every way of bracketing gives the same answer.",{"id":142,"type":143,"title":144,"problem":145,"steps":146},"we-gauss","worked_example","Adding 1 to 100 using properties","Find 1 + 2 + 3 + … + 100.",[147,148,149,150,151,152],"Write the sum forwards: 1 + 2 + 3 + … + 99 + 100.","Write it backwards underneath: 100 + 99 + 98 + … + 2 + 1. Commutativity and associativity say it has the same total.","Add the two lines column by column: each column is 101 (1 + 100, 2 + 99, …).","There are 100 columns, so twice the sum is 100 × 101 = 10,100.","The sum is 10,100 ÷ 2 = 5,050.","Legend says Carl Friedrich Gauss found this as a schoolboy in Germany in the 1780s. Whether or not the story is true, the method is pure commutative and associative reasoning.",{"id":154,"type":68,"title":155,"eyebrow":156,"navLabel":157},"ch3","The distributive law: rectangles, squares and algebra","Chapter 03","3 Distributive, deeper",{"id":159,"type":43,"markdown":160},"dist-proof","**The rectangle proof.** A rectangle a units tall and (b + c) units wide can be cut by a vertical line into an a × b rectangle and an a × c rectangle. Cutting does not change area. So **a × (b + c) = a × b + a × c** for all whole numbers. This picture is in Euclid's *Elements* (Book II, Proposition 1), written in Greece around 300 BCE, long before letters were used for numbers.\n\n**Two brackets.** Cut both sides: (a + b) × (c + d) is a rectangle split into four pieces, so (a + b) × (c + d) = a × c + a × d + b × c + b × d. Every piece of the first bracket multiplies every piece of the second. For example:",{"id":162,"type":114,"caption":163,"columns":164,"rows":168},"area-23-47","Area model for 23 × 47",[165,166,167],"×","40","7",[169,173],[170,171,172],"20","20 × 40 = 800","20 × 7 = 140",[135,174,175],"3 × 40 = 120","3 × 7 = 21",{"id":177,"type":43,"markdown":178},"dist-squares","Adding the four pieces: 800 + 140 + 120 + 21 = 1,081, and indeed 23 × 47 = 1,081.\n\n**Squaring near round numbers.** When both brackets are the same, (a + b) × (a + b) gives a × a + 2 × a × b + b × b. Written with squares: (a + b)² = a² + 2ab + b². This makes squares of numbers near a round number easy:\n\n- 21² = 400 + 2 × 20 × 1 + 1 = 441.\n- 99² = 10,000 − 2 × 100 × 1 + 1 = 9,801 (using (a − b)² = a² − 2ab + b²).\n- 105² = 10,000 + 1,000 + 25 = 11,025.\n\nAnd one more beautiful one: (a + b) × (a − b) = a² − b². So 52 × 48 = 50² − 2² = 2,500 − 4 = 2,496, and 103 × 97 = 10,000 − 9 = 9,991.",{"id":180,"type":114,"caption":181,"columns":182,"rows":186},"square-tricks","Special products that come straight from the distributive law",[183,184,185],"Identity","Example","Value",[187,191,195,199],[188,189,190],"(a + b)² = a² + 2ab + b²","31² = 900 + 60 + 1","31 × 31 = 961",[192,193,194],"(a − b)² = a² − 2ab + b²","49² = 2,500 − 100 + 1","49 × 49 = 2,401",[196,197,198],"(a + b)(a − b) = a² − b²","62 × 58 = 3,600 − 4","62 × 58 = 3,596",[200,201,202],"(10a + 5)² = 100a(a + 1) + 25","65² = 100 × 6 × 7 + 25","65 × 65 = 4,225",{"id":204,"type":77,"variant":205,"title":206,"markdown":207},"obs-ends-in-5","observation","Squaring numbers ending in 5","The last row is a famous trick: to square a number ending in 5, multiply the tens part by the next number up and write 25 after it. 35²: 3 × 4 = 12, so **1,225**. 85²: 8 × 9 = 72, so **7,225**. The reason: (10a + 5)² = 100a² + 100a + 25 = 100 × a × (a + 1) + 25. Check: 85 × 85 = 7,225.",{"id":209,"type":210,"itemId":211,"prompt":212,"check":213,"hints":217,"feedback":220},"pr-diff-squares","practice","properties-of-numbers.deepen-diff-squares","Use (a + b)(a − b) = a² − b² to work out **104 × 96**.",{"kind":214,"answer":215,"tolerance":216},"number",9984,0,[218,219],"104 = 100 + 4 and 96 = 100 − 4.","100² − 4² = 10,000 − 16.",{"correct":221,"incorrect":222},"Yes: 104 × 96 = 10,000 − 16 = 9,984.","104 × 96 = (100 + 4)(100 − 4) = 10,000 − 16 = 9,984.",{"id":224,"type":210,"itemId":225,"prompt":226,"check":227,"hints":229,"feedback":232},"pr-square-5","properties-of-numbers.deepen-square-ends-5","Use the \"ends in 5\" trick to find **45²**.",{"kind":214,"answer":228,"tolerance":216},2025,[230,231],"Multiply the tens digit by the next number: 4 × 5.","Write 25 after it.",{"correct":233,"incorrect":234},"Right: 4 × 5 = 20, then 25: 45 × 45 = 2,025.","4 × 5 = 20, then write 25 after it: 2,025.",{"id":236,"type":237,"component":238,"componentVersion":5,"config":239,"objective":272,"textAlternative":273,"help":274},"lab-sprint-deepen","interactive","arith-sprint",{"operations":240,"ranges":241,"rounds":246,"secondsTotal":247,"estimateFirst":248,"wordProblems":249},[165],{"a":242,"b":245},{"min":243,"max":244},91,109,{"min":243,"max":244},12,180,false,[250,254,258,262,265,269],{"prompt":251,"answer":252,"operation":165,"unit":253},"A square plot of land has sides of 99 m. What is its area in square metres? Use (100 − 1)².",9801,"m²",{"prompt":255,"answer":256,"operation":165,"unit":257},"A hall has 52 rows of 48 seats. Use (50 + 2)(50 − 2). How many seats?",2496,"seats",{"prompt":259,"answer":260,"operation":165,"unit":261},"A square courtyard is tiled 65 tiles by 65 tiles. How many tiles? Use the ends-in-5 trick.",4225,"tiles",{"prompt":263,"answer":264,"operation":165,"unit":257},"A stadium has 103 blocks with 97 seats in each block. How many seats? Use (100 + 3)(100 − 3).",9991,{"prompt":266,"answer":267,"operation":165,"unit":268},"A rangoli grid is 21 dots by 21 dots. How many dots? Use (20 + 1)².",441,"dots",{"prompt":270,"answer":215,"operation":165,"unit":271},"A farmer plants 104 rows of 96 saplings. How many saplings?","saplings","Multiply numbers close to 100 using (a + b)(a − b), (a ± b)² and the distributive law, then solve area and seating problems.","A 180-second sprint of 12 rounds. Plain questions multiply two numbers between 91 and 109. A fast method: write each as 100 plus or minus a small number and expand. For example 97 × 106 = (100 − 3)(100 + 6) = 10,000 + 600 − 300 − 18 = 10,282.\n\nUp to half the rounds are word problems from this list:\n1. A 99 m square plot: 10,000 − 200 + 1 = 9,801 m².\n2. 52 rows of 48 seats: 2,500 − 4 = 2,496 seats.\n3. 65 × 65 tiles: 6 × 7 = 42, then 25: 4,225 tiles.\n4. 103 blocks of 97 seats: 10,000 − 9 = 9,991 seats.\n5. A 21 × 21 rangoli grid: 400 + 40 + 1 = 441 dots.\n6. 104 rows of 96 saplings: 10,000 − 16 = 9,984 saplings.",{"hints":275},[276,277],"Write both numbers as 100 ± something.","Every part of the first bracket multiplies every part of the second.",{"id":279,"type":237,"component":280,"componentVersion":5,"config":281,"objective":290,"textAlternative":291,"help":292},"lab-orderops-deepen","order-ops",{"expressions":282,"showRuleCard":289},[283,284,285,286,287,288],"(20 + 3) × (40 + 7)","20 × 40 + 20 × 7 + 3 × 40 + 3 × 7","(100 - 1) × (100 - 1)","10000 - 200 + 1","(50 + 2) × (50 - 2)","50 × 50 - 2 × 2",true,"Evaluate each product directly and in its expanded form, and see the distributive law give the same answer both ways.","A step-by-step expression lab with three pairs.\n\n(20 + 3) × (40 + 7) = 23 × 47 = 1,081. Expanded: 20 × 40 + 20 × 7 + 3 × 40 + 3 × 7 = 800 + 140 + 120 + 21 = 1,081.\n\n(100 − 1) × (100 − 1) = 99 × 99 = 9,801. Expanded: 10,000 − 200 + 1 = 9,801.\n\n(50 + 2) × (50 − 2) = 52 × 48 = 2,496. Expanded: 50 × 50 − 2 × 2 = 2,500 − 4 = 2,496.\n\nIn each pair, the first form multiplies after adding inside brackets; the second multiplies first and then adds, following the order of operations. They agree because of the distributive law.",{"hints":293},[294,295],"Brackets first, then multiply.","In the expanded form, do every × before any + or −.",{"id":297,"type":68,"title":298,"eyebrow":299,"navLabel":300},"ch4","Every written method is a property in action","Chapter 04","4 Why methods work",{"id":302,"type":143,"title":303,"problem":304,"steps":305},"we-longmult","Long multiplication is the distributive law","Explain each line of **347 × 26** done in columns.",[306,307,308,309,310,311],"Split 26 = 20 + 6. By the distributive law, 347 × 26 = 347 × 6 + 347 × 20.","First line of the working: 347 × 6 = 2,082.","Second line: 347 × 20 = 6,940. The 0 written at the right is the × 10 part of × 20.","Add the lines: 2,082 + 6,940 = 9,022.","Inside each line, 347 × 6 itself is (300 + 40 + 7) × 6 = 1,800 + 240 + 42: the distributive law again, with carrying handling the place value.","So the column method is the area model with the pieces added as you go.",{"id":313,"type":143,"title":314,"problem":315,"steps":316},"we-shortdiv","Short division is right-distributivity","Explain why **852 ÷ 4 = 213** using properties.",[317,318,319,320,321],"Split 852 = 800 + 40 + 12 (choosing parts that 4 divides exactly).","Division distributes from the right: (800 + 40 + 12) ÷ 4 = 800 ÷ 4 + 40 ÷ 4 + 12 ÷ 4.","= 200 + 10 + 3.","= **213**. Check: 213 × 4 = 852.","The \"carry the remainder\" step in short division is just choosing a better split: 85 tens ÷ 4 = 21 tens remainder 1 ten, and that 1 ten joins the 2 ones to make 12.",{"id":323,"type":77,"variant":324,"title":325,"markdown":326},"misc-divisor-split","misconception","You cannot split the divisor","852 ÷ 4 can be split as (800 + 52) ÷ 4. But 852 ÷ 12 **cannot** be split as 852 ÷ 10 + 852 ÷ 2. That gives 85.2 + 426 = 511.2, while the true answer is 852 ÷ 12 = 71. Division distributes over a sum on its left, never over a sum on its right.",{"id":328,"type":68,"title":329,"eyebrow":330,"navLabel":331},"ch5","Divisibility tests: the distributive law in disguise","Chapter 05","5 Divisibility tests",{"id":333,"type":43,"markdown":334},"div9-prose","Why does \"add the digits\" test for divisibility by 9? Take 7,452. By place value it is 7 × 1,000 + 4 × 100 + 5 × 10 + 2. Now write each power of ten as a multiple of 9 plus 1: 1,000 = 999 + 1, 100 = 99 + 1, 10 = 9 + 1. By the distributive law:\n\n7,452 = 7 × 999 + 4 × 99 + 5 × 9 + (7 + 4 + 5 + 2).\n\nThe first three parts are multiples of 9. So 7,452 is a multiple of 9 exactly when the leftover part, the **digit sum** 7 + 4 + 5 + 2 = 18, is a multiple of 9. It is, so 7,452 ÷ 9 = 828. Since 999, 99 and 9 are also multiples of 3, the same argument gives the test for 3.",{"id":336,"type":114,"caption":337,"columns":338,"rows":342},"div-tests","Divisibility tests and the property that explains each",[339,340,341],"Divisor","Test","Why it works",[343,347,350,354,358,361],[344,345,346],"2, 5, 10","Look at the last digit","10 is a multiple of 2, 5 and 10, so every tens, hundreds, … part is too. Only the ones digit matters.",[125,348,349],"Last two digits divisible by 4","100 is a multiple of 4, so hundreds and above always are.",[351,352,353],"8","Last three digits divisible by 8","1,000 = 8 × 125.",[355,356,357],"3 and 9","Digit sum divisible by 3 or 9","10 = 9 + 1, 100 = 99 + 1, … leave the digit sum behind.",[33,359,360],"Alternating digit sum (from the right: + − + −) divisible by 11","10 = 11 − 1, 100 = 99 + 1, 1,000 = 1,001 − 1: the leftovers alternate − and +.",[362,363,364],"6","Divisible by 2 and by 3","6 = 2 × 3 and 2, 3 share no factor.",{"id":366,"type":143,"title":367,"problem":368,"steps":369},"we-div11","Testing 918,082 for 11","Is **918,082** divisible by 11?",[370,371,372,373],"Digits from the right: 2, 8, 0, 8, 1, 9.","Alternating sum: 2 − 8 + 0 − 8 + 1 − 9.","= -22. A multiple of 11 (it is −2 × 11).","So 918,082 is divisible by 11: 918,082 ÷ 11 = 83,462.",{"id":375,"type":210,"itemId":376,"prompt":377,"check":378,"hints":380,"feedback":383},"pr-digit-sum","properties-of-numbers.deepen-div9","The number 5,2□4 is divisible by 9. What digit goes in the box?",{"kind":214,"answer":379,"tolerance":216},7,[381,382],"The digit sum must be a multiple of 9.","5 + 2 + 4 = 11. What must you add to reach 18?",{"correct":384,"incorrect":385},"Yes: 5 + 2 + 7 + 4 = 18, and 5,274 ÷ 9 = 586.","5 + 2 + 4 = 11. The next multiple of 9 is 18, so the box is 7. (Adding 16 to reach 27 is impossible with one digit.)",{"id":387,"type":68,"title":388,"eyebrow":389,"navLabel":390},"ch6","Division by zero: the proof, the paradox and the history","Chapter 06","6 Dividing by zero",{"id":392,"type":43,"markdown":393},"divzero-proof","In Understand we said a ÷ 0 has no answer because nothing times 0 gives a. Here is a stronger argument: **if we insisted on giving 1 ÷ 0 a value, arithmetic would collapse.**\n\nSuppose 1 ÷ 0 = k for some number k. Division is defined so that (1 ÷ 0) × 0 = 1, so k × 0 = 1. But the zero property says k × 0 = 0. Therefore **0 = 1**. And if 0 = 1, then adding 1 to both sides gives 1 = 2, and so on: every number equals every other number. The whole of arithmetic becomes meaningless.\n\nSo the choice is stark: either keep the zero property, the distributive law and the rest, or allow division by zero. Mathematicians keep the properties and leave division by zero **undefined**. (Some specialised systems, used in parts of higher mathematics and computing, add a special symbol \"∞\" or \"NaN\", but they pay for it by breaking some ordinary rules.)",{"id":395,"type":143,"title":396,"problem":397,"steps":398,"help":406},"we-fake-proof","Spot the hidden division by zero","Here is a famous \"proof\" that 2 = 1. Find the illegal step.",[399,400,401,402,403,404,405],"Let a and b be numbers with a = b.","Multiply both sides by a: a × a = a × b.","Subtract b × b from both sides: a × a − b × b = a × b − b × b.","Factor both sides (distributive law): (a + b) × (a − b) = b × (a − b).","Divide both sides by (a − b): a + b = b.","Since a = b: b + b = b, so 2b = b, so **2 = 1**.","**The error is step 5.** Because a = b, the number (a − b) is **0**. Dividing both sides by it is division by zero. Every step before it is fine; that one step lets a false conclusion in.",{"simplerExplanation":407},"It is true that 2 × 0 = 1 × 0. But you cannot \"cancel\" the 0 to conclude 2 = 1. Cancelling a factor is dividing by it, and you cannot divide by 0.",{"id":409,"type":410,"title":411,"items":412},"timeline-zero","timeline","How people thought about zero and division by zero",[413,417,421,425,429,433,437],{"time":414,"title":415,"text":416},"c. 300 BCE","Euclid's Elements","Book II proves area results that we now read as the distributive law, using rectangles.",{"time":418,"title":419,"text":420},"628 CE","Brahmagupta","In the Brahmasphutasiddhanta, Brahmagupta gives rules for zero and negatives: a − a = 0, a × 0 = 0. He states 0 ÷ 0 = 0, which modern mathematics does not accept.",{"time":422,"title":423,"text":424},"830–850 CE","Mahavira","In the Ganita Sara Samgraha, the Jain mathematician Mahavira tries to correct Brahmagupta and writes that a number divided by zero stays unchanged — another early attempt now known to be wrong.",{"time":426,"title":427,"text":428},"1150 CE","Bhaskara II","Bhaskara II writes that a quantity divided by zero \"becomes a fraction the denominator of which is zero\" — an infinite quantity, unchanged by adding or taking away. It is commonly called khahara.",{"time":430,"title":431,"text":432},"1814","Servois names the laws","In a memoir in the Annales de mathématiques, François-Joseph Servois introduces the words \"commutative\" and \"distributive\" for these properties.",{"time":434,"title":435,"text":436},"1843–44","Hamilton's quaternions","William Rowan Hamilton invents quaternions, numbers whose multiplication is associative but not commutative. About a year later he coins the word \"associative\".",{"time":438,"title":439,"text":440},"1889","Peano's axioms","Giuseppe Peano lists basic rules for natural numbers from which the properties of + and × can be proved.",{"id":442,"type":77,"variant":78,"title":443,"markdown":444},"nuance-limits","Infinity is not a number here","Bhaskara II's idea is close to what later mathematicians called a **limit**: 12 ÷ 0.001 = 12,000 and 12 ÷ 0.000001 = 12,000,000, growing without end as the divisor shrinks towards zero. But dividing by small *negative* numbers gives huge *negative* answers. There is no single value both sides approach, so even \"infinity\" does not work as the answer to 12 ÷ 0 in ordinary arithmetic.",{"id":446,"type":68,"title":447,"eyebrow":448,"navLabel":449},"ch7","Only one identity, and why inverses need new numbers","Chapter 07","7 Inverses",{"id":451,"type":43,"markdown":452},"unique-identity","Could there be a second additive identity, some number z other than 0 with a + z = a for every a? Here is a two-line proof that there cannot be.\n\nSuppose both 0 and z are additive identities. Look at the single number 0 + z.\n\n- Because 0 is an identity, 0 + z = z.\n- Because z is an identity, 0 + z = 0.\n\nSo z = 0 + z = 0. The \"second\" identity was 0 all along. **The additive identity is unique.** The same argument with × shows 1 is the only multiplicative identity.\n\n**Inverses.** For 5, is there a whole number w with 5 + w = 0 (the identity)? No: adding a whole number never takes you below 5. We need a new number, −5, the **additive inverse**. With it, subtraction becomes adding the inverse: 8 − 5 = 8 + (−5). Similarly, 5 × w = 1 needs the fraction ⅕, the **multiplicative inverse** or reciprocal, and division becomes multiplying by it: 8 ÷ 5 = 8 × ⅕. This is how mathematicians \"fix\" subtraction and division: they are not new operations at all, just + and × with inverses. And 0 has no multiplicative inverse, since 0 × w = 0, never 1. That is division by zero, seen from a new angle.",{"id":454,"type":114,"caption":455,"columns":456,"rows":461},"inverse-table","Identities and inverses",[457,183,458,459,460],"Operation","Inverse of a","Needs","Exception",[462,468],[463,464,465,466,467],"Addition","0","−a (because a + (−a) = 0)","Integers","none",[469,470,471,472,473],"Multiplication","1","1\u002Fa (because a × 1\u002Fa = 1)","Fractions","a = 0 has no inverse",{"id":475,"type":68,"title":476,"eyebrow":477,"navLabel":478},"ch8","Even and odd, proved with letters","Chapter 08","8 Parity proofs",{"id":480,"type":481,"items":482},"formulas-parity","formulas",[483,486,489,492,495,498],{"expression":484,"caption":485},"even = 2m","Every even number is 2 times a whole number m.",{"expression":487,"caption":488},"odd = 2m + 1","Every odd number is one more than an even number.",{"expression":490,"caption":491},"2m + 2n = 2(m + n)","even + even = even (distributive law).",{"expression":493,"caption":494},"(2m + 1) + (2n + 1) = 2(m + n + 1)","odd + odd = even.",{"expression":496,"caption":497},"(2m+1)(2n+1) = 2(2mn+m+n) + 1","odd × odd = odd.",{"expression":499,"caption":500},"(2m)² = 4m²; (2m+1)² = 4(m²+m) + 1","Even squares are multiples of 4; odd squares are 1 more than a multiple of 4.",{"id":502,"type":43,"markdown":503},"parity-deep","The last line gives a surprising fact: **every square leaves remainder 0 or 1 when divided by 4**, never 2 or 3. Check: 36 = 4 × 9, 49 = 4 × 12 + 1, 81 = 4 × 20 + 1. So a number like 1,234,567 (remainder 3 when divided by 4, since its last two digits 67 leave 3) can **never** be a perfect square. You do not need a calculator to know it.\n\nAlso, odd squares are always **1 more than a multiple of 8**: m² + m = m × (m + 1) is a product of consecutive numbers, hence even, so 4 × (m² + m) is a multiple of 8. Check: 9 − 1 = 8, 25 − 1 = 24, 49 − 1 = 48, 81 − 1 = 80. All multiples of 8.",{"id":505,"type":143,"title":506,"problem":507,"steps":508},"we-parity-coins","A parity puzzle","Seven cups stand upside down. In each move you must turn over **exactly two** cups. Can you make all seven cups stand the right way up?",[509,510,511,512,513,514],"Count the upside-down cups. It starts at 7, which is odd.","A move turns two cups. Each cup either goes from down to up (−1 to the count) or up to down (+1).","So a move changes the count by −2, 0 or +2: always an even amount.","Adding or subtracting an even number never changes parity. The count stays odd for ever.","All cups up means the count is 0, which is even. It can never be reached. **Impossible.**","This kind of argument, finding a quantity whose parity never changes, is called an **invariant** argument, and it solves many olympiad problems.",{"id":516,"type":68,"title":517,"eyebrow":518,"navLabel":519},"ch9","Invent an operation: which properties survive?","Chapter 09","9 New operations",{"id":521,"type":43,"markdown":522},"newops-prose","Mathematicians often ask: *which properties does this operation have?* Let's invent some operations on whole numbers and investigate. We use a new symbol for each.\n\n- **a ⊕ b = a + b + 1** (\"add, then one more\").\n- **a ⊡ b = the larger of a and b** (write max(a, b); max(3, 7) = 7).\n- **a ⊖ b = the difference between a and b, bigger minus smaller** (5 ⊖ 8 = 3).\n- **a ⊘ b = the average of a and b** ((a + b) ÷ 2).\n- **a ⊲ b = a** (just keep the first number).\n- **a ↑ b = a multiplied by itself b times** (a power: 2 ↑ 3 = 8).",{"id":524,"type":114,"caption":525,"columns":526,"rows":530},"newops-table","Properties of the invented operations (✓ always; ✗ with a counterexample)",[457,527,528,529],"Commutative?","Associative?","Identity?",[531,536,540,544,547,552],[532,533,534,535],"a ⊕ b = a + b + 1","✓","✓ both groupings give a + b + c + 2","none in whole numbers (would need −1)",[537,533,538,539],"max(a, b)","✓ both give the largest of the three","0: max(a, 0) = a",[541,533,542,543],"a ⊖ b (difference)","✗ (5 ⊖ 3) ⊖ 1 = 1 but 5 ⊖ (3 ⊖ 1) = 3","0: a ⊖ 0 = a",[545,533,546,467],"average","✗ avg(avg(0, 4), 8) = 5 but avg(0, avg(4, 8)) = 3",[548,549,550,551],"a ⊲ b = a","✗ 3 ⊲ 5 = 3 but 5 ⊲ 3 = 5","✓ both groupings give a","none (works only on the right)",[553,554,555,556],"power a ↑ b","✗ 2 ↑ 3 = 8 but 3 ↑ 2 = 9","✗ (2 ↑ 3) ↑ 2 = 64 but 2 ↑ (3 ↑ 2) = 512","none (1 works only on the right)",{"id":558,"type":77,"variant":83,"title":559,"markdown":560},"aha-independent","The properties are independent","The table shows every combination is possible: commutative and associative (⊕, max), commutative but not associative (difference, average), associative but not commutative (⊲), and neither (power). So knowing one property tells you **nothing** about the other. That is why the syllabus lists them separately, and why each needs its own proof for + and ×.",{"id":562,"type":88,"prompt":563,"options":564,"explanation":573},"predict-plus-over-times","Multiplication distributes over addition. Does **addition distribute over multiplication**? That is, is a + (b × c) = (a + b) × (a + c) for all whole numbers?",[565,567,569,571],{"id":92,"label":566},"Yes, always",{"id":95,"label":568},"No, and 1, 2, 3 is a counterexample",{"id":98,"label":570},"Only when a = 0",{"id":101,"label":572},"Only for even numbers","**No.** Try a = 1, b = 2, c = 3: 1 + (2 × 3) = 7, but (1 + 2) × (1 + 3) = 12. It does work when a = 0 (both sides become b × c), which is why option 3 is tempting, but that is \"sometimes\", not \"always\". The distributive law runs in **one direction only** for ordinary numbers. (In the logic of sets and in computer circuits there are operations, \"and\" and \"or\", that distribute over each other both ways, so this is a special fact about + and ×, not a law of nature.)",{"id":575,"type":237,"component":576,"componentVersion":5,"config":577,"objective":649,"textAlternative":650,"help":651},"lab-sort-newops","sort-game",{"prompt":578,"bins":579,"items":592,"seconds":216},"For whole numbers, which properties does each operation have?",[580,583,586,589],{"id":581,"label":582},"both","Commutative and associative",{"id":584,"label":585},"comm","Commutative only",{"id":587,"label":588},"assoc","Associative only",{"id":590,"label":591},"neither","Neither",[593,597,601,605,609,613,617,621,625,629,633,637,641,645],{"id":594,"label":595,"bin":581,"why":596},"o1","a + b","Ordinary addition has both properties.",{"id":598,"label":599,"bin":581,"why":600},"o2","a × b","Ordinary multiplication has both properties.",{"id":602,"label":603,"bin":590,"why":604},"o3","a − b","5 − 3 ≠ 3 − 5, and (8 − 3) − 2 ≠ 8 − (3 − 2).",{"id":606,"label":607,"bin":590,"why":608},"o4","a ÷ b","6 ÷ 3 ≠ 3 ÷ 6, and (8 ÷ 4) ÷ 2 ≠ 8 ÷ (4 ÷ 2).",{"id":610,"label":611,"bin":581,"why":612},"o5","max(a, b), the larger number","The largest of a set does not depend on order or grouping.",{"id":614,"label":615,"bin":581,"why":616},"o6","min(a, b), the smaller number","The smallest of a set does not depend on order or grouping.",{"id":618,"label":619,"bin":581,"why":620},"o7","a + b + 1","Both groupings give a + b + c + 2.",{"id":622,"label":623,"bin":584,"why":624},"o8","The difference, bigger minus smaller","Order does not matter, but (5 ⊖ 3) ⊖ 1 = 1 while 5 ⊖ (3 ⊖ 1) = 3.",{"id":626,"label":627,"bin":584,"why":628},"o9","The average (a + b) ÷ 2","avg(avg(0, 4), 8) = 5 but avg(0, avg(4, 8)) = 3.",{"id":630,"label":631,"bin":587,"why":632},"o10","Keep the first number: a ⊲ b = a","Both groupings give a, but 3 ⊲ 5 = 3 and 5 ⊲ 3 = 5.",{"id":634,"label":635,"bin":590,"why":636},"o11","Power: a multiplied by itself b times","2³ = 8 but 3² = 9; (2³)² = 64 but 2 to the power 9 = 512.",{"id":638,"label":639,"bin":581,"why":640},"o12","a × b + a + b","It equals (a + 1)(b + 1) − 1, which inherits both properties from ×.",{"id":642,"label":643,"bin":590,"why":644},"o13","2 × a + b","2 × 1 + 2 = 4 but 2 × 2 + 1 = 5; groupings give 4a + 2b + c and 2a + 2b + c.",{"id":646,"label":647,"bin":587,"why":648},"o14","Keep the last number: a ⊳ b = b","Both groupings give c, but 3 ⊳ 5 = 5 and 5 ⊳ 3 = 3.","Classify ordinary and invented operations by whether they are commutative, associative, both or neither.","A sorting game with four bins and fourteen operations.\n\nCommutative and associative: a + b; a × b; max(a, b); min(a, b); a + b + 1; a × b + a + b (which equals (a + 1)(b + 1) − 1).\n\nCommutative only: the difference \"bigger minus smaller\" ((5, 3) then 1 gives 1, but 5 with (3, 1) gives 3); the average ((0, 4) then 8 gives 5, but 0 with (4, 8) gives 3).\n\nAssociative only: keep the first number; keep the last number.\n\nNeither: a − b; a ÷ b; powers (2³ = 8, 3² = 9); 2 × a + b.",{"hints":652},[653,654],"Test order with 2 and 3; test grouping with 1, 2 and 3.","Look for a counterexample for each property separately.",{"id":656,"type":68,"title":657,"eyebrow":658,"navLabel":659},"ch10","Why the properties are the rules of algebra","Chapter 10","10 Towards algebra",{"id":661,"type":43,"markdown":662},"algebra-prose","In algebra, letters stand for numbers. Everything you are allowed to do with letters comes from the properties, because the letters could be **any** numbers.\n\n- **Collecting like terms is distributive.** 5x + 3x = (5 + 3) × x = 8x. There is no separate rule; it is the distributive law read backwards.\n- **Rearranging is commutative and associative.** x + 7 + 2x + 3 = (x + 2x) + (7 + 3) = 3x + 10.\n- **Expanding brackets is distributive.** 4(x + 5) = 4x + 20, and (x + 3)(x + 2) = x² + 2x + 3x + 6 = x² + 5x + 6.\n- **Solving equations uses identities and inverses.** From x + 7 = 12, add −7 to both sides: x + 0 = 5, so x = 5. From 3x = 21, multiply by ⅓: 1 × x = 7.\n- **Factorising is distributive backwards.** 6x + 9 = 3(2x + 3).\n\nWhenever an algebra step seems mysterious, ask: *which property is this?* There is always an answer.",{"id":664,"type":143,"title":665,"problem":666,"steps":667},"we-solve","Solving an equation, naming every property","Solve **7x + 4 = 39** and name the property used at each step.",[668,669,670,671,672,673,674],"Add −4 (the additive inverse of 4) to both sides: 7x + 4 + (−4) = 39 + (−4).","Associative law and inverse: 4 + (−4) = 0, so 7x + 0 = 35.","Additive identity: 7x + 0 = 7x, so 7x = 35.","Multiply both sides by ⅐ (the multiplicative inverse of 7): ⅐ × (7 × x) = ⅐ × 35.","Associative law: (⅐ × 7) × x = 1 × x.","Multiplicative identity: 1 × x = x. And ⅐ × 35 = 5. So **x = 5**.","Check: 7 × 5 + 4 = 39. ✓ In school you write this in two lines (7x = 35, x = 5), but the properties are doing the work underneath.",{"id":676,"type":210,"itemId":677,"prompt":678,"check":679,"hints":691,"feedback":694},"pr-simplify","properties-of-numbers.deepen-simplify","Simplify **7x + 4 + 2x − 1**.",{"kind":680,"options":681,"correct":690},"choice",[682,684,686,688],{"id":92,"label":683},"9x + 3",{"id":95,"label":685},"9x + 5",{"id":98,"label":687},"12x",{"id":101,"label":689},"14x − 4",[92],[692,693],"Rearrange so like terms are together (commutative and associative laws).","7x + 2x = (7 + 2)x.",{"correct":695,"incorrect":696},"Yes: (7x + 2x) + (4 − 1) = 9x + 3.","Group the x terms: 7x + 2x = 9x (distributive law backwards). Group the numbers: 4 − 1 = 3. Answer 9x + 3.",{"id":698,"type":114,"caption":699,"columns":700,"rows":703},"algebra-table","Algebra steps and the property behind each",[701,184,702],"Step","Property",[704,708,712,716,720,724,728],[705,706,707],"Swap terms","x + 5 = 5 + x","Commutative (+)",[709,710,711],"Regroup","(2x + 3) + 7 = 2x + (3 + 7)","Associative (+)",[713,714,715],"Write numbers first","x × 4 = 4x","Commutative (×)",[717,718,719],"Expand","3(x − 2) = 3x − 6","Distributive over −",[721,722,723],"Collect like terms","7y − 2y = 5y","Distributive (backwards)",[725,726,727],"Remove + 0 or × 1","x + 0 = x, 1x = x","Identities",[729,730,731],"Zero product","if 5 × (x − 2) = 0 then x − 2 = 0","Zero-product rule",{"id":733,"type":77,"variant":324,"title":734,"markdown":735},"misc-algebra","Three algebra slips, three broken properties","- **3(x + 4) = 3x + 4** ✗ The 3 must multiply both terms: 3x + 12.\n- **(x + 3)² = x² + 9** ✗ The distributive law gives (x + 3)(x + 3) = x² + 6x + 9. Check with x = 1: 16, not 10.\n- **2 × (3x) = 6 × 2x** ✗ or 6x × 2 ✗. It is (2 × 3)x = 6x by associativity. Multiplication does not distribute over multiplication.",{"id":737,"type":237,"component":738,"componentVersion":5,"config":739,"objective":767,"textAlternative":768,"help":769},"lab-match-algebra","match-pairs",{"prompt":740,"mode":741,"pairs":742},"Match each expression to its expanded or simplified form.","connect",[743,746,749,752,755,758,761,764],{"a":744,"b":745},"4(x + 5)","4x + 20",{"a":747,"b":748},"3(x − 2)","3x − 6",{"a":750,"b":751},"5x + 3x","8x",{"a":753,"b":754},"(x + 3)(x + 2)","x² + 5x + 6",{"a":756,"b":757},"(x + 1)²","x² + 2x + 1",{"a":759,"b":760},"(x + 4)(x − 4)","x² − 16",{"a":762,"b":763},"6x + 9","3(2x + 3)",{"a":765,"b":766},"x + 7 + 2x + 3","3x + 10","Connect algebraic expressions with their expanded or simplified forms, each justified by a property of numbers.","A matching game with eight pairs of equivalent algebraic expressions.\n\n4(x + 5) = 4x + 20 and 3(x − 2) = 3x − 6 (distributive law).\n5x + 3x = 8x (distributive law backwards: collecting like terms).\n(x + 3)(x + 2) = x² + 5x + 6 and (x + 1)² = x² + 2x + 1 (every term times every term).\n(x + 4)(x − 4) = x² − 16 (the middle terms +4x and −4x cancel).\n6x + 9 = 3(2x + 3) (factorising: taking out the common factor 3).\nx + 7 + 2x + 3 = 3x + 10 (commutative and associative laws).\n\nTo check any pair, put a number in for x, for example x = 2: 4(2 + 5) = 28 and 4 × 2 + 20 = 28.",{"hints":770},[771,772],"Substitute x = 2 into both sides to check a pair.","Every term in the first bracket multiplies every term in the second.",{"id":774,"type":68,"title":775,"eyebrow":776,"navLabel":777},"ch11","Review","Chapter 11","11 Review",{"id":779,"type":780,"title":781,"terms":782},"glossary-deepen","glossary","Deeper vocabulary",[783,785,789,793,796,799,802,805,809,813,817,820,824,828],{"term":51,"meaning":784},"An argument showing a statement is true in every case it covers.",{"term":786,"meaning":787,"example":788},"Counting two ways","Proving two expressions equal by counting the same collection in two different orders.","Rows × columns = columns × rows",{"term":790,"meaning":791,"example":792},"Proof by contradiction","Assuming the opposite of what you want, and showing it leads to something impossible.","Assume 1 ÷ 0 = k; then 0 = 1.",{"term":794,"meaning":795},"General associativity","For an associative operation, every way of bracketing a long list gives the same result.",{"term":797,"meaning":798},"Additive inverse","The number that adds to a to give 0: the additive inverse of 5 is −5.",{"term":800,"meaning":801},"Multiplicative inverse (reciprocal)","The number that multiplies a to give 1: the reciprocal of 5 is ⅕. Zero has none.",{"term":803,"meaning":804},"Unique","There is exactly one. The additive identity is unique.",{"term":806,"meaning":807,"example":808},"Invariant","A quantity that never changes during a process, used to prove some outcomes impossible.","The parity of upside-down cups.",{"term":810,"meaning":811,"example":812},"Binary operation","A rule that combines two numbers to give one number.","+, ×, max, average",{"term":814,"meaning":815,"example":816},"Like terms","Terms with the same letter part, which can be collected by the distributive law.","5x and 3x",{"term":717,"meaning":818,"example":819},"Remove brackets by multiplying out with the distributive law.","3(x + 2) = 3x + 6",{"term":821,"meaning":822,"example":823},"Factorise","Write a sum as a product by taking out a common factor.","6x + 9 = 3(2x + 3)",{"term":825,"meaning":826,"example":827},"Digit sum","The sum of the digits of a number; used in the tests for 3 and 9.","Digit sum of 7,452 is 18.",{"term":829,"meaning":830},"Khahara","The traditional Sanskrit name for a quantity divided by zero, commonly attributed to Bhaskara II (c. 1150 CE), who described it as an infinite quantity.",{"id":832,"type":833,"title":834,"questions":835},"quiz-deepen","quiz","Deepen check",[836,849,862,875,888,901,913,926,939,951,964],{"itemId":837,"prompt":838,"options":839,"correct":95,"why":848},"properties-of-numbers.deepen-q-array","Rotating an a × b array a quarter turn proves that…",[840,842,844,846],{"id":92,"label":841},"a + b = b + a",{"id":95,"label":843},"a × b = b × a",{"id":98,"label":845},"(a × b) × c = a × (b × c)",{"id":101,"label":847},"a × 1 = a","The same dots are counted as a rows of b and as b rows of a.",{"itemId":850,"prompt":851,"options":852,"correct":95,"why":861},"properties-of-numbers.deepen-q-box","Counting a box of cubes by layers and by slices proves…",[853,855,857,859],{"id":92,"label":854},"the commutative law of +",{"id":95,"label":856},"the associative law of ×",{"id":98,"label":858},"the distributive law",{"id":101,"label":860},"closure under −","(a × b) × c and a × (b × c) count the same box.",{"itemId":863,"prompt":864,"options":865,"correct":92,"why":874},"properties-of-numbers.deepen-q-square","99² =",[866,868,870,872],{"id":92,"label":867},"9,801",{"id":95,"label":869},"9,981",{"id":98,"label":871},"9,999",{"id":101,"label":873},"10,199","(100 − 1)² = 10,000 − 200 + 1 = 9,801. 99 × 99 = 9,801.",{"itemId":876,"prompt":877,"options":878,"correct":92,"why":887},"properties-of-numbers.deepen-q-diffsq","53 × 47 =",[879,881,883,885],{"id":92,"label":880},"2,491",{"id":95,"label":882},"2,500",{"id":98,"label":884},"2,509",{"id":101,"label":886},"2,391","(50 + 3)(50 − 3) = 2,500 − 9 = 2,491. 53 × 47 = 2,491.",{"itemId":889,"prompt":890,"options":891,"correct":98,"why":900},"properties-of-numbers.deepen-q-fake","In the \"2 = 1\" proof, which step is illegal?",[892,894,896,898],{"id":92,"label":893},"Multiplying both sides by a",{"id":95,"label":895},"Subtracting b² from both sides",{"id":98,"label":897},"Dividing both sides by a − b",{"id":101,"label":899},"Factorising","a = b makes a − b = 0, so that step divides by zero.",{"itemId":902,"prompt":903,"options":904,"correct":98,"why":912},"properties-of-numbers.deepen-q-unique","If 0 and z are both additive identities, then 0 + z equals…",[905,907,909,911],{"id":92,"label":906},"only 0",{"id":95,"label":908},"only z",{"id":98,"label":910},"both 0 and z, so z = 0",{"id":101,"label":590},"Using each identity in turn, 0 + z = z and 0 + z = 0, so z = 0.",{"itemId":914,"prompt":915,"options":916,"correct":101,"why":925},"properties-of-numbers.deepen-q-square4","Which of these can be a perfect square?",[917,919,921,923],{"id":92,"label":918},"4,322",{"id":95,"label":920},"7,563",{"id":98,"label":922},"1,227",{"id":101,"label":924},"5,329","Squares leave remainder 0 or 1 when divided by 4. 22, 63 and 27 leave 2, 3 and 3. 5,329 leaves 1, and indeed 73 × 73 = 5,329.",{"itemId":927,"prompt":928,"options":929,"correct":95,"why":938},"properties-of-numbers.deepen-q-div9","Why does the digit-sum test for 9 work?",[930,932,934,936],{"id":92,"label":931},"Because 9 is odd",{"id":95,"label":933},"Because 10, 100, 1,000… are each 1 more than a multiple of 9",{"id":98,"label":935},"Because 9 = 3 × 3",{"id":101,"label":937},"It only works for small numbers","10 = 9 + 1, 100 = 99 + 1 … so by the distributive law a number and its digit sum differ by a multiple of 9.",{"itemId":940,"prompt":941,"options":942,"correct":95,"why":950},"properties-of-numbers.deepen-q-avg","The operation \"average of a and b\" is…",[943,945,947,949],{"id":92,"label":944},"commutative and associative",{"id":95,"label":946},"commutative only",{"id":98,"label":948},"associative only",{"id":101,"label":590},"avg(a, b) = avg(b, a), but avg(avg(0, 4), 8) = 5 while avg(0, avg(4, 8)) = 3.",{"itemId":952,"prompt":953,"options":954,"correct":98,"why":963},"properties-of-numbers.deepen-q-algebra","Which is equal to 3(2x + 5)?",[955,957,959,961],{"id":92,"label":956},"6x + 5",{"id":95,"label":958},"5x + 8",{"id":98,"label":960},"6x + 15",{"id":101,"label":962},"6x + 8","Distribute the 3 to both terms: 6x + 15.",{"itemId":965,"prompt":966,"options":967,"correct":98,"why":976},"properties-of-numbers.deepen-q-cups","Nine cups upside down; each move turns exactly two. Can all end up the right way up?",[968,970,972,974],{"id":92,"label":969},"Yes, in 5 moves",{"id":95,"label":971},"Yes, in 9 moves",{"id":98,"label":973},"No, the count stays odd",{"id":101,"label":975},"Only if you start with two","Each move changes the number of upside-down cups by an even amount, so it stays odd and never reaches 0.",{"id":978,"type":979,"prompt":980},"reflect-deepen","reflection","Choose one property (commutative, associative or distributive). Explain in your own words, with a picture described in words, why it must be true for every pair or triple of whole numbers, and say what your picture leaves out (for example, the number 0).",{"id":982,"type":983,"title":984,"points":985},"cheat-deepen","summary","Cheat sheet",[986,987,988,989,990,991,992,993,994,995,996],"**Commutativity:** count one array as rows and as columns (×), or one string of beads from both ends (+).","**Associativity:** count one box of a × b × c cubes by layers or by slices. General associativity lets you drop brackets entirely.","**Distributivity:** cut a rectangle. (a + b)(c + d) has four pieces; (a ± b)² = a² ± 2ab + b²; (a + b)(a − b) = a² − b².","**Methods:** long multiplication and short division are the distributive law with place value; you may split the dividend, never the divisor.","**Divisibility tests** come from 10 = 9 + 1 = 11 − 1 and from 100 = 4 × 25, 1,000 = 8 × 125.","**Division by zero:** defining 1 ÷ 0 = k forces k × 0 = 1, so 0 = 1. The \"2 = 1\" trick hides a division by zero.","**History:** Brahmagupta (628) first gave rules for zero; Bhaskara II (c. 1150) described a ÷ 0 as an infinite quantity; Servois (1814) named the commutative and distributive laws.","**Identities are unique;** inverses need new numbers: −a (integers) and 1\u002Fa (fractions); 0 has no reciprocal.","**Parity with letters:** 2m, 2m + 1. Squares leave remainder 0 or 1 on division by 4; odd squares are 1 more than a multiple of 8.","**Properties are independent:** an operation may be commutative without being associative (average), or the reverse (keep-first).","**Algebra** is arithmetic with properties: collect like terms, expand, factorise and solve all use them.",{"id":998,"type":999,"conceptId":1000,"relation":1001,"explanation":1002},"conn-order","connection","order-of-operations","related_to","Expanded forms like 20 × 40 + 20 × 7 + 3 × 40 + 3 × 7 rely on the order of operations: all × before +.",{"id":1004,"type":999,"conceptId":1005,"relation":1001,"explanation":1006},"conn-hcf","hcf-and-lcm","Factorising 6x + 9 = 3(2x + 3) takes out the highest common factor, 3, using the distributive law backwards.",{"id":1008,"type":999,"conceptId":1009,"relation":1001,"explanation":1010},"conn-prime","prime-and-composite","Divisibility tests for 2, 3, 5, 9 and 11, proved here with the distributive law, are the first tools for finding factors and primes.",{"id":1012,"type":999,"conceptId":1013,"relation":1001,"explanation":1014},"conn-patterns","patterns","Squares as sums of odd numbers, and the square tricks here, connect algebra identities to number patterns.",{"id":1016,"type":1017,"sourceIds":1018},"sources-deepen","sources",[1019,1020,1021,1022,1023,1024,1025,1026,1027,1028,1029,1030,1031],"properties-of-numbers-ncert-class6-whole-numbers","properties-of-numbers-ncert-class7-integers","properties-of-numbers-ncert-class8-rational-numbers","properties-of-numbers-mathsisfun-properties","properties-of-numbers-mathsisfun-divide-by-zero","properties-of-numbers-wiki-commutative","properties-of-numbers-wiki-distributive","properties-of-numbers-wiki-division-by-zero","properties-of-numbers-wiki-parity","properties-of-numbers-wiki-brahmagupta","properties-of-numbers-wiki-associative","properties-of-numbers-mactutor-servois","properties-of-numbers-mactutor-bhaskara",[1019,1020,1021,1022,1023,1024,1025,1026,1027,1028,1029,1030,1031],"needs_review",{"generatedBy":1035,"notes":1036},"claude-code","Draft generated with Python generators; every stated number computed and asserted. Pending owner review.","ff94dc503a184484b3b6000d924b9d369cdb885a7cb3aef234ba5e0644852e7b",{"logic:practice":1039,"component:arith-sprint@1":1040,"component:order-ops@1":1041,"component:sort-game@1":1042,"component:match-pairs@1":1043,"source:properties-of-numbers-mactutor-bhaskara":1044,"source:properties-of-numbers-mactutor-servois":1045,"source:properties-of-numbers-mathsisfun-divide-by-zero":1046,"source:properties-of-numbers-mathsisfun-properties":1047,"source:properties-of-numbers-ncert-class6-whole-numbers":1048,"source:properties-of-numbers-ncert-class7-integers":1049,"source:properties-of-numbers-ncert-class8-rational-numbers":1050,"source:properties-of-numbers-wiki-associative":1051,"source:properties-of-numbers-wiki-brahmagupta":1052,"source:properties-of-numbers-wiki-commutative":1053,"source:properties-of-numbers-wiki-distributive":1054,"source:properties-of-numbers-wiki-division-by-zero":1055,"source:properties-of-numbers-wiki-parity":1056},"3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","c0c63ed40e1bca5ba6d446d43886b887a3a7709e5cd6679419f64fd3ba62afe6","3d630ebc066352df791b020d545e1876ce4620d55c8c1c082414d6b4717255c5","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","15e1e4b78e54c7b997ae4a6e46751eb2c569305beff15e50785b978ed58ed2d6","133d7f2c86a7206aa5b120449a1da91a2eabc993909ecc13fb46dcbe780bed2f","f8a42fd8c82c266f4310450d74f1ae0449b3dba177afcf000b24aafa2839b629","a37b85a36742e06f34cdc083710eacdd662e2f7493395766e4b8c5632bfade3a","56b97dbf2988a58e1aa10f693ec9dd1039e1de14e795b1bdfa19980b6b18c1fa","aab8fe53a32660ef6307b4dbd6f5af02319957d9aecf3c0396895bc01ab46e2e","acb8b92cad8387724d72528e70ef68195351e5b2f03b580c5da191fde82dc67a","a1f428dd8d96495a8192f873a0fb33e394560cac9c9a0abbf24a1d58ef2371a9","32e7279091a5877afe54d6da75cdd06a04f2ff8677338777578592d6dd98a338","a3e0fb47a2115f5962c9795b138bf38cd9c2c4874bc66e1b834717b9a01908c1","befe26bd7ec4e1c3911549d33a91844ae3eb5bd63388711f61822d398d3d576f","046256cc76cdb1b7cca72a1b391bef686855afdfb6a68d18b45a38723d1fe93d","01c5bcf03886aec6f9c2c18c9ec5333b19ce1d2ca6bfb4f8544c4ca9f1578e27",{"state":1058,"reviewer":1059,"selfReview":289,"reviewedAt":1060,"method":1061},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899597142]