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Most students meet the binomial theorem as a formula with factorials, and mathematical induction as a rigid three-step template. Both ideas are older and more natural than they look. A thousand years ago in Baghdad, al-Karaji built the binomial coefficients by simple addition and argued from one case to the next in a way that historians see as an early form of induction. This post covers what he did, two worked examples, and the mistakes students make with both ideas.
Who was al-Karaji?
The MacTutor biography from the University of St Andrews gives his birth as 13 April 953 in Baghdad and his death as about 1029. Even his name is uncertain: "al-Karaji" would link his family to the city of Karaj in Iran, "al-Karkhi" to a suburb of Baghdad, and historians still disagree. He wrote his main algebra book, *Al-Fakhri*, in Baghdad and dedicated it to the city's ruler. Later he moved to the "mountain countries" and turned to engineering, including the drilling of wells.
MacTutor describes him as possibly the first person to free algebra completely from geometrical operations. He defined the powers x, x², x³ and so on, and their reciprocals 1/x, 1/x², and gave rules for multiplying any two of them. He studied the *Arithmetica* of Diophantus, which had been translated into Arabic around 870, and tried to generalise its methods.
How did al-Karaji build the binomial table?
His own text on this is lost. What survives is a description by al-Samaw'al, writing more than a century later. The MacTutor biography of al-Samaw'al says his book *al-Bahir* attributes the method to al-Karaji and is "the only surviving account of this remarkable work".
In the translation by Rashed and Ahmad quoted by MacTutor, the instruction is to place "one" on a table and "one" below it, then move the first "one" to a second column and add it to the "one" below to get "two". That gives the column 1, 2, 1. Repeat the move-and-add step and you get 1, 3, 3, 1, then 1, 4, 6, 4, 1. Al-Samaw'al explains that 1, 2, 1 means: to square a number split into two parts, multiply each part by itself once and the two parts by each other twice. That is (a + b)² = a² + 2ab + b². The account continues up to the coefficients for (a + b)⁵.
Worked example 1: expand (a + b)⁵ from the table.
Row for power 4: 1, 4, 6, 4, 1.
Add neighbours, keeping a 1 at each end: 1, 1 + 4 = 5, 4 + 6 = 10, 6 + 4 = 10, 4 + 1 = 5, 1.
So (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵.
Check: put a = b = 1. The coefficients should add to 2⁵ = 32, and 1 + 5 + 10 + 10 + 5 + 1 = 32.
Our binomial theorem guide for JEE covers the general term and middle term questions built on this table.
What was his early form of induction?
MacTutor explains that al-Karaji would prove a result for n = 1, then use it to prove n = 2, use that for n = 3, and continue to about n = 5 before remarking that the process can go on indefinitely. MacTutor is careful to add that this "is not induction proper", but calls it "a major step towards understanding inductive proofs". A study by Roshdi Rashed, L'induction mathématique: al-Karajī, as-Samaw'al (1972), examines this reasoning in detail.
His clearest example is the sum of cubes. He wrote, in Rashed and Ahmad's translation, that to add the cubes of the numbers in their natural order "we multiply their sum by itself". He showed that (1 + 2 + ... + 10)² equals (1 + 2 + ... + 9)² + 10³, then applied the same step to the 9, then the 8, all the way down.
Worked example 2: check al-Karaji's step with numbers.
1 + 2 + ... + 10 = 55 and 1 + 2 + ... + 9 = 45.
55² − 45² = (55 − 45)(55 + 45) = 10 × 100 = 1000 = 10³. The step works.
Unwinding it all the way: 1³ + 2³ + ... + 10³ = 55² = 3025.
In general, write Tₙ = n(n + 1)/2. Then Tₙ² − Tₙ₋₁² = (Tₙ − Tₙ₋₁)(Tₙ + Tₙ₋₁) = n × n² = n³, because Tₙ − Tₙ₋₁ = n and Tₙ + Tₙ₋₁ = n². That single line is the inductive step al-Karaji kept repeating.
What mistakes do students make here?
Off-by-one rows. The row 1, 4, 6, 4, 1 belongs to power 4, not 5. Count the entries: power n has n + 1 of them.
Checking a few cases and stopping. Al-Karaji's chain of cases is persuasive, but a proof needs the general step, like line 4 above, which works for every n at once.
Forgetting the base case. The step Tₙ² − Tₙ₋₁² = n³ needs a starting point: 1³ = 1².
Syllabus surprise. The JEE (Main) 2026 syllabus lists the binomial theorem but not mathematical induction, so induction matters more for olympiad proof writing and university maths. Check the syllabus for your own exam year.
If you can follow a proof but get lost writing your own inductive step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
What can a student take from al-Karaji's way of working?
He trusted arithmetic. Instead of a picture, he built the binomial table from one simple rule, and instead of a clever trick for the cubes, he found one step that could be repeated. He also learned by extending what he read: MacTutor notes that he took many problems from Diophantus but always tried to generalise them. When you finish a textbook example, ask what one step would make it work for every n.
Try it now: use al-Karaji's step to show that 1³ + 2³ + ... + 12³ = 78², then write the general inductive proof in three lines.
