PrepShorts · Study sheet · Class 8 Mathematics · Chapter 2, The Baudhāyana-Pythagoras Theorem
Chapter 2 · The Baudhāyana-Pythagoras Theorem
Why √2 is neither a terminating decimal nor a fraction
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The third side has a square of exactly 2. That is a complete answer — and it is a name, not a value.
The idea
Squaring candidates gets you closer to √2 forever and never lands on it, and no amount of further squaring can tell you whether that is the method's fault or the number's. Two arguments settle it, and each works by counting something the answer would have to have and cannot. The decimal argument looks at the last digit: a decimal that stops has a last digit that is not zero, squaring it leaves a last digit that is still not zero, and 2 has no digits after the point at all. The fraction argument looks at how often the prime 2 appears: in a square, every prime is used an even number of times, so 2n² = m² would need the prime 2 an odd number of times on one side and an even number on the other. Neither argument searches. Both refuse.
What you should be able to do
- Bound √2 between 1 and 2 by comparing the areas of squares, and name 1 the lower bound and 2 the upper bound
- Tighten the bound to tenths, hundredths and thousandths by squaring candidates, and read the chapter's three bounding boxes
- Explain why the tightening never terminates, using the last-digit argument
- State the conclusion that the decimal expansion of √2 does not stop
- Set up the fraction assumption, reach 2n² = m², and derive the contradiction from the parity of the exponent of the prime 2
- Say which of the two results is the stronger one, and why the chapter still gives both
- Distinguish a bound from a value, and a truncated decimal from the number itself
- Attribute the fraction proof as the chapter attributes it, to Euclid's Elements, dated in the chapter to about 300 BCE
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| lower bound | a number known to be below the value being pinned down | printed in this chapter (Part II, §2.3, p.38) |
| upper bound | a number known to be above it | printed in this chapter (Part II, §2.3, p.38) |
| terminating decimal | a decimal whose digits come to an end | printed in this chapter (Part II, §2.3, p.38) |
| non-terminating decimal | a decimal whose digits never come to an end | printed in this chapter (Part II, §2.3, p.39) |
| decimal representation | the way a number is written in digits with a decimal point | printed as a subheading in this chapter (Part II, §2.3, p.38) |
| fraction | a number written as one counting number over another | printed in this chapter (Part II, §2.3, p.39) |
| counting numbers | 1, 2, 3, … — the chapter's phrase for the numerator and denominator allowed | printed in this chapter (Part II, §2.3, p.39) |
| prime factorization | the expression of a number as a product of primes, spelt this way in this book | printed in this chapter (Part II, §2.3, p.39) |
| square number | a number that is some whole number times itself | printed in this chapter (Part II, §2.3, p.39) |
| Elements | Euclid's book, dated in the chapter to about 300 BCE, credited with the fraction proof | printed in this chapter (Part II, §2.3, p.39) |
| proof by contradiction | assuming the opposite of what you want and driving it into an impossibility | an added term; the chapter runs the method twice and does not name it |
| irrational number | the standard name for a number that is no fraction of counting numbers | not printed in this chapter at all; do not use it when explaining it |
Where people slip up
- "√2 = 1.41421356." That is a truncation and it is the number the calculator had room for. Its square is not 2. Compute it and show the shortfall.
- "Non-terminating means it is not a fraction." The single most important correction in this topic, and the reason the chapter needs two separate arguments. One third is a fraction whose decimal never stops. Section 11 exists to fix this.
- "So the two proofs are the same proof twice." They are not, and the relationship runs one way: any decimal that stops can be written as a fraction — 1.414 is 1414 over 1000 — so the fraction result already rules out a stopping decimal. The decimal argument is the weaker of the two. The chapter runs it first anyway, because it is the one a student can reach with the bounding table already in hand, and that is a fair reason.
- "If we compute far enough we will find the pattern and it will repeat." The chapter does not raise repetition at all, and neither should the explanation — repeating decimals are a later topic. What it establishes is that the expansion does not stop, and that no fraction produces it.
- "Every prime appears an even number of times." Only in squares. Twelve is two twice and three once. The student needs the fact in the right form or the proof of the fraction result evaporates.
- "1.414 < √2 < 1.415 means √2 = 1.414 to three places." True but beside the point. The bound is a statement about where the number is, and it does not become a value however tight it gets.
- "You can't have a length you can't write down." You can draw this one with a ruler in one stroke — it is the diagonal of a square of side 1. Being unwritable as a decimal or a fraction is a fact about notation, not about existence. This is worth saying explicitly, because the two proofs can otherwise leave a student feeling the number has been argued away.
- "2 has a last digit, namely 2, so the last-digit argument fails." The argument is about digits after the decimal point. Written as a decimal, 2 has none — or, equivalently, all zeros. Make the decimal point visible when this step is made.
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Worked answers to this chapter’s exercises
Transcript1,334 words
Last time we found a length by going after an area instead. A right triangle with two equal sides of one has a third side whose square is exactly two. That is an answer, and it is a complete one. But it is a name, not a value. If someone asks how long that side actually is, in digits, we still cannot say. So let us try. We will hunt for it the obvious way, by squaring candidates and closing in.
And the hunt will fail, twice over, in two completely different ways. Each failure turns out to be worth more than the number would have been. Start as wide as possible, and do it with areas rather than lengths, because areas are the thing we can compare. A square of side one has area one. The square we are after has area two. And a square of side two has area four.
One is under two, and four is over two, so our side sits somewhere between one and two. Notice that nothing was measured. A longer side makes a bigger square, so comparing the areas compares the sides. And both ends were checked, separately. That matters more than it sounds, and in a moment you will see why. Those two numbers have names. One is the lower bound: a number we know the answer is above.
Two is the upper bound: a number we know it is below. A bound is not a guess. A guess is a number somebody likes. A bound is a number with a reason attached. And a bound has two halves. Drop either one and what is left says nothing at all. One to two is a very loose pen, a whole unit wide. Let us tighten it. Between one and two, take the candidates with one decimal place, and square each one.
One point one squared is one point two one. Under. One point two squared is one point four four. Under. One point three squared is one point six nine. Under. One point four squared is one point nine six. Still under, but only just. One point five squared is two point two five. Over. So the answer lies between one point four and one point five. Five multiplications, and the pen is ten times tighter than it was.
Now the same thing again, one place further in. One point four one squared is one point nine eight eight one. Under. One point four two squared is two point nought one six four. Over. So it lies between one point four one and one point four two. Once more. One point four one one, one point four one two, one point four one three and one point four one four all come out under.
One point four one five squared is two point nought nought two two two five, which is over. Three rounds of this, and the pen has gone from a whole unit wide down to a thousandth. And now the uncomfortable question, which is exactly the right one to ask. Does this ever land? Every round has taken us closer. Not one round has hit. And more squaring cannot answer it, because more squaring only ever hands you another bound.
If we ran it over and over and never landed, we would have learned nothing at all about whether the next round would. The question is not how close we can get. It is whether there is anything there to land on. Here is the argument that settles it, and it does not search at all. It counts one digit. Suppose some decimal that stops squares to exactly two. A decimal that stops has a last digit, and that last digit is not zero. If it were, it would not be the last one.
Now square it. The last digit of a square depends only on the last digit you started with. One squared ends in one. Two squared ends in four. Three in nine, four in six, five in five, six in six, seven in nine, eight in four, nine in one. Nine digits, nine last digits, and not one of them is zero. So the square of our stopping decimal also ends in a digit that is not zero.
And now look at two, written as a decimal. After the point it has nothing at all. No digits, or if you prefer, zeros forever. So the square has to end in something that is not zero, and it has to be a number ending in nothing but zeros. It cannot be both. There is no such decimal. So the expansion does not stop. Not at three decimal places, not at any number of them, not ever.
And notice what kind of argument that was. It never went looking for the number. It counted something the answer would have to have and cannot have. That is a refusal, not a search, and it is why one line can settle infinitely many candidates at once. Now a different question, and it really is a different one. Forget decimals. Could our length be a fraction, one counting number over another?
That is not the same question, and here is why. A fraction can perfectly well have a decimal that never stops. One third is a fraction. Its decimal is nought point three three three, going on forever. So what we just proved does not answer this. Never stopping is not the same as not being a fraction. It needs an argument of its own, and the argument is again a count.
Suppose our length is m over n, with m and n counting numbers. Square both sides. Two is m squared over n squared, so two times n squared is m squared. Now count how many times the prime two appears on each side, and nothing else. In any square, every prime appears an even number of times, because squaring a number doubles every one of its counts. Twelve is two twice and three once, so twelve is not a square. Thirty-six is two twice and three twice, and it is.
So on the right, m squared has an even number of twos. On the left, n squared has an even number, and then there is one more two sitting in front of it. Even plus one is odd, and an odd count cannot equal an even one. No such m and n exist anywhere. That argument comes down to us from Euclid's Elements. So we have two results, and they are not one result told twice.
Which of them is stronger? Any decimal that stops can be written as a fraction. One point four one four is one thousand four hundred and fourteen over a thousand. So ruling out fractions has already ruled out stopping decimals. The fraction result is the stronger of the two. And it does not run the other way. One third is the standing witness: never stops, and is a fraction all the same.
So why bother with the weaker argument first? Because it is the one you can reach with the squaring table already in your hand, and an argument you got to yourself is worth more than a better one you were handed. So what is the number, then? It is one point four one four two one three five six, and then more, and then more. That is where a calculator stops, and it is not the number.
Square it and you fall short of two, exactly and measurably. Eight digits go in, sixteen come out, and the last of them is not a zero. But do not let the two proofs talk you out of the length. You can draw it in one stroke. It is the diagonal of a square of side one, and it is exactly as long as it is. Being unwritable as a decimal, and unwritable as a fraction, are facts about writing. They are not facts about the line.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- The isosceles right triangle: why its hypotenuse must be a√2Class 8 · Ch 2, The Baudhāyana-Pythagoras Theorem
Either side of this one
- Combining two different squares: a² + b² = c²Class 8 · Ch 2, The Baudhāyana-Pythagoras Theorem