General Mathematics · Class 9-10 · Chapter 1 · Real numbers
√2 is irrational.
There is one real proof in this chapter, and this is it. The proof is not hard, but its method is new: instead of showing the thing directly, you assume the opposite and walk to something impossible. Once the method is in hand the proof is six lines long, and at the end of it one question is left over whose answer is a picture.
1. Begin by assuming the opposite
The method can be put in one sentence: assume the opposite is true, work from there to something impossible, and the assumption must have been wrong. It is what a court does when it takes a claim as true for a moment and shows that it cannot be squared with the rest.
One small fact is needed first, and the book asks for it separately: the square of any odd integer is odd. It is easy to see. Any odd number can be written as n = 2x − 1. Squaring gives n² = 4x² − 4x + 1, which rearranges to 2(2x² − 2x) + 1. Divide that by 2 and 1 is left over, so n² is odd.
n = 2x − 1 ⇒ n² = 2(2x² − 2x) + 1
Now turn it round, because the proof needs it the other way about: if p² is even then p must be even. Why? Because if p were odd, then by the line above p² would be odd too. That single line is step five of the proof, and getting stuck there means getting stuck on all of it.
2. The proof, in eight steps
In the widget below the steps move along on one side while a ledger fills in two columns above: on the left what has been assumed, on the right what has been worked out. At step seven the two columns come face to face, and that facing off is the proof.
step by stepgo to the next step
√2 is irrational: the proof in eight steps
The method: proof by contradiction. Assume the opposite and walk to something impossible.
Suppose √2 is a rational number.
Then √2 can be written as p/q, with p and q natural numbers, q ≠ 0, and p and q co-prime, that is the fraction in its lowest terms.
Square both sides: 2 = p²/q², so p² = 2q².
2q² is even, so p² is even. And the square of an odd number is odd, so p is even.
Since p is even, write p = 2m. Substituting gives 4m² = 2q², that is q² = 2m², so q is even as well.
But p and q both even means they share the factor 2, so their greatest common divisor is not 1. And at step 3 we assumed they were co-prime.
Both cannot hold at once. So the assumption at step 2 was wrong. Therefore √2 is an irrational number.
Where the contradiction bites is worth saying separately, because this is where most students point at the wrong thing. There is no mistake in the algebra; p² = 2q² is exact. The contradiction bites at step three, the step we wrote ourselves. Taking the fraction in its lowest terms was our own choice, and the algebra then proved that it was not in its lowest terms.
Memorising the steps collapses on a question that has been altered slightly, because the steps hang together in a chain of dependence. The six lines below are shuffled; tap them in order and join the chain up.
which firsttap them in the right order
Put the proof in order
The proof opens with an assumption. Which line does that?
Yes. And then?
Not yet. Which one follows straight from the line before?
That is the order. Notice that every step stands on the one before it, and the last one collides with the first. The proof is not a circle, it is a trap.
check yourself
One question
In this proof that √2 is irrational, where exactly does the contradiction bite?
The equation is exact. There is no error anywhere in the algebra, and that is the strength of the proof: the error is in the assumption, not the working.
Right. At the start we chose the fraction in its lowest terms ourselves, that is with gcd 1. At the end both p and q turn out even, so the gcd is at least 2. Both cannot hold, and that is where the assumption breaks.
√2 does have a decimal value, 1.414213562…. The proof says nothing against that; it says that value cannot be got by dividing one whole number by another.
q = 0 was ruled out at the start, since a zero denominator is not a fraction at all. The proof never raises the question.
3. Which square roots are irrational
The book now makes the statement general: the square root of any natural number that is not a perfect square is irrational. So √3, √5, √6, √7, √8 and √10 are all irrational, and the shape of the proof barely changes; for √5 the word “even” is simply replaced by “divisible by 5”.
But having the rule still leaves a question: without a calculator, how do you tell whether a number is a perfect square? A method learned in class 7 does the job here. Keep subtracting the odd numbers 1, 3, 5, 7 in turn. Stop exactly at 0 and the number is a perfect square, with the count of subtractions as its square root. Drop below 0 and it is not, so its square root is irrational.
1 + 3 + 5 + … + (2n − 1) = n²
test it by handpick a number, then tap 1, 3, 5 …
Successive subtraction: perfect square or not?
check yourself
Another question
Subtracting successive odd numbers from 40 leaves 4 after six goes. What does that tell you?
Six subtractions happened, but it did not land on 0; 4 was left. The count only gives a square root when it stops exactly at 0.
The 4 is what was left over, not the root. And 4² is 16, not 40.
Right. 40 − 1 − 3 − 5 − 7 − 9 − 11 = 4, and subtracting the next odd number, 13, would go negative. It did not stop at 0, so 40 is not a perfect square. Since 6² = 36 and 7² = 49, √40 lies between them: not a whole number, and not a fraction either.
40 = 2 × 2 × 2 × 5, so it is not prime. And primeness is not what this test is about; the test is about perfect squares.
4. And yet it has a place
When the proof is over nearly every student is left with a doubt, though nobody says it aloud: does a number that cannot even be written as a fraction really exist? It does, and a ruler and a compass can show exactly where it sits.
The trick is Pythagoras'. Draw a right triangle with both legs one unit and its hypotenuse is √(1² + 1²) = √2. Lift that hypotenuse with a compass, swing it down onto the number line, and the point it lands on is √2. Tap the steps below one at a time and put the construction up.
ruler and compasstap the steps one at a time
Put √2 on the number line
The picture settles something important: the proof never said √2 has no value. It has one, and an exact one, 1.414213562…. All the proof said was that this value cannot be got by dividing one whole number by another. The geometric construction is outside the textbook, but nothing settles the doubt better.
One last trap, and nearly everyone falls into it once. Having learned that √2 is irrational, it starts to feel as if anything an irrational touches turns irrational. It does not. Tap the eight expressions below and see what each gives. (These use the rule √a × √b = √(ab), which is set out properly in the chapter on indices.)
judge themtap a chip
What happens when you do arithmetic on an irrational?
= 2√2, irrational. Multiplying by a non-zero rational leaves an irrational irrational.
= 2, rational. The product of two irrationals can perfectly well be rational. Here the root and the square have cancelled each other.
= 0, rational. The additive inverse works for irrationals exactly as it does for anything else.
= √16 = 4, rational. 2 × 8 = 16, and 16 is a perfect square. So the product has come out a whole number.
= √6, irrational. 6 is not a perfect square, so nothing cancels the way it did on the chip before.
Irrational. A rational plus an irrational is always irrational. If the sum were rational, subtracting 3 would make √2 rational too, which is impossible.
= 1, rational. Any non-zero number divided by itself is 1, irrational or not.
= 0, rational. Zero is the one rational multiplier that turns an irrational rational. With any other, an irrational stays irrational.
All eight seen. The rule to keep: a rational plus or minus an irrational is always irrational, and a non-zero rational times an irrational is always irrational. But an irrational plus an irrational, or times an irrational, carries no guarantee at all; either can come out.
check yourself
Last question
Which of these is a rational number?
= 2√3, irrational. Multiplying by the non-zero rational 2 leaves it irrational.
Right. 3 × 12 = 36, a perfect square, so √3 × √12 = √36 = 6, a rational number. The product of two irrationals can be rational, and this is the case.
A rational plus an irrational, so irrational. If it were rational, subtracting 2 would make √3 rational too.
A non-zero rational times an irrational, so irrational.
The lines from this note you will need for writing the proof and for judging expressions are cut onto the cards below.
cut from the notetap to turn over
Four cards
frontWhat are the three moves of a proof by contradiction?
backAssume the opposite, reach something impossible, so the assumption was wrong.
frontWhich step does the contradiction bite at?
backAt step 3, where they were assumed co-prime. At the end both turn out even.
frontHow do you spot a perfect square without a calculator?
backSubtract 1, 3, 5, 7 in turn. Stopping exactly at 0 means a perfect square, and the count is the root.
frontWhat happens when you mix rational and irrational?
backRational ± irrational is always irrational, and so is non-zero rational × irrational. Irrational with irrational gives no guarantee.