How to Prove That Root 2 Is Irrational (Step by Step)

If you’ve ever wondered how to prove that √2 is irrational, good news: it’s one of the most famous proofs in maths, and it asks for very little. If you know what an even number is and can square a number, you’ve got everything you need.

We’ll go through it slowly, look at why each step works, point out where students usually slip, and finish with a short version for exam day. Read it once, then close the page and try writing it out yourself. That’s what makes it stick.

☰ Table of Contents

What does “irrational” actually mean?

A quick word on the terms first.

A rational number is anything you can write as a fraction a/b, where a and b are whole numbers and b isn’t zero. 1/2, 7/3, 5 (which is 5/1) and 0.25 (which is 1/4) are all rational.

An irrational number is one you can’t write that way. As a decimal, it goes on forever without ever settling into a repeating pattern.

√2 is the positive number that gives 2 when multiplied by itself. Your calculator will show something like 1.41421356…, but that’s just a rounded slice of it. What we want to show is that no fraction, no matter how huge its top and bottom numbers, equals √2 exactly.

The idea: proof by contradiction

The classic approach is called proof by contradiction. It sounds fancier than it is. Imagine someone tells you, “There’s a large whole number.” You say, “Okay, let’s suppose that’s true,” and then you add 1 to it. Now you’ve got something bigger, so the claim couldn’t have been true in the first place.

We’ll pull the same move here:

Pretend the opposite is true (that √2 is rational).
Follow the logic one step at a time.
Arrive at something impossible.
Conclude that the starting assumption was wrong, so √2 must be irrational.

This is the version most schools and universities teach, so it’s the one your examiner is most likely hoping to see.

The full proof

Take it slowly and make sure each line makes sense before you move on.

Step 1: Make the assumption. Suppose √2 is rational. Then we can write

√2 = a/b

where a and b are whole numbers, b isn’t zero, and the fraction is in its lowest terms, meaning a and b share no common factor other than 1. (Any fraction can be reduced like this, so assuming it costs us nothing.)

Step 2: Square both sides.

2 = a²/b²

Step 3: Rearrange. Multiply both sides by b²:

a² = 2b²

Step 4: Spot that a² is even. The right side is 2 times a whole number, so a² has to be even.

Step 5: Conclude that a is even. This leans on a small fact: if a² is even, then a is even. Here’s why. Suppose they were odd. Then a = 2k + 1 for some whole number k, and

a² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1

which is odd. So an odd number always has an odd square. Since our a² is even, a can’t be odd, which means it’s even.

Step 6: Write as 2k. Because a is even, we can say a = 2k for some whole number k.

Step 7: Substitute back. Put a = 2k into a² = 2b²:

(2k)² = 2b²
4k² = 2b²

Divide both sides by 2 and you get:

2k² = b²

Step 8: Spot that b is even too. Same reasoning as Step 5. b² equals 2 times a whole number, so b² is even, and therefore b is even.

Step 9: Find the contradiction. Now both a and b are even, so both can be divided by 2. That means a/b was not in lowest terms after all, even though that’s exactly what we assumed at the start. Something’s gone wrong.

Step 10: Conclude. The assumption led us to a dead end, so it has to be false. √2 can’t be written as a fraction, so it’s irrational. ∎

The short version

Pressed for time, or is the question only worth a few marks? Here’s the compact version:

Suppose √2 = a/b in lowest terms. Squaring gives a² = 2b², so a² is even, which makes an even. Write a = 2k. Then 4k² = 2b², so b² = 2k² and b is even too. But now a and b share a factor of 2, which contradicts lowest terms. So √2 can’t be rational.

Those few lines are enough, as long as you don’t skip the “lowest terms” idea or the reason an even square means an even number. Those two points are where most marks are won or lost, so don’t rush them.

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Why this actually works

Understanding the “why” matters more than memorising the steps, so let’s zoom out for a second.

The whole proof rests on one clash. A fraction in lowest terms can’t have an even top and an even bottom. Yet the equation a² = 2b² forces both to be even. Those two facts can’t live together, so something has to give. The only thing we chose freely was our starting assumption, so that’s what collapses.

It also shows why the trick doesn’t work for √4. Try it with a² = 4b² and you just get a = 2b, with no contradiction anywhere. That makes sense, because √4 = 2 is perfectly rational. What makes 2 special is that it’s a prime appearing just once on the right-hand side.

The result is also very old. Ancient Greek mathematicians discovered that the diagonal of a square can’t be expressed as a fraction of its side, and the argument above is the modern form of that discovery. It changed the way people thought about numbers.

Common mistakes

Students tend to trip over the same few things:

Forgetting “lowest terms.” Without it there’s no contradiction, so state it right at the start.
Skipping why an even a² means an even a. Examiners usually want to see the odd-number check from Step 5, even if it’s only one sentence.
Treating the calculator’s decimal as evidence. A long, random-looking decimal proves nothing. Only logic settles this.
Stopping too early. Showing a is even is only half the job. You still need b to be even to get the clash.
Writing “this is impossible” without saying why. Spell out what clashes: a and b share a factor of 2, but they were supposed to share none.

Other ways to prove it

The method above is the most popular, but it’s not the only one.

Prime factorisation. Every whole number greater than 1 breaks into primes in exactly one way. In a² = 2b², count how many times the prime 2 appears on each side. On the left, a² has an even number of 2s, because squaring doubles every exponent. On the right, 2b² has an odd number. An even count can’t equal an odd count, so the equation fails. The same idea works neatly for √3, √5, √6 and others.

The geometric method. Picture the smallest whole-number square whose area is exactly double that of another whole-number square. A clever construction shows you can always find a smaller pair, which contradicts the word “smallest.” This idea is known as infinite descent.

Different roads, same destination. That’s one of the nice things about maths.

Conclusion

Learning to prove that √2 is irrational is a small step that teaches a big lesson: in maths, we don’t trust patterns or approximations, we trust reasoning. Once this proof feels comfortable, you’ll start to see that the proofs for the square roots of other primes follow the same pattern.

Here’s my suggestion: close this page, grab a blank sheet of paper, and write out the ten steps in your own words. If you can do it without peeking, you’ve really got it.

Frequently Asked Questions

Is root 2 a real number?

Yes. Every irrational number is real, so √2 has its own spot on the number line, at roughly 1.414.

Can I use the same proof for root 3?

Yes. Swap 2 for 3 and use the fact that if a² is divisible by 3, then a is too. The rest of the argument follows the same pattern.

Is 1.41421356 the exact value of root 2?

No, it’s just a rounded approximation. The true decimal never ends and never falls into a repeating pattern.

Why can we trust proof by contradiction?

In logic, a statement is either true or false. If assuming it’s true leads to something impossible, it can’t be true.