‹ Class 8 · Ch 9
The Baudhāyana-Pythagoras Theorem · Principle 7 of 17

Is √2 a fraction?

If √2 = m/n then 2n² = m². Count the 2s.

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NCERT: 2.3 Hypotenuse of an Isosceles Right Triangle

Think

Square numbers and primes

Could √2 be a fraction mn, with m and n counting numbers? If √2 = mn, squaring gives 2 = m²n² and so 2n² = m².

Look at a square number: 36 = 2 × 2 × 3 × 3. The prime 2 appears twice and the prime 3 appears twice.

In the prime factors of any square number, how often does each prime appear?

What this lesson covers

The idea

√2 cannot be written as m/n with counting numbers m and n: 2n² = m² would have the prime 2 an odd number of times on the left and an even number of times on the right.

Square numbers and primes

Could √2 be a fraction m/n, with m and n counting numbers? If √2 = m/n, squaring gives 2 = m²/n² and so 2n² = m².

Look at a square number: 36 = 2 × 2 × 3 × 3. The prime 2 appears twice and the prime 3 appears twice.

In the prime factors of any square number, how often does each prime appear?

  • an even number of times
  • an odd number of times
  • it can be anything

Count the 2s

If √2 were m/n, then 2 × n × n would be equal to m × m. Choose n and m. The 2s are orange. How many 2s are on each side?

Odd against even

In the prime factorization of a square number, each prime occurs an even number of times. In 2n² = m² the prime 2 would occur an odd number of times on the left and an even number of times on the right. This is impossible. So √2 cannot be written as a fraction.

Why odd on the left? In n × n the 2s come in pairs. The extra 2 in front has no partner.

This proof was given by Euclid in his book *Elements* (about 300 BCE).

Notes

If √2 = m/n, then 2n² = m². The prime 2 would occur an odd number of times on the left and an even number of times on the right. That is impossible: √2 is not a fraction.

Check yourself

Which of these can be a square number?

Take n = 6. Then 2 × 6 × 6 = 72. How many 2s are in the prime factors of 72?

Answer: 3

72 = 2 × 2 × 2 × 3 × 3 has three 2s: one from each 6 and the extra 2 in front. Three is odd.

Why can 2n² = m² never be true for counting numbers?

Take m = 10. Then m × m = 100. How many 2s are in the prime factors of 100?

Answer: 2

100 = 2 × 2 × 5 × 5 has two 2s. Two is even, as it must be for a square number.

  • 2 × 2 × 2 × 3 × 3. The prime 2 appears 3 times, an odd number of times. A square number needs every prime an even number of times. (It is 72, between 8² = 64 and 9² = 81.)
  • 2 × 2 × 3 × 3 × 3 × 3 — correct. Yes! 2 appears twice and 3 appears four times. It is 324 = 18 × 18.
  • 2 × 3 × 3 × 5 × 5. The prime 2 appears once, an odd number of times. (It is 450, not a square.)
  • The number of 2s is odd on the left and even on the right — correct. Yes! 2 × n × n has one extra 2 that is not paired, and m × m has its 2s in pairs.
  • The left side is always bigger. Not always: n = 2, m = 3 gives 8 on the left and 9 on the right. The real reason is the count of 2s.
  • m must be an odd number. m can be even, like 8. The problem is the count of 2s: odd on the left, even on the right.
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