The argument that no fraction can do it

Trying fractions one by one never finishes. This argument shows that every fraction fails. The lab tries each odd or even pair.

One fact first

An odd number squared is odd. An odd number is 2k + 1, where k is an integer. Its square is 4k² + 4k + 1. That equals 2(2k² + 2k) + 1, which is odd. So if a square is even, its number is even.

The argument

Take a fraction p/q, in lowest terms, that squares to 2. Here p and q are integers, and q is not 0.

• (p/q)² = 2, so p² = 2q².
• So p² is even, because it is 2 times q².
• So p is even, by the fact above. Write p = 2m, where m is an integer.
• Then 4m² = 2q², so q² = 2m².
• So q² is even, and q is even too.

Both p and q are even. So they share the factor 2. But lowest terms says they share no factor above 1. That is a clash: two facts that cannot both be true. So no such fraction exists. The argument works for any p/q, so it covers every fraction.

The same argument for 3

Suppose p/q, in lowest terms, squares to 3. Then p² = 3q², so 3 divides p². "3 divides p²" means p² is a multiple of 3. A prime that divides a square divides the number. So 3 divides p. Write p = 3m. Then 9m² = 3q², so q² = 3m². So 3 divides q. Both share a factor of 3. That is the same clash.

Only one step changes. For 2, we use odd and even. For 3, we use that 3 is a prime.

Try the lab to finish this lesson.