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.
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- Suppose (assume for now) that p/q, in lowest terms, squares to 2. Here p is the top and q is the bottom.
- Then p² = 2q², so p² is even.
- An even square means p is even.
- Write p = 2m, where m is an integer. Then q² = 2m², so q is even.
- So p and q are both even.
- But lowest terms says they share no factor above 1. That is a clash: two facts that cannot both be true.
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.