[數撚圍爐區] 一唔開post就成班白咭衝哂出去派膠 (82)
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If p and q are natural numbers, then (p/q)^2≠2 aka p^2 ≠ 2q^2
Trivially this is true for odd p by mod 2 (case 1)
Trivially this is true for even p and odd q by mod 4 (case 2)
For even p and q, let p’=p/gcd(p,q) and q’=q/gcd(p,q)
- for odd p’, by case 1 (p’/q’)^2≠2 => (p/q)^2≠2
- for even p’ and odd q’, by case 2 (p’/q’)^2≠2 => (p/q)^2≠2
Thus the contrapositive: (p/q)^2=2 => p and q cannot both be natural numbers
Trivially this is true for odd p by mod 2 (case 1)
Trivially this is true for even p and odd q by mod 4 (case 2)
For even p and q, let p’=p/gcd(p,q) and q’=q/gcd(p,q)
- for odd p’, by case 1 (p’/q’)^2≠2 => (p/q)^2≠2
- for even p’ and odd q’, by case 2 (p’/q’)^2≠2 => (p/q)^2≠2
Thus the contrapositive: (p/q)^2=2 => p and q cannot both be natural numbers
研究C++可以證明到啲咩
