→ JCwbear :假設根號2是有理數,可以寫成q/p gcd(q,p)=1 03/31 11:49
→ JCwbear :之後在證出gcd(q,p)>1 矛盾 03/31 11:52
推 suhorng :設 √2 = q/p \in Q, (p,q)=1 03/31 19:15
→ suhorng :then 2 = q^2/p^2 => 2p^2 = q^2 => 2 | q^2 => 2|q 03/31 19:15
→ suhorng :設 q = 2k. 得 2p^2 = q^2 = 4k^2 => p^2 = 2k^2 03/31 19:15
→ suhorng :與上面同理, 推出 2 | p. 但是這與 (p,q)=1 矛盾. 03/31 19:16