標題:
Discrete Math 問題
發問:
1) (P-->Q)^(R-->S) PvR --------------------- so QvS 2) Prove that m^3 + 2n^2 = 36 has no solution in positive integers. 呢2條點 proofs呀 唔該
最佳解答:
1 From (P-->Q)^(R-->S) and PvR is true => P-->Q...(1), R-->S...(2), PvR...(3) is true If P is true, then Q should be true by (1) If P is False. Then R should be true by (3) and so S should be true by (2) So QvS should be true 2 Consider, m^3 + 2n^2 = 36. If there are positive integers m and n satisfy this equation, then m should be less than 4. when m=1, 36-m^3=35=> contradiction when m=2,36-m^3=28=>n^2=14 contradiction when m=3,36-m^3=9=> contradiction So, m^3 + 2n^2 = 36 has no solution in positive integers
其他解答:
Discrete Math 問題
發問:
1) (P-->Q)^(R-->S) PvR --------------------- so QvS 2) Prove that m^3 + 2n^2 = 36 has no solution in positive integers. 呢2條點 proofs呀 唔該
最佳解答:
1 From (P-->Q)^(R-->S) and PvR is true => P-->Q...(1), R-->S...(2), PvR...(3) is true If P is true, then Q should be true by (1) If P is False. Then R should be true by (3) and so S should be true by (2) So QvS should be true 2 Consider, m^3 + 2n^2 = 36. If there are positive integers m and n satisfy this equation, then m should be less than 4. when m=1, 36-m^3=35=> contradiction when m=2,36-m^3=28=>n^2=14 contradiction when m=3,36-m^3=9=> contradiction So, m^3 + 2n^2 = 36 has no solution in positive integers
其他解答:
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