標題:
Pure Mathematics - Proof
發問:
1. Determine whether the following statements are correct. Write down the converse of these statements and prove/disprove them.(a) If a and b are even integers, then a + b is an even integer.(b) If a, b and c are real numbers such that a + b = c, then (a+b)^2 = c^2.2. Prove that (a) if n is a positive integer... 顯示更多 1. Determine whether the following statements are correct. Write down the converse of these statements and prove/disprove them. (a) If a and b are even integers, then a + b is an even integer. (b) If a, b and c are real numbers such that a + b = c, then (a+b)^2 = c^2. 2. Prove that (a) if n is a positive integer and n^2 is divisible by 3, then n is divisible by 3; (b) sqrt (3) is irrational. 更新: Thanks you two answer my questions! ^^ To myisland8132: for 2(a), did you reverse the question? as the question asks n^2 is divisible by 3, not n is divisible by 3. To 雞尾包: for 2(a), did you use the method of contrapositive? i don't know the chinese name of contrapositive... 更新 2: To myisland8132 again: Is your method in 2(a) contradiction? Or just contradiction = contrapositive? Sorry for asking some silly questions.
最佳解答:
1. Determine whether the following statements are correct. Write down the converse of these statements and prove/disprove them. (a) If a and b are even integers, then a + b is an even integer. (b) If a, b and c are real numbers such that a + b = c, then (a+b)^2 = c^2. (a) The statement is true Since a and b are even integers, let a=2m, b=2n then a + b=2(m+n) is an even integer The converse statement is If a + b is an even integer, then a and b are even integers The statement is false for example 1+3=4 but 1 and 3 are odd integers (b) The statement is true Since a+b=c (a+b)^2=(a+b)(a+b)=c*c=c^2 The converse statement is if (a+b)^2 = c^2 then a + b = c The statement is false for example (-2)^2=2^2 but 2 is not equal to -2 2 (a) Using contradiction Assume that n is not divisible by 3 Then n=3k+1 or 3k+2 if n=3k+1 n^2=9k^2+6k+1 which is not divisible by 3, contradict the hypothesis. if n=3k+2 n^2=9k^2+12k+4 which is also not divisible by 3, contradict the hypothesis. So n should equal to 3k, n is divisible by 3. (b) Assume √3 is rational let √3=p/q (where p and q are coprime) 3=p^2/q^2 3q^2=p^2 So 3| p^2 that is 3|p let p=3k, then 3q^2=9k^2 q^2=3k^2 So 3|q^2 that is 3|q but this is a contradiction since we assume that p and q are coprime We just prove that √3 is irrational 2007-09-17 22:30:35 補充： 條題目完全不是數學歸納法啦 2007-09-18 15:29:12 補充： 2(a)無問題已知n^2 is divisible by 3﹐想證n is divisible by 3策略是假設n is not divisible by 3 (即n=3k+1 or 3k+2)﹐再推出這2個可能性都會得到n^2 is not divisible by 3, 與已知條件矛盾。因而可得出n is divisible by 3 2007-09-18 15:33:47 補充： 2(a)用的方法(和雞尾包一樣)正是contraposition
其他解答:
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(a) If a and b are even integers, then a + b is an even integer. let a = 2m, b = 2n a + b = 2m + 2n = 2(m+n) is an even integer statement are correct ------------------------------------------------------------ (b) If a, b and c are real numbers such that a + b = c, then (a+b)^2 = c^2. consider (a+b)^2 = (a+b)(a+b) = (c)(c) = c^2 statement are correct ---------------------------------------------- (a) if n is a positive integer and n^2 is divisible by 3, then n is divisible by 3; let n is not divisible by 3 n = 3k + 1 or n = 3k + 2 when n = 3k + 1, n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = (3)(3k^2 + 2k) + 1 三 1 mod 3 when n = 3k + 2, n^2 = (3k + 2)^2 = 9k^2 + 12k + 4 = (3)(3k^2 +4k + 1) + 1 三 1 mod 3 呢個係反正法, 兩個情況都有予盾, 所以 if n^2 is divisible by 3, then n is divisible by 3 ------------------------------------------------- (b) sqrt (3) is irrational suppose sqrt (3) is rational let sqrt (3) = p/q, p and q are integers q sqrt (3) = p p - q sqrt (3) = 0 imply p = q = 0 有予盾 又一反正法, 所以 sqrt (3) is irrational 2007-09-17 22:37:28 補充： 慢左, 不過請多多支持
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