题目
Problem
(i) Prove by counter example that the statement
“If n is a prime number then 3n+2 is also a prime number.”
is false.
(2)
(ii) Use proof by exhaustion to prove that if m is an integer that is not divisible by 3, then
m2−1
is divisible by 3
(4)
题目中文翻译
(i) 用反例证明以下命题是错误的:(2)
“如果 n 是质数,那么 3n+2 也是质数。”
(ii) 使用穷尽法证明:如果 m 是不能被 3 整除的整数,那么 m2−1 能被 3 整除。(4)
解答
(i)
解法一
思路
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反例只需要找一个 prime n,使 3n+2 不是 prime。选 n=5 很方便。
答题过程
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Take
n=5.
This is prime. But
35+2===243+22455×49.
So 35+2 is not prime. Therefore, the statement is false.
(ii)
解法一
思路
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一个整数不被 3 整除,只可能形如 3k+1 或 3k+2。proof by exhaustion 就是把这两种情况都检查完。
答题过程
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If m is not divisible by 3, then either
m=3k+1
or
m=3k+2,
where k is an integer.
Case 1:
m=m2−1====3k+1(3k+1)2−19k2+6k+1−19k2+6k3(3k2+2k).
So m2−1 is divisible by 3.
Case 2:
m=m2−1====3k+2(3k+2)2−19k2+12k+4−19k2+12k+33(3k2+4k+1).
So m2−1 is divisible by 3.
In all possible cases where m is not divisible by 3, m2−1 is divisible by 3.
解法二
思路
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先将 m2−1 因式分解。仍穷尽两个余数类别,分别找出可被 3 整除的那个因子。
答题过程
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If m is not divisible by 3, then m=3k+1 or m=3k+2, where k∈Z.
Factorise:
m2−1=(m−1)(m+1).
If m=3k+1, then
m2−1==(3k)(3k+2)3[k(3k+2)].
If m=3k+2, then
m2−1==(3k+1)(3k+3)3[(3k+1)(k+1)].
In each case the bracket is an integer, so m2−1 is divisible by 3. These exhaust all integers not divisible by 3, hence the result is proved.