Pure Mathematics - Proof

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

(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
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(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
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(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
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(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