急!!!!!! pure math 嘅proof

用戶95851

2009-09-15 6:54 pm

1. A,B,C are 3 sets such that A∪C = B∪C and C﹨A = C﹨B
prove that A= B

2. By completing the square, prove that x^2 + xy + y^2 = 0 <=> x=0 and y=0

3a. Suppose p and q are integers satisfying p^3 + p(q^2) + q^3 =0
i) prove that p is even if and only if q is even.
ii) Using (i), deduce that p and q are both even
3b. Hence show that if x is a real number for which x^3 + x + 1=0

更新1:

1. A,B,C are 3 sets such that A∪C = B∪C and C﹨A = C﹨B prove that A= B

更新2:

3b. Hence show that if x is a real number for which x^3 + x + 1=0 then x is irrational.

更新3:

myisland8132::: 我補充左3b題目係問x is irrational 我睇唔明第一題ge proof... 3aii 點做?

更新4:

第一題我明明地na... 不過第一行ge Remember that C﹨A=C∩A' 有咩用? 仲有... 點樣唸用 if x∈A but x /∈ C 同埋 if x∈A and x ∈ C 黎prove?

更新5:

(II) otherwise both p and q are odd From p^3+pq^2+q^3=p(p^2+q^2)+q^3 we know that this number is odd which is a contradiction since 0 is even. 係咪用prove by contradition? p and q are both even 嘅negation 唔係 p or q is odd 嗎?

回答 (1)

myisland8132

2009-09-15 10:06 pm

✔ 最佳答案

1 Remember that C﹨A=C∩A'
if x∈A but x /∈ C, then since x∈ A∪C = B∪C => x∈ B
if x∈A and x ∈ C, then x/∈C﹨A= C﹨B => x∈ B
So A ⊂ B
Similarly, we can show that B ⊂ A. So A=B
2 => if x^2 + xy + y^2 = 0
x^2+y^2+[(x+y)^2-(x-y)^2]/4=0
Since x^2,y^2,(x+y)^2,(x-y)^2>=0
We conclude that x=y=0
<= Sub. x=y=0 into x^2 + xy + y^2
3(a) => if p is even, p=2a, p^3 + p(q^2) + q^3 =0
8a^3+2a(q^2)+q^3=0
since 8a^3+2a(q^2) is an even number. We conclude that q^3 is even and so q is even
<= if q is even, q=2b, p^3 + p(q^2) + q^3 =0
p^3+p(4b^2)+8b^3=0
since p(4b^2)+8b^3 is even. We conclude that p^3 is even and so p is even.
3b Here q=1 and p=x, so x should not be even by the result of (a)

2009-09-15 22:05:09 補充：
第一題是試出來的﹐而且分CASE是數學上常數的方法。 C﹨A=C∩A' 是幫自己嘗試找關係﹐在這題其實用不上

2009-09-15 22:10:17 補充：
(II) otherwise both p and q are odd From p^3+pq^2+q^3=p(p^2+q^2)+q^3
we know that this number is odd which is a contradiction since 0 is even.

(III) if x is rational, let x=p/q (in simplest form), then x^3+x+1=0
p^3+pq^2+q^3=0=>x=2m/2n which is a contradiction.

2009-09-17 12:39:13 補充：
I only use (i) p is even if and only if q is even bor

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