a^2 b^2 prove a b

題目出錯, 因為你無指名a, b是正數還是負數.
for example, (-3)^2 > (2)^2 (9>4) but -3<2

要令statement成立 a must be greater than 0

如果你想簡單些, assume 埋b是正數, 第一位答你的人就是你想要的答案.

但真正的數學是要consider b can be taken any value

Method 1:

a^2>b^2
a^2-b^2>b^2-b^2
a^2-b^2>0
(a-b)(a+b)>0


we have case (1) a-b>0 and a+b>0 or case (2) a-b < 0 and a+b<0
case (1): a>b and b>-a
ie a > lbl (absolute value of b)
so a > b (since a is positive)

case (2): b>a and b<-a
-a>b>a which is impossible (since a>0)

summing up the 2 cases a>b or no solution, which means a>b

Method 2: (I prefer most)
You should learn something about logic first.
Let x, y are the statements
If x, then y.
the same meaning as
If not y, then not x.

use this result,
That means we need to prove if a is less than or equal to b, then a^2 is less than or equal to b^2. And the original statement will be true.

The proof:

Suppose a is less than or equal to b,
because a is positve, so b is positive,
so we have a^2 is less than or equal to b^2.
so by the contrapositive result, if a^2>b^2, then a>b.

This proof is easy because I only need to write 4 sentence.

2006-12-01 18:48:12 補充：
myisland8132
的答案只是針對b>0的情況, 所以他的proof不是嚴謹.

a^2>b^2
a^2-b^2>0
(a+b)(a-b)>0
(a+b)>0
(a-b)>0
a>b

we should assume that
a and b both greater than 0
then
a^2>b^2
a^2-b^2>b^2-b^2
a^2-b^2>0
(a-b)(a+b)>0
since a+b>0
a-b should be greater than 0
that is
a-b>0
a>b