m******n
发帖数: 462 | 1 Positive real numbers: X1 Y1 X2 Y2 Q1 Q Q2;
X1>Y1 and X2>Y2(May not need this condition);
0
P1=X1*Q1/Y1;
P2=X2*Q2/Y2;
and P1
X=X1+X2;
Y=Y1+Y2;
P=X*Q/Y;
If all of above stay true, how can I verify "P1<=P<=P2" is not necessary?
Thanks. |
z****f
发帖数: 484 | 2 小学数学题，花5分钟就构造出来了。
【在 m******n 的大作中提到】
: Positive real numbers: X1 Y1 X2 Y2 Q1 Q Q2;
: X1>Y1 and X2>Y2(May not need this condition);
: 0: P1=X1*Q1/Y1;
: P2=X2*Q2/Y2;
: and P1: X=X1+X2;
: Y=Y1+Y2;
: P=X*Q/Y;
: If all of above stay true, how can I verify "P1<=P<=P2" is not necessary?
|
m******n
发帖数: 462 | 3
I have waited for one hour to see your next reply.
【在 z****f 的大作中提到】
: 小学数学题，花5分钟就构造出来了。
|
z****f
发帖数: 484 | 4 我找出来之后才回的帖。。。
X1 Y1 X2 YX Q1 Q: 10 1 2 1 3/8 1/2
then P< P1.
【在 m******n 的大作中提到】
: Positive real numbers: X1 Y1 X2 Y2 Q1 Q Q2;
: X1>Y1 and X2>Y2(May not need this condition);
: 0: P1=X1*Q1/Y1;
: P2=X2*Q2/Y2;
: and P1: X=X1+X2;
: Y=Y1+Y2;
: P=X*Q/Y;
: If all of above stay true, how can I verify "P1<=P<=P2" is not necessary?
|
m******n
发帖数: 462 | |
