Math Quickies

[jay_shark]

Here are two problems that have really short solutions. See if you can figure them out .

1) x1, x2, ... , xn are non-negative reals. Let s ={sum of all i<j} xixj. Show that at least one of the xi has square not exceeding 2s/(n2 - n).

2)Given any distinct real numbers x, y, show that we can find integers m, n such that mx + ny > 0 and nx + my < 0.

[jay_shark]

It should be 2s/(n^2-n) for problem A .

[jay_shark]

Solution 1) If this isn't true , then each term exceeds 2s/(n2-n) but there are only n(n-1)/2 terms . QED

Solution 2) if x>y , take m=1 , n=-1 then x-y>0 . If x<y , take m=-1 , n=1 then y-x>0 QED

Two problems that seem complicated are actually really simple . With a bit of patience , many difficult problems can be reduced to something trivial .

[jogsxyz]

Hard to read. 2+2 needs to support size.

Xi

[BruceZ]

[ QUOTE ]
Solution 1) If this isn't true , then each term exceeds 2s/(n2-n) but there are only n(n-1)/2 terms . QED

Solution 2) if x>y , take m=1 , n=-1 then x-y>0 . If x<y , take m=-1 , n=1 then y-x>0 QED

Two problems that seem complicated are actually really simple . With a bit of patience , many difficult problems can be reduced to something trivial .

[/ QUOTE ]

I did these the same way.

You post solutions too fast. You should wait at least 24 hours. 48 would be better.

[jay_shark]

Bruce , I believe you . You seem to solve everything .

I will however wait a bit longer to post solutions .