Solving a Floor Value System

This video is about a floor value system
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EXPLORE:

[Qermaq]

Never worked with fractionals. Was very insightful!

[pwmiles]

With a slight change of notation, put x1=n1+s1 etc with n1 integer, 0<=s1<1
The conditions can be written as a linear system
s1+s3 = 3.9 - n1 - n2
s2+s1 = 3.5 - n2 - n3
s3+s2 = 2 - n3 - n1

Solve
s1 = 2.7 - n2
s2 = 0.8 - n3
s3 = 1.2 - n1

You can read off the unique solution
s1=0.7 n2=2
s2=0.8 n3=0
s3=0.2 n1=1

[ramaprasadghosh]

Adding one gets
x + y + z = 4.7
so floor (z) + fraction (y) = 0.8
floor (x) + fraction (z) = 1.2
floor (y) + fraction (x) = 2.7
so x = 1.7, y = 2.8, z = 0.2

[nicogehren]

great solution sir thank u keep it up!

[IlTrojo]

Even more simmetrically, let rho = floor(x) + fractional(z), theta = floor(y) + fractional(x) and phi = floor(z) + fractional(y) and work from there.

[eu]

But a = floor(a) +{a} only if a is positive right? And if a is negative we should use ceil(a) instead of floor. Doesn't this give us other sollutions or an unsolvable problem since the problem doesn't say if x, y and z are positive?

[srijanbhowmick]

DON'T CLICK READ MORE IF YOU DON'T WANT SPOILERS :----


























Add all three equations, you will get the following :----
x + y + z = 4.7
Now subtract the 3 given equations from the above equation and you will get the following 3 equations :---
[z] + {y} = 0.8
[x] + {z} = 1.2
[y] + {x} = 2.7
Using the 2 facts, that 0 <= {p} < 1 and [p] is an integer, you will get the following :---
[z] = 0 ; {z} = 0.2 => z = 0.2
[x] = 1 ; {x} = 0.7 => x = 1.7
[y] = 2 ; {y} = 0.8 => y = 2.8
Q.E.D.