Greatest integer and factorial part equation

Calculation of Real (x,y,z)(x,y,z)$ in

x[x]+z{z}−y{y}=0.16x[x]+z\{z\}-y\{y\} = 0.16

y[y]+x{x}−z{z}=0.25y[y]+x\{x\}-z\{z\} = 0.25

z[z]+y{y}−x{x}=0.49z[z]+y\{y\}-x\{x\} = 0.49

where [x]=[x] = Greatest Integer of xx and {x}=\{x\} = fractional part of xx

#Algebra #MathProblem #Math

Note by Jagdish Singh
7 years, 6 months ago

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Comments

(POST#1)

  • Let [x],{x},[y],{y},[z],{z}[x],\{x\},[y],\{y\},[z],\{z\} be a,b,c,d,e,fa,b,c,d,e,f respectively.
  • {a2+ab+ef+f2−cd−d2=0.16—(1)c2+cd+ab+b2−ef−f2=0.25—(2)e2+ef+cd+d2−ab−b2=0.49—(3)\left\{\begin{array}{lclr}a^2+ab+ef+f^2-cd-d^2&=&0.16&\mbox{---(1)}\\c^2+cd+ab+b^2-ef-f^2&=&0.25&\mbox{---(2)}\\e^2+ef+cd+d^2-ab-b^2&=&0.49&\mbox{---(3)}\end{array}\right.
  • (1)+(2)+(3): a2+ab+c2+cd+e2+ef=0.9—(4)a^2+ab+c^2+cd+e^2+ef = 0.9 \mbox{---(4)}
  • (4)-(2): a2−b2+(e+f)2=0.65—(5)a^2-b^2+(e+f)^2=0.65 \mbox{---(5)}
  • (4)-(3): c2−d2+(a+b)2=0.41—(6)c^2-d^2+(a+b)^2=0.41 \mbox{---(6)}
  • (4)-(1): e2−f2+(c+d)2=0.74—(7)e^2-f^2+(c+d)^2=0.74 \mbox{---(7)}
  • Since 0≤b2,d2,f2<10\le b^2,d^2,f^2<1,
  • (5): a2+(e+f)2=0.65+b2<1.65a^2+(e+f)^2=0.65+b^2<1.65 ---(8)
  • (6): c2+(a+b)2=0.41+d2<1.41c^2+(a+b)^2=0.41+d^2<1.41 ---(9)
  • (7): e2+(c+d)2=0.74+f2<1.74e^2+(c+d)^2=0.74+f^2<1.74 ---(10)
  • Since 0≤b,d,f<10\le b,d,f<1,
  • (8): a2+e2<1.65a^2+e^2<1.65 ---(11)
  • (9): c2+a2<1.41c^2+a^2<1.41 ---(12)
  • (10): e2+c2<1.74e^2+c^2<1.74 ---(13)
  • Since aa, cc, and ee are integers, they can only be 00 or ±1\pm1 according to (11), (12) and (13).
  • Moreover, there can be at most one of aa, cc, or ee which is ±1\pm1, or else (11), (12) or (13) will not be satisfied.
  • However, of all three of them are zero, (4) will not be satisfied.
  • In the next post, I shall explore all the three cases.

Kenny Lau - 6 years, 6 months ago

(POST#2) - In post #1, we have discovered that one and only one of [x][x], [y][y] and [z][z] must be 11 or −1-1.

  • Case 1: [x]=1[x]=1,[y]=0[y]=0,[z]=0[z]=0
  • (1): x+z2−y2=0.16x+z^2-y^2=0.16 ---(4)
  • (2): x(x−1)−z2=0.25x(x-1)-z^2=0.25 ---(5)
  • (3): y2−x(x−1)=0.49y^2-x(x-1)=0.49 ---(6)
  • (4)+(5)+(6): x=0.9x=0.9, which contradicts the prerequisite.
  • Case 2: [x]=−1[x]=-1,[y]=0[y]=0,[z]=0[z]=0
  • (1): −x+z2−y2=0.16-x+z^2-y^2=0.16 ---(4)
  • (2): x(x+1)−z2=0.25x(x+1)-z^2=0.25 ---(5)
  • (3): y2−x(x+1)=0.49y^2-x(x+1)=0.49 ---(6)
  • (4)+(5)+(6): x=−0.9x=-0.9 ---(7)
  • (7) into (6): y=0.4y=\sqrt{0.4}
  • (7) into (5): z=−0.34z=\sqrt{-0.34}, which is not real.
  • Case 3: [x]=0[x]=0,[y]=1[y]=1,[z]=0[z]=0
  • (1): z2−y(y−1)=0.16z^2-y(y-1)=0.16 ---(4)
  • (2): y+x2−z2=0.25y+x^2-z^2=0.25 ---(5)
  • (3): y(y−1)−x2)=0.49y(y-1)-x^2)=0.49 ---(6)
  • (4)+(5)+(6): y=0.9y=0.9, which contradicts the prerequisite.
  • In fact, I expect case 5 to fail as well, because it would yield z=0.9z=0.9.
  • Therefore, I will skip case 5.
  • Case 4: [x]=0[x]=0,[y]=−1[y]=-1,[z]=0[z]=0
  • (1): z2−y(y+1)=0.16z^2-y(y+1)=0.16 ---(4)
  • (2): −y+x2−z2=0.25-y+x^2-z^2=0.25 ---(5)
  • (3): y(y+1)−x2=0.49y(y+1)-x^2=0.49 ---(6)
  • (4)+(5)+(6): y=−0.9y=-0.9 ---(7)
  • (7) into (6): x=±−0.56x=\pm\sqrt{-0.56}, which is not real.
  • Case 5: [x]=0[x]=0,[y]=0[y]=0,[z]=1[z]=1 (skipped)
  • Case 6: [x]=0[x]=0,[y]=0[y]=0,[z]=−1[z]=-1
  • (1): z(z+1)+y2=0.16z(z+1)+y^2=0.16 ---(4)
  • (2): x2−z(z+1)=0.25x^2-z(z+1)=0.25 ---(5)
  • (3): −z+y2−x2)=0.49-z+y^2-x^2)=0.49 ---(6)
  • (4)+(5)+(6): z=−0.9z=-0.9 ---(7)
  • (7) into (5): x=0.34x=\sqrt{0.34}
  • (7) into (4): y=0.25y=\sqrt{0.25}
  • x=0.34,y=0.5,z=−0.9x=\sqrt{0.34}, y=0.5, z=-0.9.

Kenny Lau - 6 years, 6 months ago
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