Solving the Equation: x+[2x]+[3x]=7

  • Context: MHB 
  • Thread starter Thread starter solakis1
  • Start date Start date
Click For Summary
SUMMARY

The equation x + [2x] + [3x] = 7 can be analyzed by categorizing real numbers into distinct classes based on their integer properties. The discussion identifies five specific classes: integers, non-integers where 2x is an integer, non-integers where 3x is an integer, non-integers where neither 2x nor 3x is an integer but 6x is, and all other real numbers. The tool Desmos was referenced, indicating that the solution x = 4/3 does not satisfy the equation, as it yields a left-hand side of 8 instead of the required 7.

PREREQUISITES
  • Understanding of floor functions in mathematics
  • Familiarity with real number classifications
  • Basic algebraic manipulation skills
  • Experience using graphing tools like Desmos
NEXT STEPS
  • Explore the properties of floor functions in mathematical equations
  • Research integer and non-integer classifications in real analysis
  • Learn how to use Desmos for solving piecewise functions
  • Investigate similar equations involving floor functions and their solutions
USEFUL FOR

Mathematicians, students studying algebra, and anyone interested in solving equations involving floor functions and real number classifications.

solakis1
Messages
407
Reaction score
0
An easy one:

x+[2x]+[3x]=7
 
Mathematics news on Phys.org
Yes, that is an easy problem! Why did you post it?

I would start by dividing the real numbers into classes:
1) The set of integers.
2) The set of non-integer, x, such that 2x is an integer
3) The set of non-integers, x, such that 2x is not an integer but 3x is.
4) The set of non-integers, x, such that neither 2x nor 3x is an integer but 6x is.
5) The set all other real numbers.
 
Because RHS is integer so LHS is integer

As $\lfloor 2x \rfloor$ and $\lfloor 3x \rfloor$ are integers so x is integer so $\lfloor 2x \rfloor = 2x $ and $\lfloor 3x \rfloor = 3x$

so x + 2x + 3x = 6x = 7 so $x = \frac{7}{6}$ which is not integer so NO solution
 
Beer soaked ramblings follow.
solakis said:
An easy one:

x+[2x]+[3x]=7
Desmos somehow gives the impression that $x + \lfloor 2x \rfloor + \lfloor 3x \rfloor = 7$ is basically the line $x = \frac{4}{3}$.
https://www.desmos.com/calculator/stz3o2wn2h
 
The answer given by Kaliprasad is the right one 7\6 is not an integer
4/3 is not a solution of the above equation LHS is 8 RHS Is 7
 
Last edited:

Similar threads

MHB Absolute Value Equation |3x - 2|/|2x - 3| = 2
  • · Replies 10 ·
Replies
10
Views
2K
MHB Equation 2: Prove that ##x^2+2x\sqrt x+3x+2\sqrt x+1=0##
  • · Replies 4 ·
Replies
4
Views
2K
MHB Solving $\tan^2x + \tan^2{2x} + \cot^2{3x} = 1$ in R
  • · Replies 6 ·
Replies
6
Views
1K
MHB Solve √(x^2+3x+7)-√(x^2-3x+9)+2=0
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Using an array to solve a system of equations
  • · Replies 7 ·
Replies
7
Views
2K
MHB Solve Inequality & Simul Eqns: Alg Workings Q (a,b,c)
  • · Replies 1 ·
Replies
1
Views
1K
MHB Proving Equivalence of f(x) and g(x)
  • · Replies 2 ·
Replies
2
Views
1K
MHB Solve System of Inequalities: 2x+3y-4, 3x-4y+5
  • · Replies 8 ·
Replies
8
Views
2K
MHB Quadratic equations ( solving for zeros/roots) trickey
  • · Replies 1 ·
Replies
1
Views
1K
MHB Prove Trig Identity: $\sin^7 x=\dfrac{35\sin x-21\sin 3x+7\sin 5x-\sin 7x}{64}$
  • · Replies 1 ·
Replies
1
Views
1K