Floor Equation 6 Proven: Solvable for All $n\geq 1$

[solakis1]

Given:
x+[x]+[2x]+......[nx}
Prove that there exists an A such that the equation: x+[x]+[2x]+......[nx]=A has a solution for all $n\geq 1$

 

[jvicens]

Dear forum members,

I would like to provide a proof for the statement that there exists an A such that the equation x+[x]+[2x]+......[nx]=A has a solution for all n≥1.

First, let's define [x] as the greatest integer less than or equal to x. This means that [x] is a whole number, which we can represent as an integer. For example, [3.7] = 3, [5.2] = 5, and so on.

Now, let's consider the equation x+[x]+[2x]+......[nx]=A. We can rewrite this equation as x+[x]+[2x]+......[nx] = [x] + [2x] + ...... [nx] + (x - [x]) + ([2x] - [x]) + ...... ([nx] - [x]) = A.

From this, we can see that A is equal to the sum of [x], [2x], [3x], ..., [nx], plus the remainder of each term (x - [x], [2x] - [x], [3x] - [x], ..., [nx] - [x]). Since [x], [2x], [3x], ..., [nx] are all whole numbers, the sum of these terms will also be a whole number. Therefore, A must also be a whole number.

This means that for any value of n, we can always find a value of x such that the equation x+[x]+[2x]+......[nx]=A has a solution. For example, if n=3, we can choose x=2.5 and A=10, which satisfies the equation: 2.5+[2.5]+[5]+[7.5]=10.

In conclusion, since A is always a whole number and we can always find a value of x that satisfies the equation for any value of n, we can say that there exists an A that makes the equation x+[x]+[2x]+......[nx]=A a solution for all n≥1. This proves the statement. Thank you for reading.