1. ## Equations

Hello , help please I need to solve those equations :

1) $\displaystyle [x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345$

2) $\displaystyle [x^2]=[x]^2$

3) $\displaystyle [\frac{x^2-2x}{3}]=\frac{x}{2}$

2. Originally Posted by linda2005
Hello , help please I need to solve those equations :

1) $\displaystyle [x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345$

2) $\displaystyle [x^2]=[x]^2$

3) $\displaystyle [\frac{x^2-2x}{3}]=\frac{x}{2}$
What is being represented by the square brackets?

3. Originally Posted by Mush
What is being represented by the square brackets?
whole part of number

$\displaystyle \left[ x \right]or\left\lfloor x \right\rfloor = \max \left\{ {\left. {n \in \mathbb{Z}} \right|{\text{ }}n \leqslant x} \right\}.$

4. Originally Posted by DeMath
whole part of number

$\displaystyle \left[ x \right]or\left\lfloor x \right\rfloor = \max \left\{ {\left. {n \in \mathbb{Z}} \right|{\text{ }}n \leqslant x} \right\}.$
Ah, right. Are you familiar with the following property, for the first problem:

$\displaystyle \lfloor kx \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{k}\right\rfloor +\dots+\left\lfloor x+\frac{k-1}{k}\right\rfloor$

In other words:

$\displaystyle \lfloor 2x \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{2}\right\rfloor$

$\displaystyle \lfloor 4x \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{4}\right\rfloor+ \left\lfloor x+\frac{2}{4}\right\rfloor + \left\lfloor x+\frac{3}{4}\right\rfloor$

Followed by the rule that:

$\displaystyle \lfloor x + n \rfloor = \lfloor x \rfloor + n$

Hence:

$\displaystyle \lfloor 2x \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x\right\rfloor +\frac{1}{2}$ etc...

5. For the 2nd one, think of it like this:

$\displaystyle x = k + p$ where k is an integer, and p is a non-integer between 0 and 1.

In mathematical terms: $\displaystyle k \in Z \text{, } p \notin Z \text{ and } 0 < p<1$

Hence:

$\displaystyle \lfloor (k + p)^2 \rfloor = \lfloor k + p \rfloor^2$

$\displaystyle \lfloor k^2+ 2pk+ p^2 \rfloor = \lfloor k + p \rfloor^2$

On the RHS, we know that k is the integer part of the number so:

$\displaystyle \lfloor k^2 +2pk+ p^2 \rfloor = k^2$

In other words, the NON-integer part of the LHS must be $\displaystyle 2pk+p^2$

So now you need to come up with values of k and p such that:

$\displaystyle k \in Z \text{, } p \notin Z \text{, } 0 < p < 1\text{ and } 0 < 2pk+p^2 < 1$

For example, p = 0.1, k = 1 is a solution. (x = 1.1, in other words).

6. A fairly obvious solution to the last one is x = 0.

7. Originally Posted by Mush
A fairly obvious solution to the last one is x = 0.
hello thanks Mush , but with x=4 it's ok? so you are wrong

8. Originally Posted by linda2005
hello thanks Mush , but with x=4 it's ok? so you are wrong
Eh? x = 4 is a solution to the equation, yes. But so is x = 0. I am not wrong, there are several solutions.

9. i'm not sure about 1) but i guess there is no solution. my solutions to 2) and 3):

Originally Posted by linda2005

2) $\displaystyle [x^2]=[x]^2$
clearly any integer is a solution. so we'll assume that $\displaystyle x \notin \mathbb{Z}.$ let $\displaystyle [x]=n.$ then $\displaystyle n < x < n+1$ and $\displaystyle [x^2 - n^2]=0,$ which is equivalent to $\displaystyle 0 < x^2 - n^2 < 1. \ \ \ \ (1)$

case 1. $\displaystyle x < 0$: in this case, since $\displaystyle x > n,$ we'll have: $\displaystyle x^2 < n^2.$ but from (1) we have $\displaystyle x^2 > n^2.$ this contradiction means there's no solution in this case.

case 2. $\displaystyle x > 0$: then the conditions $\displaystyle n < x < n+1$ and (1) are equivalent to $\displaystyle n < x < \sqrt{n^2 + 1}.$ conversely if $\displaystyle n < x < \sqrt{n^2 + 1},$ for some positive integer $\displaystyle n,$ then

$\displaystyle n < x < n+1$ and $\displaystyle n^2 < x^2 < n^2 + 1.$ thus $\displaystyle [x]=n$ and $\displaystyle [x^2]=n^2=[x]^2.$

so the complete set of solutions of the equation is: $\displaystyle \bigcup_{n \in \mathbb{N} \cup \{0 \}} (n , \sqrt{n^2 + 1}) \cup \mathbb{Z}.$

3) $\displaystyle [\frac{x^2-2x}{3}]=\frac{x}{2}$
the equation is equivalent to these two conditions together: $\displaystyle \frac{x}{2} \in \mathbb{Z}$ and $\displaystyle \frac{x^2 - 2x}{3} - 1 < \frac{x}{2} \leq \frac{x^2 - 2x}{3}. \ \ \ \ (2)$

let $\displaystyle x = 2n,$ where $\displaystyle n \in \mathbb{Z}.$ solving the inequalities in (2) for $\displaystyle n$ gives you: $\displaystyle \left[\frac{7-\sqrt{97}}{8}, 0 \right] \cup \left[\frac{7}{4}, \frac{7 + \sqrt{97}}{8} \right],$ which gives us $\displaystyle n = 0,2.$ thus $\displaystyle x=2n=0,4.$

10. Hello, Thanks you NonCommAlg Your solutions is , so you can help me with 1) thanks

11. Originally Posted by linda2005
Hello, Thanks you NonCommAlg Your solutions is , so you can help me with 1) thanks
as i said i'm not sure about 1) right now but i have some reasons to believe that the equation has no solution. so tell me, if you know the answer, am i right?

12. There is no solution

Using the following two properties:

$\displaystyle \lfloor kx \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x+\frac{1}{k}\right\rfloor +\dots+\left\lfloor x+\frac{k-1}{k}\right\rfloor$

$\displaystyle \lfloor x + n \rfloor = \lfloor x \rfloor + n$

It can be deduced that:

$\displaystyle \displaystyle \lfloor kx \rfloor = k \lfloor x \rfloor + \sum_{n = 1}^{k-1} \frac{n}{k}$

Hence:

$\displaystyle \displaystyle \lfloor 2x \rfloor = 2\lfloor x \rfloor + \sum_{n = 1}^{1} \frac{n}{2} =2\lfloor x \rfloor + \frac{1}{2}$

$\displaystyle \displaystyle \lfloor 4x \rfloor = 4 \lfloor x \rfloor + \sum_{n = 1}^{3} \frac{n}{4} = 4\lfloor x \rfloor + \frac{3}{2}$

And in general $\displaystyle \displaystyle \lfloor kx \rfloor = k\lfloor x \rfloor + \frac{k-1}{2}$

Continuing like this, the equation can be reduced to:

$\displaystyle 63 \lfloor x \rfloor + \frac{1+3+7+15+31}{2} = 12345$

$\displaystyle 63\lfloor x \rfloor + \frac{57}{2} = 12345$

$\displaystyle 63\lfloor x \rfloor = \frac{24633}{2}$

$\displaystyle \lfloor x \rfloor = \frac{391}{2}$

It is non-integer, and so there are no solutions.

13. Originally Posted by Mush

It can be deduced that:

$\displaystyle \displaystyle \lfloor kx \rfloor = k \lfloor x \rfloor + \sum_{n = 1}^{k-1} \frac{n}{k}$
this is not true! why? see below for example:

Hence:
$\displaystyle \displaystyle \lfloor 2x \rfloor = 2\lfloor x \rfloor + \sum_{n = 1}^{1} \frac{n}{2} =2\lfloor x \rfloor + \frac{1}{2}$
$\displaystyle \lfloor 2x \rfloor - 2\lfloor x \rfloor$ is an integer but $\displaystyle \frac{1}{2}$ is not. so they can never be equal. the function $\displaystyle \lfloor 2x \rfloor - 2\lfloor x \rfloor$ is always either 0 or 1.

14. Originally Posted by NonCommAlg
this is not true! why? see below for example:

$\displaystyle \lfloor 2x \rfloor - 2\lfloor x \rfloor$ is an integer but $\displaystyle \frac{1}{2}$ is not. so they can never be equal. the function $\displaystyle \lfloor 2x \rfloor - 2\lfloor x \rfloor$ is always either 0 or 1.
Oh very, true, my apologies.

This is because the 2nd property I used only works if n is an integer. My mistake.

15. look this
When he plugs $\displaystyle x = 196$, the result is $\displaystyle 12348$.We're going to need to take off three. Now imagine we decrease that $\displaystyle 196$ slightly. Every term on the lefthand side will decrease by AT LEAST one (for example, $\displaystyle [32\cdot 196] = 6272, [32\cdot 195.99] = 6271$), therefore the sum will decrease by AT LEAST six, and that's too much.

some one can more explain this way

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