# Equations

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• April 6th 2009, 03:42 PM
linda2005
Equations
Hello , help please I need to solve those equations :

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

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

3) $[\frac{x^2-2x}{3}]=\frac{x}{2}$
• April 6th 2009, 05:02 PM
Mush
Quote:

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

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

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

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

What is being represented by the square brackets?
• April 6th 2009, 05:14 PM
DeMath
Quote:

Originally Posted by Mush
What is being represented by the square brackets?

whole part of number

$\left[ x \right]or\left\lfloor x \right\rfloor = \max \left\{ {\left. {n \in \mathbb{Z}} \right|{\text{ }}n \leqslant x} \right\}.$
• April 6th 2009, 05:26 PM
Mush
Quote:

Originally Posted by DeMath
whole part of number

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

Ah, right. Are you familiar with the following property, for the first problem:

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

In other words:

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

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

Followed by the rule that:

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

Hence:

$\lfloor 2x \rfloor=\left\lfloor x\right\rfloor + \left\lfloor x\right\rfloor +\frac{1}{2}$ etc...
• April 6th 2009, 05:50 PM
Mush
For the 2nd one, think of it like this:

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

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

Hence:

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

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

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

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

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

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

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

For example, p = 0.1, k = 1 is a solution. (x = 1.1, in other words).
• April 6th 2009, 05:58 PM
Mush
A fairly obvious solution to the last one is x = 0.
• April 7th 2009, 03:18 AM
linda2005
Quote:

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
• April 7th 2009, 06:18 AM
Mush
Quote:

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.
• April 7th 2009, 10:37 AM
NonCommAlg
i'm not sure about 1) but i guess there is no solution. my solutions to 2) and 3):

Quote:

Originally Posted by linda2005

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

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

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

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

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

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

Quote:

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

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

let $x = 2n,$ where $n \in \mathbb{Z}.$ solving the inequalities in (2) for $n$ gives you: $\left[\frac{7-\sqrt{97}}{8}, 0 \right] \cup \left[\frac{7}{4}, \frac{7 + \sqrt{97}}{8} \right],$ which gives us $n = 0,2.$ thus $x=2n=0,4.$
• April 7th 2009, 01:17 PM
linda2005
Hello, Thanks you NonCommAlg Your solutions is (Yes) , so you can help me with 1) thanks
• April 7th 2009, 01:54 PM
NonCommAlg
Quote:

Originally Posted by linda2005
Hello, Thanks you NonCommAlg Your solutions is (Yes) , 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?
• April 7th 2009, 02:32 PM
Mush
There is no solution

Using the following two properties:

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

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

It can be deduced that:

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

Hence:

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

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

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

Continuing like this, the equation can be reduced to:

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

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

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

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

It is non-integer, and so there are no solutions.
• April 7th 2009, 02:50 PM
NonCommAlg
Quote:

Originally Posted by Mush

It can be deduced that:

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

this is not true! why? see below for example:

Quote:

Hence:
$\displaystyle \lfloor 2x \rfloor = 2\lfloor x \rfloor + \sum_{n = 1}^{1} \frac{n}{2} =2\lfloor x \rfloor + \frac{1}{2}$
$\lfloor 2x \rfloor - 2\lfloor x \rfloor$ is an integer but $\frac{1}{2}$ is not. so they can never be equal. the function $\lfloor 2x \rfloor - 2\lfloor x \rfloor$ is always either 0 or 1.
• April 7th 2009, 03:03 PM
Mush
Quote:

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

$\lfloor 2x \rfloor - 2\lfloor x \rfloor$ is an integer but $\frac{1}{2}$ is not. so they can never be equal. the function $\lfloor 2x \rfloor - 2\lfloor x \rfloor$ 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.
• April 7th 2009, 03:17 PM
linda2005
look this
When he plugs $x = 196$, the result is $12348$.We're going to need to take off three. Now imagine we decrease that $196$ slightly. Every term on the lefthand side will decrease by AT LEAST one (for example, $[32\cdot 196] = 6272, [32\cdot 195.99] = 6271$), 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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