Fractional part real numbers

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August 11th 2012, 03:10 AM
Idiotinabox

Fractional part real numbers

Essentially the question asks if x^3 = [(x + 1)^3] then find all real solutions of x. The square brackets in this case represent the fractional part of the number they enclose.

So far I have figured out that [z] is the number minus the largest integer smaller than itself. I have no idea how to go about finding all real solutions. Any help would be appreciated
August 11th 2012, 03:30 AM
Prove It

Re: Fractional part real numbers

Quote:

Originally Posted by Idiotinabox

Essentially the question asks if x^3 = [(x + 1)^3] then find all real solutions of x. The square brackets in this case represent the fractional part of the number they enclose.

So far I have figured out that [z] is the number minus the largest integer smaller than itself. I have no idea how to go about finding all real solutions. Any help would be appreciated

$\displaystyle \begin{align*} x^3 &= (x + 1)^3 \\ x^3 &= x^3 + 3x^2 + 3x + 1 \\ 0 &= 3x^2 + 3x + 1 \end{align*}$

The discriminant of this quadratic is negative, so there are not any real solutions.
August 11th 2012, 03:37 AM
Idiotinabox

Re: Fractional part real numbers

Is this taking into account that it is only the fractional part of (x+1)^3?
August 11th 2012, 03:38 AM
Prove It

Re: Fractional part real numbers

What do you mean by the fractional part?
August 11th 2012, 03:41 AM
Idiotinabox

Re: Fractional part real numbers

For example the fractional part of 5/2 is 2 because that is the largest integer in that number (4/2) and the fractional part is what is left, that is, 1/2.
August 11th 2012, 03:42 AM
Prove It

Re: Fractional part real numbers

There are not any real solutions to that equation - whole, fractional or irrational.
August 11th 2012, 03:48 AM
Idiotinabox

Re: Fractional part real numbers

(x+1)^3 does not equal x^3. The fractional part of it does.
August 11th 2012, 04:41 AM
Educated

Re: Fractional part real numbers

Quote:

Originally Posted by Idiotinabox

For example the fractional part of 5/2 is 2 because that is the largest integer in that number (4/2) and the fractional part is what is left, that is, 1/2.

So you're saying the fractional part of the number 5/2 is 2 and 1/2?

Quote:

Originally Posted by Idiotinabox

Essentially the question asks if x^3 = [(x + 1)^3] then find all real solutions of x. The square brackets in this case represent the fractional part of the number they enclose.

Are you trying to say that your question is: $\displaystyle x^3 &= \frac{1}{(x + 1)^3}$

Then it would become:

$\displaystyle \begin{align*}x^3 &= \frac{1}{x^3 + 3x^2 + 3x + 1} \\ \\ (x^3)(3x^2 + 3x + 1) -1 &= 0 \end{align*}$
August 11th 2012, 04:50 AM
Idiotinabox

Re: Fractional part real numbers

Fractional part - Wikipedia, the free encyclopedia
I'm finding it hard to say
August 11th 2012, 05:24 AM
Plato

Re: Fractional part real numbers

Quote:

Originally Posted by Educated

So you're saying the fractional part of the number 5/2 is 2 and 1/2
Are you trying to say that your question is: $\displaystyle x^3 &= \frac{1}{(x + 1)^3}$

Then it would become:
$\displaystyle \begin{align*}x^3 &= \frac{1}{x^3 + 3x^2 + 3x + 1} \\ \\ (x^3)(3x^2 + 3x + 1) -1 &= 0 \end{align*}$

The fractional part is $\left\{ x \right\} = x - \left\lfloor x \right\rfloor$ . Example: $\{-3.4\}=-3.5-(-4)=0.6$ .

Because $\left\lfloor x \right\rfloor \le x < \left\lfloor x \right\rfloor + 1\; \Rightarrow \;0 \le x - \left\lfloor x \right\rfloor < 1$ , we get $0 \le \left\{ x \right\} < 1$ .

If there are solutions to this they are in $[0,1]$ .
August 11th 2012, 05:31 AM
HallsofIvy

Re: Fractional part real numbers

Quote:

Originally Posted by Educated

So you're saying the fractional part of the number 5/2 is 2 and 1/2?

NO! He is saying that the "fractional part" of the number 5/2 is 1/2. The "fractional part of x" is the same as "x modulo 1".

Quote:

Are you trying to say that your question is: $\displaystyle x^3 &= \frac{1}{(x + 1)^3}$

NO! He is trying to say that $x^3$ is equal to the fractional part of $(x+ 1)^3$ .

First, notice that that the "fractional part" of any number lies between 0 and 1. That means that whatever x is $x^3$ is between 0 and 1 and then x itself is between 0 and 1. That, in turn, means that x+1 is between 1 and 2 so that $(x+1)^3$ is between 1 and 8.
I think the most direct way to do this is to look at each possiblility in turn:
$x^3= (x+1)^3- 1$
$x^3= (x+1)^3- 2$
$x^3= (x+1)^3- 3$
$x^3= (x+1)^3- 4$
$x^3= (x+1)^3- 5$
$x^3= (x+1)^3- 6$
$x^3= (x+1)^3- 7$
$x^3= (x+1)^3- 8$

Quote:

Then it would become:

$\displaystyle \begin{align*}x^3 &= \frac{1}{x^3 + 3x^2 + 3x + 1} \\ \\ (x^3)(3x^2 + 3x + 1) -1 &= 0 \end{align*}$
August 11th 2012, 05:03 PM
Idiotinabox

Re: Fractional part real numbers

Do I have to solve each of those equations for x? Because when I try to solve the first I find either x = 0 or x= -1. The latter is obviously not a solution to the original problem.
August 12th 2012, 12:24 AM
Idiotinabox

Re: Fractional part real numbers

Oh yes, it does seem to produce solutions. Are these all of the solutions?
August 12th 2012, 01:12 AM
Idiotinabox

Re: Fractional part real numbers

So a solution would be x = (-3 + rt21)/6 ? Using the quadratic formula after expanding and simplifying the above equations?
August 12th 2012, 05:56 AM
HallsofIvy

Re: Fractional part real numbers

If "rt21" means "square root of 21", then, yes, that is a solution. Of course, you should check that any "purported" solution is, in fact, a solution.

A more general method would be to look at $x^3= (x+1)^3- b$ where b is an integer from 1 to b. That gives $3x^2+ 3x+ 1- b= 0$ which, by the quadratic formula, has solutions $x= \frac{-3\pm\sqrt{9- 12(1- b)}}{6}= \frac{-3\pm\sqrt{12b- 3}}{6}$ .