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

Re: Fractional part real numbers

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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

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

The discriminant of this quadratic is negative, so there are not any real solutions.

Re: Fractional part real numbers

Is this taking into account that it is only the fractional part of (x+1)^3?

Re: Fractional part real numbers

What do you mean by the fractional part?

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.

Re: Fractional part real numbers

There are not any real solutions to that equation - whole, fractional or irrational.

Re: Fractional part real numbers

(x+1)^3 does not equal x^3. The fractional part of it does.

Re: Fractional part real numbers

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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: $\displaystyle \displaystyle x^3 &= \frac{1}{(x + 1)^3}$

Then it would become:

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

Re: Fractional part real numbers

Re: Fractional part real numbers

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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: $\displaystyle \displaystyle x^3 &= \frac{1}{(x + 1)^3}$

Then it would become:

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

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

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

If there are solutions to this they are in $\displaystyle [0,1]$.

Re: Fractional part real numbers

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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".

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

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

First, notice that that the "fractional part" of any number lies between 0 and 1. That means that whatever x is $\displaystyle x^3$ 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 $\displaystyle (x+1)^3$ is between 1 and 8.

I think the most direct way to do this is to look at each possiblility in turn:

$\displaystyle x^3= (x+1)^3- 1$

$\displaystyle x^3= (x+1)^3- 2$

$\displaystyle x^3= (x+1)^3- 3$

$\displaystyle x^3= (x+1)^3- 4$

$\displaystyle x^3= (x+1)^3- 5$

$\displaystyle x^3= (x+1)^3- 6$

$\displaystyle x^3= (x+1)^3- 7$

$\displaystyle x^3= (x+1)^3- 8$

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Then it would become:

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

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.

Re: Fractional part real numbers

Oh yes, it does seem to produce solutions. Are these all of the solutions?

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?

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