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

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

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

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What do you mean by the fractional part?

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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.

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

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

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:

Then it would become:

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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.

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

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So a solution would be x = (-3 + rt21)/6 ? Using the quadratic formula after expanding and simplifying the above equations?

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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 where b is an integer from 1 to b. That gives which, by the quadratic formula, has solutions .