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If $[x]$ represents the floor of $x$, solve the equation:


Choose the correct answer:

a) $(2,7)\cup(9,15)$

b) $(-3,2)\cup[3,4)\cup(6,14)$

c) $(-1,1]\cup[2,3)\cup(5,8)$

d) $(-5,-3)\cup(1,3]\cup[5,7)$

e) $[1,\frac32]\cup(2,4)\cup[5,7)$

f) $[0,2]\cup[4,7]\cup(9,+\infty)$

Can someone please explain how I can solve this type of equation? Thank you very much!

P.S. I'm not sure if $[x]$ is called whole part of $x$ in English, it's not my native language.

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This looks really strange to me, the LHS is always an integer, so the RHS has to be an integer two. Hence $x$ has to be an odd integer. So why are all options intervals? – Simon Markett May 24 '12 at 10:31
$$\frac{x-1}{2} \in \mathbb Z$$, so $x$ is an odd number. Seems "e" is the correct answer. Umm, except for $6$! – Gigili May 24 '12 at 10:32
@Grozav Please comment if you would like more help with some notation that you don't fully understand. – Ronald May 24 '12 at 10:38
up vote 4 down vote accepted

So as $k:=[\frac{x+1}3] \in \mathbb Z$, we have $\frac{x-1}2 =k \in\mathbb Z$ also. So $x = 2k+1$ is an odd integer. We have to solve $[\frac{2k+2}3] = k$, we now consider three cases.

  • $k = 3\ell$ for some integer $\ell$: Then $$ \left[\frac{2k+2}3\right] = \left[\frac{6\ell+2}3\right] = \left[2\ell + \frac 23\right] $$ Now $2\ell$ is an integer, therefore the integer part of this is $\left[2\ell + \frac 23 \right] = 2\ell$. This equals $k = 3\ell$ for $\ell = 0$, which yields $k=0$, hence $x=1$.

  • $k = 3\ell + 1$ for some $\ell \in \mathbb Z$, here $$\left[\frac{2k+2}3\right] = \left[\frac{6\ell+4}3\right] = \left[2\ell + 1 + \frac 13\right] = 2\ell + 1,$$ which equals $k = 3\ell + 1$ for $\ell = 0$, giving $k = 1$, hence $x = 3$.

  • $k = 3\ell + 2$ for some $\ell \in \mathbb Z$, here $$\left[\frac{2k+2}3\right] = \left[\frac{6\ell+6}3\right] = \left[2\ell + 2\right] = 2\ell + 2,$$ which equals $k = 3\ell + 2$ for $\ell = 0$, giving $k = 2$, hence $x = 5$.

The only solutios are there fore $1$, $3$ and $5$, hence it is (e) (intersected with $2\mathbb Z+1$).

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"e" includes $6$ which is not an answer to the equation. – Gigili May 24 '12 at 10:36
Yes. So I wrote "intersected $2\mathbb Z + 1$". At least (e) doesn't contain wrong odd numbers. – martini May 24 '12 at 10:38
Right, didn't notice that part. – Gigili May 24 '12 at 10:40
Thank you very much, this is similar to their explanation, but it contains all the steps into getting the results. – Grozav Alex Ioan May 24 '12 at 11:19
There's one thing I didnt exactly understand. When giving k the values of 3l, 3l+1, and 3l+2, why do we equal the floor equation with 2l, 2l+1, 2l+2? – Grozav Alex Ioan May 24 '12 at 13:09

First of all, notice that $\frac{x-1}{2}$ has to be a whole number (an integer).

Therefore, $x-1$ has to be even, and $x$ has to be an odd integer.

We can use the options to eliminate any number which includes an even number. I guess we are talking only about integers, $\mathbb{Z}$, since any non-integer solutions are definitely out.

For example, the region $(2,7)$ in part (a) includes 4 and 6, so (a) is not correct.

This on its own is not enough to come to a solution, but it should allow us to quickly eliminate wrong options.

In fact, we can go through like this and eliminate all the options that you've given - none of them are right, as it stands.

Perhaps there is a typo in (e)? Since two of the three regions are representing an odd number, but $[5,7)$ incorrectly includes 6. I guess that (e) is supposed to be the correct answer, but maybe it should say $[5,6)$.

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You put too much effort into improving your answer, +1. – Gigili May 24 '12 at 11:01
Thank you for your help, martini's answer contained the practical way into solving the ecuation, however, thank you very much! – Grozav Alex Ioan May 24 '12 at 11:20
@Gigli Haha thanks! Answer first, edit later. – Ronald May 24 '12 at 11:38
@GrozavAlexIoan You are right to choose that answer, but I think using multiple choice options is an important skill. :) – Ronald May 24 '12 at 11:39

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