MHB Floor Function: Real Value of $k$ & $x$

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The discussion focuses on two mathematical problems involving the floor function. The first problem seeks the real values of \( k \) such that \( k\lfloor x \rfloor = \lfloor kx \rfloor \), concluding that \( k \) can only be 0 or 1. The second problem finds values of \( x \) for which \( \lfloor \frac{x}{2011} \rfloor = \lfloor \frac{x}{2012} \rfloor \), identifying intervals where this equality holds, specifically \( x \in [2012, 4021] \) and \( 4024 \leq x \leq 6032 \). The largest value of \( x \) for which the floors are equal is \( 4044120 \), where both floor functions yield 2010. The analysis emphasizes the integer nature of \( k \) and the behavior of \( x \) under the floor function.
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(1) find real value of $k$ for which $k\lfloor x \rfloor = \lfloor kx \rfloor$

(2) find value of $x$ for which $\displaystyle \lfloor \frac{x}{2011}\rfloor = \lfloor \frac{x}{2012}\rfloor$

where $\lfloor x \rfloor = $ floor function
 
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jacks said:
(1) find real value of $k$ for which $k\lfloor x \rfloor = \lfloor kx \rfloor$

(2) find value of $x$ for which $\displaystyle \lfloor \frac{x}{2011}\rfloor = \lfloor \frac{x}{2012}\rfloor$

where $\lfloor x \rfloor = $ floor function

For (2), 2012 since the floor of 2012/2011 = 1 and 2012/2012 = 1.

If you want values though, it would be $x\in [2012,4021]$
 
in general \lfloor x \rfloor = a \Rightarrow a \leq x < a+1

\lfloor x k \rfloor = k \lfloor x \rfloor

k \lfloor x \rfloor \leq x k < k \lfloor x \rfloor +1

0 \leq x k - k \lfloor x \rfloor <1

0 \leq k(x - \lfloor x \rfloor) <1

x - \lfloor x \rfloor = {0,1} if x integer 0 if x is not integer 1

0 \leq k < 1
But
\lfloor kx \rfloor , \lfloor x \rfloor are both integers which leave k to be integer so k=0 and k=1
 
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dwsmith said:
jacks said:
(1) find real value of $k$ for which $k\lfloor x \rfloor = \lfloor kx \rfloor$

(2) find value of $x$ for which $\displaystyle \lfloor \frac{x}{2011}\rfloor = \lfloor \frac{x}{2012}\rfloor$

where $\lfloor x \rfloor =$ floor function

For (2), 2012 since the floor of 2012/2011 = 1 and 2012/2012 = 1.

If you want values though, it would be $x\in [2012,4021]$
Those are the numbers for which $\displaystyle \left\lfloor \frac{x}{2011}\right\rfloor = \left\lfloor \frac{x}{2012}\right\rfloor = 1.$ There are also the numbers $0\leqslant x\leqslant 2010$ for which $\displaystyle \left\lfloor \frac{x}{2011}\right\rfloor = \left\lfloor \frac{x}{2012}\right\rfloor = 0.$ Then there are the numbers $4024 \leqslant x\leqslant 6032$ for which $\displaystyle \left\lfloor \frac{x}{2011}\right\rfloor = \left\lfloor \frac{x}{2012}\right\rfloor = 2,$ and a whole lot of other intervals going right up to the number $X = 4044120 = 2012\times 2010.$ This has the property that $\displaystyle \left\lfloor \frac{X}{2011}\right\rfloor = \left\lfloor \frac{X}{2012}\right\rfloor = 2010,$ and it is the largest number for which $\displaystyle \left\lfloor \frac{X}{2011}\right\rfloor$ and $\left\lfloor \dfrac{X}{2012}\right\rfloor$ are equal.
 
Amer said:
in general \lfloor x \rfloor = a \Rightarrow a \leq x < a+1

\lfloor x k \rfloor = k \lfloor x \rfloor

k \lfloor x \rfloor \leq x k < k \lfloor x \rfloor +1

0 \leq x k - k \lfloor x \rfloor <1

0 \leq k(x - \lfloor x \rfloor) <1

x - \lfloor x \rfloor = {0,1} if x integer 0 if x is not integer 1

0 \leq k < 1
But
\lfloor kx \rfloor , \lfloor x \rfloor are both integers which leave k to be integer so k=0 and k=1
Or you can solve it in other way if x=1

k = \floor k \rfloor so k should be integer
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the trivial solution is k={0,1}, suppose k is not 0 or 1

choose x = 1/k, if k>0 \lfloor \frac{1}{k} \rfloor = 0 if k <0 \lfloor \frac{1}{k} \rfloor = -1

k \lfloor \frac{1}{k} \rfloor = {0,-1}

\floor k \left(\frac{1}{k}\right) \rfloor = 1
 
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