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Q74E

Expert-verifiedFound in: Page 155

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Prove or disprove each of these statements about the floor and ceiling functions.**

**a) $\lfloor \lceil \mathbf{x}\rceil \rfloor =\lceil \mathbf{x}\rceil $**** ****for all real numbers x****.**

**b) $\lfloor \mathbf{x}+\mathbf{y}\rfloor =\lfloor \mathbf{x}\rfloor +\lfloor \mathbf{y}\rfloor $**** ****for all real numbers x ****and y****.**

**c) **** $\lceil \lceil x/2\rceil /2\rceil =\lceil x/4\rceil $ ****for all real numbers x****.**

**d) $\lfloor \sqrt{[\mathbf{x}\rceil}\rfloor =[\sqrt{\mathbf{x}}\rfloor $**** ****for all positive real numbers x****.**

**e) **** $\lfloor \mathbf{x}\rfloor +\lfloor \mathbf{y}\rfloor +\lfloor \mathbf{x}+\mathbf{y}\rfloor \le \lfloor \mathbf{2}\mathbf{x}\rfloor +\lfloor \mathbf{2}\mathbf{y}\rfloor $ ****for all real numbers x ****and y****.**

(a) $\lfloor \lceil \mathbf{x}\rceil \rfloor =\lceil \mathbf{x}\rceil $ holds true.

(b) $\lfloor \mathbf{x}+\mathbf{y}\rfloor =\lfloor \mathbf{x}\rfloor +\lfloor \mathbf{y}\rfloor $ does not holds true.

(c) $\lceil \lceil x/2\rceil /2\rceil =\lceil x/4\rceil $ holds true.

(d) $\lfloor \sqrt{[\mathbf{x}\rceil}\rfloor =[\sqrt{\mathbf{x}}\rfloor $ does not hold true.

(e) $\lfloor \mathbf{x}\rfloor +\lfloor \mathbf{y}\rfloor +\lfloor \mathbf{x}+\mathbf{y}\rfloor \le \lfloor \mathbf{2}\mathbf{x}\rfloor +\lfloor \mathbf{2}\mathbf{y}\rfloor $ holds true.

**The function is described as the relationship amongst one or more variable. The functions are mainly used to formulate the physical relationsxhips with the help of the outputs and the inputs.**

The given equation is expressed as:

$\lfloor \lceil x\rceil \rfloor =\lceil x\rceil $

Here, the “ceiling function” mainly yields the smallest possible integer, that is equal to or greater than the function . The floor functions also yields the smallest possible integer when it is applied to a particular integer. Hence, both of the sides becomes similar.

Thus, $\lfloor \lceil x\rceil \rfloor =\lceil x\rceil $ holds true.

** **The given equation is expressed as:

$\lfloor x+y\rfloor =\lfloor x\rfloor +\lfloor y\rfloor $

Here, the “floor function” mainly yields the greatest possible integer, that is equal to or lesser than the inserted value. The particular equality mainly doesn’t hold as the the numbers’ sum are different. For example, Let x be 2.4 and y 2.7 be, then the above equation can be expressed as:

$\begin{array}{r}\lfloor 2.4+2.7\rfloor =\lfloor 2.4\rfloor +\lfloor 2.7\rfloor \\ 5\ne 4\end{array}$

Thus, $\lfloor x+y\rfloor =\lfloor x\rfloor +\lfloor y\rfloor $ does not holds true.

The given equation is expressed as:

$\lceil \lceil x/2\rceil /2\rceil =\lceil x/4\rceil $

Here, the “ceiling function” mainly yields the smallest possible integer, that is equal to or greater than the inserted value. As the function x/2 mainly gets applied to the function x/2 , then the immediate integer that is greater than the function that is also divided by the number and also it is given as inputs as the ceiling function. For example, Let x be 1.2 , then the above equation can be expressed as:

$\begin{array}{r}\lceil \lceil 1.2/2\rceil /2\rceil =\lceil 1.2/4\rceil \\ 1=1\end{array}$

Thus, $\lceil \lceil x/2\rceil /2\rceil =\lceil x/4\rceil \rceil $ holds true.

The given equation is expressed as:

$\lfloor \sqrt{[x\rceil}\rfloor \rfloor =\lfloor \sqrt{x}\mid $

Here, in the left side, as the “ceiling function” is mainly applied to the inserted value , the square root of that function will be unequal to the right side. Taking an example, if the value that is less than the number or more than the number that eventually becomes the number in the ceiling function. The number comes when the square root of 36 is being obtained and the number 5 comes when it is mainly applied to the floor function.

Thus, $\lfloor \sqrt{[x\rceil}\mid =\lfloor \sqrt{x}\rfloor $ does not hold true.

The given equation is expressed as:

$\lfloor x\rfloor +\lfloor y\rfloor +\lfloor x+y\rfloor \le \lfloor 2x\rfloor +\lfloor 2y\rfloor $

Here, if the fractional part’s value is either or comes more than the number 0.5 , then the left hand side will be lesser than the number 1 from the right hand side. Apart from that, the right and the left hand side mainly yields the equal answers

Thus , $\lfloor x\rfloor +\lfloor y\rfloor +\lfloor x+y\rfloor \le \lfloor 2x\rfloor +\lfloor 2y\rfloor $ holds true.

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