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I'm interested in the graph of the function $[\frac{1}{x}]$. may seems simple, also I graphed it myself. But there is difference between my solution and the manual solution. Here is the solution of the solution manual: enter image description here
But I think it's wrong about $x<-1$.
any response appreciated.

shapoor
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    If by $[x]$ you mean the floor function, the book is wrong and all the lines for $x \lt 0$ should move down $1$. If you mean integer part, see how it is defined for negative $x$. – Ross Millikan Feb 11 '14 at 17:41

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If by $\left[\frac{1}{x}\right]$ you mean "the integer part of $x$", then the plot is correct.

Note that if $x < -1$, then $\left|\frac{1}{x}\right| < 1$, so $\left[\frac{1}{x}\right]$ has integer part $0$.

Edit: This answer is based on the following convention: http://mathworld.wolfram.com/IntegerPart.html

Emily
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  • if $x=-1.5 \Longrightarrow \frac{1}{x} = -\frac{2}{3} \Longrightarrow -1<\frac{1}{x}<0 \Longrightarrow [\frac{1}{x}]=-1$ – shapoor Feb 11 '14 at 17:42
  • The integer part of $-\frac23$ is $0$. – Emily Feb 11 '14 at 17:44
  • Note that the most common convention is for $\lceil x\rceil$ to mean "ceiling of $x$", $\lfloor x \rfloor$ to mean "floor of $x$", and $\left[ x\right]$ to mean "integer part of $x$." This nomenclature is by no means universal, however, and different individuals have different intentions for ceiling/floor as to whether it rounds in a signed sense, or in an absolute sense. – Emily Feb 11 '14 at 17:46
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    In the "caculus, spivak, 3rd ED", in the heading of problem 17 chapter 4, he says:" The symbol $[x]$ denotes the largest integer which is $\leq x$. Thus, $[2.1] =[2]=2$ and $[-0.9]=[-1]=-1$ ..." is it true? – shapoor Feb 11 '14 at 17:47
  • It is true within his book, but maybe not other books. We must also be very clear what is meant by "largest integer", because of course, $0 \ge -1$, but $|0| \le |-1|$. – Emily Feb 11 '14 at 17:51
  • Better use the notation in @Arkamis' comment, floor/ceiling are mnemonic. "Integer part" is ill-defined, better avoid it. – vonbrand Feb 11 '14 at 17:52
  • So could we say his answer in his solution manual is wrong, considering his definition in his text? – shapoor Feb 11 '14 at 17:53
  • @shapoor What problem/page is this on? I'll check my 4th ed copy when I get home to see if it persists or was fixed. – Emily Feb 11 '14 at 17:57
  • page 72, problem 17, chapter 4 – shapoor Feb 11 '14 at 17:59
  • @Arkamis! The integer part of $1/2$ is $-1$, not $0$. – Mikhail Katz Feb 11 '14 at 18:01
  • @user72694 Some people explicitly define the integer part of a number as the thing that comes before the decimal point! $[-0.5]$ can be either $0$ or $-1$ depending on your convention. – Emily Feb 11 '14 at 18:08
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When $x\to-\infty$ the reciprocal $1/x$ is strictly between $-1$ and $0$ and therefore the integer part (the largest integer smaller than $x$) is $-1$. The graph you reproduced shows the value $0$ for large negative $x$, which is incorrect.

Mikhail Katz
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  • From Mathworld: http://mathworld.wolfram.com/IntegerPart.html "The integer part function satisfies $\textrm{int}(-x) = -\textrm{int}(x)$." So $\textrm{int}(-0.5) = -\textrm{int}(0.5) = -0 =0$. – Emily Feb 11 '14 at 18:09
  • As my other comments have indicated, this is far from a universal convention, however! – Emily Feb 11 '14 at 18:11
  • @Arkamis, as mathworld points out, this is the case " In many computer languages". I have never seen a math textbook using such a convention. Notice that the OP used the standard notation found in math books rather than mathworld's "int(.)" – Mikhail Katz Feb 11 '14 at 18:31