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In my real analysis text book there is a question that says:

Decide whether $f(x)=[x]$ is bounded above or below on the interval $[0,a]$ where $a$ is arbitrary, and whether the function takes on it's maximum or minimum value within that same interval.

This question is very straightforward, assuming $[x]=x$. But if that is the case, then the choice of notation is very strange.

Is there another way to interpret the notation's meaning?

Xander Henderson
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objectivesea
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    I have seen $[x]$ denote the fractional part of $x$, but that doesn't seem to be the desired meaning in this context. – bzc May 01 '11 at 22:44
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    I've seen that notation to be the nearest integer function before. – yunone May 01 '11 at 22:44
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    Could you specify what textbook you're using? I'm guessing there should be a place where the notation used is explained... – J. M. ain't a mathematician May 01 '11 at 22:47
  • [x] is often used as the Iverson Bracket, 1 if is x is true, though that's clearly not the usage here. I've also seen it used for either floor or ceiling, when the right brackets (missing either the top or bottom) were difficult to produce. – wnoise May 01 '11 at 23:00
  • The question is from Michael Spivak's "Calculus" - 4th Edition (also 3rd edition). It's part (xii) of chapter 7 question 1. – objectivesea May 01 '11 at 23:20

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It had been fairly standard for $[x]$ to represent "the greatest integer not greater than $x$" (aka, the "floor" function). With fancier type-setting options allowing for $\lfloor x \rfloor$ for a more suggestive "floor" notation ---as well as $\lceil x \rceil$ for the counterpart "ceiling" ("smallest integer not smaller than $x$")--- I've seen $[x]$ taking on the role of "nearest integer" (that is, the "rounding" function) although $\lfloor x \rceil$ is also available for this, freeing up $[x]$ for author's discretion.

With regard to @Brandon's "fractional part", I've seen that more often represented as $\{ x \}$, usually in conjunction with the floor interpretation of $[x]$, so that one would write $x = [x] + \{x\}$ (at least for non-negative $x$).

Blue
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Late to the party, I know ...

I've gone back to my own copy of Spivak's "Calculus" (I seem to have the 1967 edition) and looked it up in the Glossary of Symbols at the back, and there it is on page 573, referring me back to its definition on page 70.

It is indeed defined, in Problem 15 of Part 2 Foundations, chapter 4 Graphs, as the greatest integer $\le x$.

Beware, though, because he makes the egregious howler of presenting as an example: $$[-0.9] = [-1.2] = -1$$ which is particularly bad because this is precisely the mistake that beginners are likely to be confused over and need clarification on.

Of course $[-1.2] = -2$ as $-1 > -1.2$. The greatest integer smaller than $-1.2$ is $-2$.

Prime Mover
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  • The Answer Book for the 3rd edition (published nearly 30 years later) contains similar errors: https://math.stackexchange.com/a/3210910/754927 and https://math.stackexchange.com/q/3840158/754927. I don't have 4th. Wonder if it's since been corrected :) – Ben May 12 '21 at 15:34
  • @Ben Hm. Good job I've never used it except to occupy a place on my bookshelf. I may go through and give it a full and frank assessment, but life's already short enough as it is. Thanks. – Prime Mover May 12 '21 at 16:25
  • Unfortunately, square brackets are sometimes used elsewhere in the text simply to indicate order of operations, for example exercise 11-8: "Prove that if $f$ is convex, then $f([x + y]/2) < [f(x) + f(y)]/2$." I stared at this problem for a long time before figuring out Spivak didn't mean to invoke the floor function. – Ben May 12 '21 at 16:36
  • It's a wonderful book, but there's a bit of extra difficulty for the reader created (I'm sure intentionally ;) ) by all the errors. – Ben May 12 '21 at 16:38
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It's certainly not $[x]=x$.

Rather, it's undoubtedly the greatest integer function, aka floor function (also denoted $\lfloor x\rfloor$); that is, the greatest integer less than or equal to $x$. The ceiling function ($\lceil x\rceil$) is, correspondingly, the smallest integer greater than or equal to $x$..

According to this, Spanier and Oldham called it the "integer value" in $1987$. (Incidentally, I took a course from Spanier, and I consider him to have been an outstanding mathematician. Of course, that puts me in a fairly large club. But I digress.)

The formula would be: $\lfloor x\rfloor =n$, where $n$ is the integer such that $n\le x\lt n+1$.

Another way of describing it, would be to take the number's decimal representation, and truncate it. That is, set the part after the decimal to zero.